Andreescu - Old and New Inequalities Vol 1

Prévia do material em texto

and
New
Titu Andreescu
Gabriel Dospinescu
Vasile Cirtoaje
Mircea Lascu
Inequalities
VOLUME 1
GIL
I
Old and New
Inequalities
propriety of
GIL Publishing House 
published by
/■ EDITORA
CAtSMtA,
TITU ANDREESCU GABRIEL DOSPINESCU 
VASILE CIRTOAJE MIRCEA LASCU
©GIL Publishing House
51(075,35)
I.Cirtoaje.Vasile
II Dospionescu Gabriel
III Lascu Mircea
ISBN 973-9417-35-3
Copyright © 2004 by GIL. All rights reserved.
OLD AND NEW INEQUALITIES
Authors: Titu Andreescu, Vasile Cirtoaje, Gabriel Dospinescu Mircea Lascu
National Library of Romania CIP Description 
ANDREESCU, TITU
Old and New Inequalities/ Titu Andreescu, Vasile
Cirtoaje -Zalau: GIL, 2004
p.; cm.
Bibliogr.
ISBN 973-9417-35-3
GIL Publishing House
P.O. Box 44, Post Office 3, 450200, Zalau, Romania, 
tel. (+40) 260/616314 
fax.: (+40) 260/616414 
e-mail: comenzi@gil.ro
www.gil.ro
mailto:comenzi@gil.ro
http://www.gil.ro
Preface
The authors
3
This work blends together classic inequality results with brand new problems, 
some of which devised only a few days ago. What could be special about it when so 
many inequality problem books have ahead}' been written? We strongly believe that 
even if the topic we plunge into is so general and popular our book is very different. 
Of course, it is quite easy to say this, so we will give some supporting arguments. This 
book contains a large variety of problems involving inequalities, most of them difficult, 
quest ions that became famous in competitions because of their beauty and difficulty. 
And, even more importantly, throughout the text we employ our own solutions and 
propose a large number of new original problems. There arc memorable problems 
in this book and memorable solutions as well. This is why this work will clearly 
appeal to students who are used to use Cauchy-Schwarz as a verb and want to further 
improve their algebraic skills and techniques. They will find here tough problems, 
new results, and even problems that could lead to research. The student who is not 
as keen in this field will also be exposed to a wide variety of moderate and easy 
problems, ideas, techniques, and all the ingredients leading to a good preparation 
for mathematical contests. Some of the problems we chose to present are known, 
but we have included them here with new solutions which show the diversity of ideas 
pertaining to inequalities. Anyone will find here a challenge to prove his or her skills. If 
we have not convinced you, then please take a look at the last problems and hopefully 
you will agree with us.
Finally, but not in the end, we would like to extend our deepest appreciation 
to the proposers of the problems featured in this book and to apologize for not 
giving all complete sources, even though we have given our best. Also, we would like 
to thank Marian Tetiva, Dung Tran Nam, Constantin Tanasescu, Calin Popa and 
Valentin Vornicu for the beautiful problems they have given us and for the precious 
comments, to Cristian Baba, George Lascu and Calin Popa, for typesetting and for 
the many pertinent observations they have provided.
Contents
3Preface
7Chapter 1. Problems
25Chapter 2. Solutions
121Glossary
127Further reading
5
CHAPTER 1
Problems
7
Problems8
1. Prove thal the inequality
Komal
Gazeta Matematica
Tournament of the Towns, 1993
> 4 (a + b + c)’
5. Find the maximum value of the expression x3 -I- y3 4 z3 — 3xyz where x2 4- y2 4- 
z2 = 1 and x, y, z are real numbers.
4. If lhe equation .r4 4- ax3 + 2x2 + bx 4- 1 = 0 has al least one real rool, then 
a2 + b2 > 8.
3. [ Mircea Lascu ] Let a,b,c be positive real numbers such that abc = 1. Prove 
that
2. | Dinu Sjerbanescu ] If a, b, c € (0,1) prove that
y/abc + \/(l — o)(l - b)(l - c) 0. Prove that
x/a’ 4 a2b2 4 b4 + y/b4 + b2c2 + c4 4 x/c4 4 c2a2 4 a4 >
> ay/2a2 I be. I by/2b2 4- ca I- c\/2c2 4- ab.
Gazeta Matematica
6. Let a,b,c,x,y,z be positive real numbers such that x 4- y 4- z = 1. Prove that ‘ • 
ax 4- by 4- cz 4- 2 y/(xy + yz + zx)(ab 4 be 4 ca) ^+y/b+y/~c + 3. ■
v/C
. . r. 4- n
4- —;— 4 
x/b
y/a2 4- (1 - b)2 4- y/b2 + (1 — c)2 4- x/c2 4- (1 - a)2 > —y- 
holds for arbitrary real numbers a, b, c.
9. If a, b, c are positive real numbers such that abc = 2, then
4- b^ 4“ > ci\/b 4" c + 6\/c *}■ 4* c^/o. 4~ b.
+ (a 4 b)2H---------------- 2
(c 4- a)
7. [ Darij Grinberg ] If a, b, c are three positive real numbers, then 
a b c 9
(b+c)2
b 4- c
a
Old and New Inequalities 9
When does equality hold?
JBMO 2002 Shortlist
Gazeta MatematicA
, for all
> I--I- I-
1 4- >
4-
JBMO 2002 Shortlist
15. | Vasile Cirtoaje, Mircea Lascu | Let a,b, c, x,y,z be positive real numbers 
such that a + x > b+y > c+z and a4-b4-c = x + y + z. Prove that ay + bx > ac+xz.
13. [ Adrian Zahariuc ] Prove that for any a,b,c € (1,2) the following inequality 
holds
16. [ Vasile Cirtoaje, Mircea Lascu ] Let a, b, c be positive real numbers so that 
abc = 1. Prove that
11. [ Mihai Piticari, Dan Popescu | Prove that
5(a2 I b2 I c2) 0 with a 4- b 4- c = 1.
X2 + ■ • ■ + xn
i € {1,2, ...,n}.
b\/a 
4by/c — cy/a
cVb 
4cy/a - aVb
10. [ Ioan Tomescu ] Let x^y,z > 0. Prove that
___________ xyz___________ 
(1 4- 3i)(x + 8y)(y 4- 9z)(z + 6)
When do wc have equality?
6
ab 4- ac + be'
Junior TST 2003, Romania
3
a 4- b + c
a\/c 
4a\/b - b>/c.
14. For positive real numbers a, b,c such that abc a 4- b 4- c. b c a
■ 12. | Mircea Lascu ] Let xj,X2, ...,xn 6 /?, n > 2 and a
= a and x? 4- x? 4- ... 4- x2 0 such that xi i 
2a
n
Problems10
> 1.4- ••• 4-
Russia, 2004
21. [ Florina Carlan, Marian Tetiva ] Prove that if x, y, z > 0 satisfy the condition
x 4- y 4- z = xyz
then xy + xz 4- yz > 3 4- y/x2 + 1 4- ^/y2 4- 1 + y/z2 4- 1.
>2,+4-
1 + x h y2
JBMO, 2003
>2.
Kvant, 1988
20. | Marius Oltcanu ] Let X1.X2.X3, 14,15 € R so that ii -fio 1 X;j 4 X4 4-15 = 0.
Prove that
24. Let a,b,c > 0 such that a4 + b4 + c4 1.
Gazeta MatematicS
19. | Marian Tetiva | Let x, y, z be positive real numbers satisfying the condition 
x2 + y2 4- z2 + 2xyz = 1.
22. [ Laurenpu Panaitopol ] Prove that
1 + x2 1 4- y2
1 4 y I- z2 ' 1 4 z I x2 
for any real numbers x,y,z > -1.
Prove that
1
xy: 3 and Xi,X2,...,xn > 0 have product 1, then
1 1 1 
------------------------- f. -------------------------j- •. • + ---------------------- 
1 I Xj I T1X2 1 I r-2 I X2X3-------------- 1 !• Xn I XnXj
23. Let a, b, c > 0 with a + b + c = 1. Show that 
a2 + b b2 + c c2 + a 
—------- 4-------------- 1----------7-b I c c4- a a + 0
Old and New Inequalities 11
+ ... ++ rn +1998
Prove that
> 1998.
Vietnam, 1998
Russia, 2002
2a + b c 2b + c + a
India, 2002
3(ab + be -r ca)
>4-+b2 — be + c2
x'j 4- x'2 + • • • 4- i2 > iii2 + Z2I3 + ■ ■ • + znii 4- 2n - 3.
31. [ Adrian Zahariuc ) Consider the pairwise distinct integers xi.xj, ... ,x„, 
n > 0. Prove that
26. [ Marian Tetiva ] Consider positive real numbers x,y, z so that 
x2 + y2 + z2 = xyz.
xn be positive real numbers satisfying
1 , 1 1 
x2 + 1998 + "■ + xn + 1998 ” 1998 ‘
25. Let n > 2 and X\,..
1
xi + 1998
4
30. Let a, b, c be positive real numbers. Prove that 
a3 . b3 c3
c2 - ac + a2 a2 - ab 4- b2 ~ a + b + c
Proposed for the Balkan Mathematical Olympiad
tfx1x2...xn
n- 1
27. Let x,y,z be positive real numbers with sum 3. Prove that 
\/x + y/y + \/z > xy + yz ~ zx.
29. For any positive veal numbers a, b, c show that the following inequality holds 
a b c c + a a + b b + c
b c a ~ c + b a 4- c b + aa)2.
AMM
1. Prove that 
d 
27
n" nd + O + 
which is the desired inequality.
11 _
4 ’ 9 + 27
1 4- Qabc.
4
60. Let a, b, c, d > 0 such that a 4- b 4- a =
(1 4-b2)(l 4-c2) 
(1 4- a2)(b - c)2
Of course, this is true for any other variable, so we can add all these inequalities to 
obtain that
Solution:
Suppose the inequality is false. Then we have, taking into account that abc a3 4- b3 4- c3 - . We may assume that abc It is sufficient to 
•1
prove that
Solution:
The inequality can be also written as Y' 7——~rr? — 1 (°f course, we may 
' (14 cq(a — b)z 
assume that a, b,c are distinct). Now, adding the inequalities
(1 4-a2)(l 4-b2) (1 4- b2)(l 4- c2) 1 + b2
(1 4 c2)(n - b)2 (1 4-a2)(b - c)2 |a-b||c-b|
(which can be found using the AM-GM Inequality) we deduce that 
v—' (1 4- a2)(l 4- b2) 1 4- b~
^(14 c2)(a - b)2 ~ [(b-a)(b-c)|
a3 + + c3 _ 1
------ i-------------— abc 4- a3 4- b3 4- c3 > |.
A-abc 4
27
But this inequality is equivalent to 4 q3 4- 15abc > 1. We use now the identity 
a3 = 3abc-f-1 — 3 anc^ reduce the problem to proving that ab y'
2
|(x(y+z) + y(z+x) + z(x+y)]]
zx).+ z
+ .’/
and. similarly,
I
From Bernoulli’s Inequality, we get
— [] + (x - 1)]
and so it is enough to prove that the last quantity is at least 1. But it follows from 
1 + b2 
|(b-a)(fr-c)| 
and the problem is solved.
3(1-a)
2
1 + b2 
(b - a)(b - c)
T+ a 
~T~X
62. [ Titu Andreasen, Mircea Lascu ] Let q,x, y, z be positive real numbers such 
that xyz = 1 and n > 1. Prove that
xa ya za 3
z + x x + y 2
X^ +1/^+2^
First solution:
Wc may of course assume that x > y > z. Then we have
xys 
------ > —— > ------ y I z z I x x |- y
> z0-1. Using Chebyshev’s Inequality we infer that
Now, all wc have to do is to observe that this follows from the inequalities xQ-1 > 3
z 3
(from the AM-GM Inequality) and - > -
i + q, v / x+ —2~(x + y + z) - (x + y + z) =
, 1 + a, 1 - a
> 1 + —(x-1)- —+
Second solution:
According to the Cauchy-Schwarz Inequality, we have: 
xa ya za -------- f.------- -|-------- 
y + z z + x x + y
Thus it remains to show that
(x1^2 + y1^2 + z1^ > 3(xy + yz +
Since (x + y + z)2 > 3(xy + yz + zx), it is enough to prove that 
x1^1 + t/1’2 + z2^2 > x + y + z.
1-n l + o no 1 - a 1 + a
- 2 + 2 1/12 ~ 2 + ‘2 Z‘
za
x + y
65Old and New Inequalities
(3 tyxyz - 3) = 0.
+ 1 (0 > 2) and x = y =
+
Korea, 2001
n
and the problem is solved.
al + a2 + ■ ■ ' + an —
64. | Lauren^iu Panaitopol ] Let ai,a2,...,a,t be pairwise distinct positive inte­
gers. Prove that
Solution:
We clearly have the inequality
63. Prove that for any real numbers Xi,..., xn, yi,...,yn such that xf d----- Fx2 =
l/i + ••• + !/?, = 1,
2n + 1
3
a — 1
2
/ n
(xij/2 - X21/1)2 
TST Romania
/ n
Equality holds for x — y = z — 1.
/ n \2
-152 ]
Remark:
Using the substitution 0 = a 
inequality becomes as follows 
1 1 . 1 3
a°(b 4- c) bd(c + a) + ^(a I b) ~ 2
For 0 = 3, we obtain one of the problems of IMO 1995 (proposed by Russia).
i z — - (abc - 1) the 
b c
52(J2^
H \ (
i - 52 ) (1
(111/2-X21/1)2
Cy/abyfc
^>~r
With the substitution .r.
becomes ,ry | yz I zx
E
E >
I
I 
I
I
n 
E*
Now, using the fact that «j+i > a, we infer that &i (n I 1)£> >
3v/3
4 ’
2n 4- 1
3
2n+ 1
3
•T
v/3y + yz
But, by applying the Cauchy-Schwarz Inequality we obtain
(E*r
>/3 I 3zi/c
where we used the inequalities
(Ex) 11/2-^5-
■r2 
y/?>xy t- xyz
■Y,bi +
and (he conclusion is immediate. Also, from the above relations we can see imme­
diately that we have equality if and only if ai,a2,...,an is a permutation of the 
numbers 1,2.... ,n.
65. [ CSJin Popa ] Let a,b,c be positive real numbers such that a + b + c = 1. 
Prove that
by/c Cy/a a\/b
a(y/3c I- Vab) b(\/3a + x/bc) c(\/3b -I- y/ca)
Solution:
Without loss of generality, we may assume that dj i for all t. Thus, we may take b 0 and the inequality becomes
-A .2 a V"' 1 n(n + l)(2n + 1) 2n I- 1 v~'. , n(n + l)(2n + 1)
Eb>+ 2 E lb
- 4 ■
[be
1 and the inequality turns into
>
“ 4 ’
Solution:
Rewrite the inequality in the form
l3c.a^~ + a
[ca [ab , ,
= —,z = y — the condition a + b + c = I
3E*y 
"73 + E 
\/3
3\/3
4
67Old and New Inequalities
67. Prove that
APMO, 2004
+ 1 >
4- 1
+ 1 >
>
>* + E >
66. [ Tilu Andreescu, Gabriel Dospinescu ] Let a, b, c, d be real numbers such that 
(1 4- a2)(l 4- b2)(l + c2)(l + d2) = 16. Prove that
—3 9(ab + bc~ ca)
for any positive real numbers a, b, c.
(a 4- b 4- c)2
9
a2 — 1- 1 + E 3 - g
ii) If two of the three numbers are at least 1, let them be a and b, then we have
n(^ 
and the proof is complete.
o»2
b2 -
+ 3
a2 - 1 b2 - 1\ (c2 +2\1 + 3 + 3 J Q 3 J
(a2 + b2 + 1) (I2 + I2 + c2) > (a 4- b 4- c)2 
9 9 ~ 9
by the Cauchy-Schwarz Inequality.
iii) If all three numbers are at most 1, then by Bernoulli Inequality we have
(z»2
9
First solution:
We will prove even more: (a2 I- 2)(b2 4 2)(c2 I 2) > 3(a I b I c)2. Because 
(a 4- b + c)2 1 4- x 4- y- So, 
we write the inequality in the form
n(^
and we have three cases
n(^
Solutions68
Third solution:
In the same manner as in the Second solution, we reduce the problem to proving 
that
Second solution:
Expanding everything, we reduce the problem to proving that 
(aba)2 4- 2 y a2b2 4- 4 y a2 4- 8 > 9 y ab.
Because 3 y a2 > 3 y; ab and 2 y a2t>2 + 6 > 4 y ab, we are left with the 
inequality (nbc)2 4- y a2 + 2 > 2y ab. Of course, we can assume that a,b,c are 
non-negative and we can write a = x2,b = y2,c = z2. In this case
2y^ob - yy2 = (r + y + z)(r + y- z)(y4- z - j)(z + x-y').
It. is clear that if x,y,z are not side lengths of a triangle, then the inequality • • 
is trivial. Otherwise, we can take x - u + v,y = v 4- w,z = w 4- u and reduce the 
inequality to
163(u 4- v)4(v + w)4(w + u)4 > —^-(uvw)3(u 4- v + w)3.
v
But this follows from the known inequalities
g
(u 4- v)(v + w)(w 4- u) > ~(u + V + w)(uv + VW 4- wu),
g
(uv -I- VW + wit)4 > 34(wvw)3, 11 + v + w > 3\/uvw.
(abc)2 4- 2 > 2 y ab - y a2.
Now, using Schur’s Inequality, we infer that 
v—' , v-' o tycibc.2 y ab - y a2 3 (/(abc)2,
the problem is solved. But the last assertion follows from the AM-GM Inequality.
((u + v)(v 4- w)(w + it))4 + 2 > 16(u + v + w)uvw,
We have ((u + v)(v + w)(w 4- u))4 4- 1 + 1 > 3 (u 4- v)*1 (v 4- w)4 (u 4- w)4 and 
it remains to prove that the last quantity is at least 16(u 4- v 4- w\uvw. This comes 
down t.o
Old and New Inequalities G9
+
are true.
TST 2001. USA
and similarly
(1 - yz)(l - zx) = (1 - z)2 > 0, (1 - zx)(l - xy) - (1 - x)2 > 0.
Solution:
a) We have
(1 -xy)(l-yz) = 1 - xy - yz + xy2z - 1 - xy - yz + y(x 4 y + z - 2) - (y - I)2 > I)
y
232
27
So the expressions 1 - xy, 1 - yz and 1 - zx have the same sign.
b) We rewrite the relation x + y + x = xyz+ 2 as(l—x)(l —y) + (l—x)(l-xy) = 0. 
If x > 1 then z > y > x > 1 and so (1 - x)(l - y) + (1 - z)(l - xy) > 0, impossible. 
So we have x 1.
32
1) xy 1. From y > ^/xy we get y > 1. Next we rewrite the relation x + y + z = 
xyz + 2asx + y- 2 = (xy - l)z. Because z > y gives x + y - 2 > (xy - l)y, 
(y - 1)(2 - x — xy) > 0 so 2 > x(l + y). Using the AM-GM Inequality, we have
1 + y > 2yy and l+y=l+|+|>3
r~^ &
2 > 3xy —, which means that x2y 2xyy and
32
- 27'
y - 1 and the equality x’y2
68. [ Vasile Cirtoaje ] Prove that ifO 0;
b) x2y 0 and xy + yz + zx > 1 and we have to prove that at least two of the 
inequalities 2x+3y + 6c > 6, 2y+3x + 6x > 6, 2c + 3x+6y > 6 are true. Suppose this is 
not the case. Then we may assume that 2x+3y + 6z Thus, 12 > -—+ 9y + 8z
69. [ Titu Andrccscu ] Let a, b, a be positive real numbers such that a -i- 6-t c > abc. 
Prove that at least two of the inequalities
236,236„236„
- -r r + - > 6, -r + - + ->6, - + - + 7 > 6.abc b c. a cab
3
a
Solutions70
70. | Gabriel Dospincscu, Marian Tetiva | Let x,y, z > 0 such that
x + y + z = xyz.
Prove that
(i - 1) (y - 1) (z - 1) 0, this yields
 SVabacbc = 3x2,
which is the same as 12(y + c) > 5 + 9y2 + 8z2 + 12yz (2z -1)2 + (3y + 2z- 2)2 yz > 1 (and the similar relations xz > 1, xy > 1) at most . 
one of the three numbers can be less than 1. In any of these cases (x 1, 
z > 1 or the similar ones) the inequality to prove is clear. The only case we still have 
to analyse is that when x > 1, y > 1 and z > 1.
In this situation denote
\/abc - x
x-l = a, y-l = b, z - 1 = c.
abc (x + 1) (x2 + 2x - 2) 
«• (x + 1) (x + 1 + x/5) (x + 1 - \/3) 0.
is
- 1/'(t) = 3 (t2 - (73 
and has only one positive zero. It is 75 — 1 and it’s easy to see that this is a minimum 
point for f in the interval [0,co). Consequently
/(*)>/(75-1) =0,
a + b > 2 Tab
and 675 - 10 > 0, so it suffices to show that
a2b2 + (675 — io) (2>/ab + ab) - 2ab > 0.,
The substitution t = Vab > 0 reduces this inequality to 
t4 + ^6x/3 — 12) I2 + 2 (gT5 - io) 1 > 0,
t3 + (675 - 12)
The derivative of the function
f (t) = t3 + (gT5 - 12) t + 2
t + 2 (gT5 - 10)
(gT5 - io) , t > 0
The relation to prove is
(x - 1) (y - 1) (z - 1) 
 2xyz — (xy + xz + yz) . (xy - x - y)2 + (675 - 10) xy 0.
Second solution-.
Like in the first solution (and due to the symmetry) we may suppose that x > 1, 
y > 1; we can even assume that x > 1, y > 1 (for x = 1 the inequality is plain). Then 
we get xy > 1 and from the given1 condition we have
x + y
x — -i •
. xy - 1
and we are done.
A final observation: in fact we have
/ (t) = (t - 73 +1)2 (t + 2J3 - 2),
Solutions72
which shows that / (t) > 0 for t > 0.
 >1/+ 77x +
a + 7 >
and of course
1 
a + c
b
2
a
2‘
a2 + ac + c2
a + c
b3 - c3 
b + c
= (a‘-b3)( —k V (b3-c3)(-^~-
ka + b a+cj \b + c
r’-a3
4“-------------
c + a
a+ £ >
Ing the AM-GM Inequality, we reduce this inequality to the following one 
£a»3*|n b > c and in this case it 
becomes 8
(a - b)(a - c)(b - c) '
——----F
a -F b
 b > c > 0. The idea is to use the inequality 
y^ + ^ + y2 >v + £ 
2 “ x + y ~y 2'
which is true if x > y > 0. The proof of this inequality is easy and we won’t insist.
Now, because a > b > c > 0, we have the three inequalities 
b a2 + ab + b2 a c b2 + be + c2 • 
a + 7 > ----------------> b+ -,b+ - >----—------- >2 a 4* b 2 2 ofc
First solution’.
First of all, we observe that right hand side can be transformed into
.. .. / 1 1 X
' \a + b a + cj
(a - b)(c - b)(a - c) (£ab)
(a + b)(a + c)(b + c) 
and so we have to prove the inequality
|(a - b)(b - c)(c - a)|(ab + bc-F ca) 
(a + b)(b + c)(c + a)
It is also easy to prove that (a + b)(b + c)(c -Fa) > |(a + b + c)(ab + be + ca) and 
' we are left with .
s I (£«*-£■»)•
8
9
73OI(a - b)
>
and the conclusion follows.
3a+a3+ 2
II(a3 + 2)> (Ea)3'
which is what, we wanted.
Prove that
>
Wc write two similar inequalities and then add up all these relations. We will find 
that
Solution'. ■ ■
We start with the inequality a5 — a2 + 3 > a3 + 2 (a2 - l)(a3 — 1) > 0. Thus, 
it remains to show that
E^ 
,k=l
72. [ Titu Andreescu ] Let a, b,c be positive real numbers. Prove that
(a5 - a2 + 3)(b5 - b2. + 3)(c° - c2 + 3) > (a + b + c)3.
USAMO, 2004
2
n(n-l)'
E(a-fe)2
4
n i 
e? fc=l Xk
n (°3+2) (e a) •
Using the AM-GM Inequality, one has 
a3 1 1
^4-2
ac I c2 
a + c
73. [ Gabriel Dospinescu ] Let n > 2 and xi, Z2,..., zn > 0 such that 
/ n \ / n - \ 
(E*» (E^ ="2 + ‘- 
\fc=l / \fc=l /
b2 4 be. f- c2 , . a2 I
—c -(°-e)
e) I e + 0 - (a - e) ('
y, a3 - b3 
a + b
That is why we can write 
a2 I tib + b2 
a + b
(a - b) (b F I- (b-
E(q ~6)2 
4 ' '
In the same manner we can prove that
E^ 0
2
>
=n2 + l.
Xk.*:=! */
. Xj .Xi Xj ] 
+ -4 — 4 -----------4 . -i + b =
Xi Xj Xi I
5i-2 
Xi
E (r +
E (-+
Thus we’could find from the inequality
'fi + fl _ 2
 «2 + 4 1- 
' \i=l X /
which is what we wanted to prove. Of course, we should prove that we cannot have 
equality. But equality would imply that = x2 = ••• s= xn, which contradicts the 
assumption
) (E?) ^n2+4- 
\i-i / \>=i I 0.
This is clearly true, because the discriminant of the quadratic equation is —4(a + 
l)(a-l)2sitive real numbers a, b, c,
a2 + b2 + c2 + 2abc + 3 > (1 + a)(l + b)( 1 + c).
75. [ TitU Andreescu, Zuming Feng ] Let a,b, c be positive real numbers. Prove 
that
4a + 3
2+ (a - I)2
4a + 4
3 '
(2a + b + c)2 
2a2 + (b + c)2
(2b + a + c)2 
2b2 + (a + c)2
(2c + a + b)2 
2c2 + (a + b)2
 x2y2 + y2z2 + z2t2 + t2x2 + x2z2 + y2t2
r,x = z2 and using
First solution:
Let /(a, b, c) = a2 + b2 + c2 + abc + 2 - a - b - c — ab - be - ca. We have to prove 
that all values of f arc nonnegative. If a, b, c > 3, then we clearly have - + - + - a2 + b2 + c2 + 2 — a — b - c > 0. So, we may assume 
d *1“ cthat a 0 and so it remains to prove that /(a, m, m) > 0, which is the same
(2a + b + c)2
' 2a2 -I- (b -|- a)2
First solution:
Because the inequality is homogeneous, we can assume that a -l b I c. - 3. Then 
(2a + b 4- c)2 a2 + 6a + 9 1 / 
2a2 + (b + c)2 “ 3a2 - 6a + 9 “ 3
Austrian-Polish Competition, 1995
>
l,n+
4-
t
rnn(x -y)(xm*n-' -ym*H~
.m+n-l
/ -,C 
2
>4
It remains to prove that
2(x2 + y2 + z2 4- 2ii/ 4- 2yz f 2zx - 6x - 6y - 6z 4- 9) > x2 + y2 4- z2 4- 6 
 x2 4- y2 + z2 + 4(xy 4- yz 4- zx) — 12(x 4-y 4- z) + 12 > 0.
Now xy +:yz 4- zx > 3 \Jx2y2z2 > 12 (because xyz > 8), so we still have to prove that 
(x 4- y 4- zj2 + 24 - 12(x 4- y 4- z) 4-12 > 0, which is equivalent to (x 4- y 4- z - 6)2 > 0, 
clearly true. • .
Solution:
We consider the standard substitution
z . V z , a — ~ c = -,a
76. Prove that for any positive real numbers x, y and any positive integers m, n, 
(n -1) (m -1) (xm+,‘ 4- ym+n) 4- (m 4- n -1) (xmyn 4- xnym) > mn(xm+n"1 y 4- y’n+n~1 x).
77. Let a,b,c,d,e be positive real numbers such that abede = 1. Prove that 
a 4- abc b 4- bed c 4- cde d 4- dea . e 4- eab 
1 4- ab 4- abed 1 4- be 4- bede
Solution:
We transform the inequality as follows:
’’) > (ni 4- n - l)(xm - y,n)(xn - y") 
xmTn-i _ ytn-t-n-i xm_ym Xn - yn 
(mln- l)(x - y) ~ m(x - y) n(x - y) 
(we have assumed that x > y). The last relation can also be written
(x - y) f tm+n~2dl > F tm~'dt ■ [X tn~ldt
Jy Jy Jy
and this follows from Chebyshev’s Inequality for integrals.
. t --------- > 10
1.4- de 4- deab 1 4- ea 4- eabc ~1 3
Crux Mathematicorum
I u
U X
4----------------------F
1 4- cd 4- odea
Second solution:
b+c c + a a + b XU x. , >Denote x =------, y = —;, z =------ . We have to prove that
a b c
(^ T 2)2 o _ v-' 2x T * 0. It is clear that
>0.
Solution:
It is clear that it suffices to prove the following inequalities 
i- y7 °2^2 - n3b y7n^3
79. Prove that if a,b, c are positive real numbers then,
\/a4 4- b4 4- c4 4- \/a2b2 4- b2c2 + c2a2 > \/a3b + b3c + c3a 4- y/ab3 4- be’ 4- ca3.
KMO Summer Program Test, 2001
78. [ Titu Andrccscu ] Prove that for any a,b,c, e (fl, the following inequality 
holds -
sin a • sin (a - b) ■ sin (a - c) sin b • sin(b - c) ■ sin(b - a) sin c ■ sin(c - a) ■ sin(c - b) 
sin(b 4- c) + sin(c4-a) sin(a 4-b)
TST. 2003, USA
Solution: ~ • ’
Let x = sin a, y = sin b, z = sin c. Then we have x, y, z > 0. It is easy to see that 
the following relations are true:
sin a ■ sin(a - b) • sin(a — c) • sin(a + b) ■ sin(a + c) = r(r2 - y2)(x2 - z2)
Using similar relations for the other terms, we have to prove that:
^2x(x2 - y2)(x2 - y2) > 0.
With the substitution x = y/u, y = \/v, z = y/w the inequality becomes ~
v)(u - io) > 0. But this follows from Schur’s Inequality.
a 4- abc 
1 + ab 4- abed ■ 1 + 1 1
x z u
Writing the other relations as well, and denoting - = ai, — a2, | = an, y = a,t, - = 
a5, we have to prove that if a, > 0, then
y-A 02 + 04 >
ai 4- a3 4- as ~ 3
Using the Cauchy-Schwarz Inequality, we minor the left-hand side with
____________________________4S2____________________________
2S2 - (a2 4- a4)2 - (ai 4- a^)2 — (a3 4- as)2 - (02 4- as)2 - (a1 4- a3)2 ’
5
where S = By applying the Cauchy-Schwarz Inequality again for the 
denominator of the fraction, we obtain the conclusion.
1
z
1
u
1 1— + *7 
y__ L
1
z
1
X
Solutions78
and
and the fact that
 •F
2,1-
>2.
Clearly,
1.
(5>‘)
The first, one follows from Schur’s Inequality .
^2 fl4 + abc 57a - 57 + 57 ab3
n
y~' a2b2 > abcy^a.
The second one is a simple consequence of the Cauchy-Schwarz Inequality: 
(a3b 4- b3c + c3a)2 (1 + x)2
for all x > 0. Clearly, this implies k„ > n-2. Let us prove that n - 2 is a good 
constant and the problem will be solved.
First, we will prove that (x2 4- y)(y2 4- x) > xy( 1 4- x)(l + y). Indeed, this is the 
same as (x F y)(x - y)2 > 0. So, it suffices to prove that
■: i ’ i
—------------------4-----------------------
(1 4- ai)(l 4- n2) (1 4-a2)(l 4-aa)
Now, we take ai = an = — and the above inequality becomes 
x2 ii
n 
E 
’ . k=l
________ Xk+lXk+2._______ 
(xk 4-xfc+i)(ifc+i 4- x*+2)
which can be also written in the following from
V' Xk 4.______ _____
“Xfc-Fifc+i (xfc 4-ifc+i)(xfc+i 4-ifc+2)
80. [ Gabriel Dospinescu, Mircea Lascu ] For a given n > 2 find the smallest 
constant'A",, with the property: if ai, .>. .a,, > 0 have product 1, then
________ Uia2________ ________ Q2Q3__________ _________________ QnQl
(a2 4-a2)(a2 4-aj (a2 4-a3)(a?t 4-a2) (a2 4- ai )(a? 4- an)
n
>y------ it------
“ Xl + + ■ ■ ■ + Zn
K= !
79Old and New Inequalities
> I-
2
tp, y — tq and
Since
81, ( Vasile Cirtoaje ] For any real numbers a,b,c,x,y, z prove that the inequality 
holds
Solution:
Let us denote t = 
z =' trwhich imply
n n
> 2 ^2 XfcZfc-H + x^Xk+2
{t-k
But this one is equivalent to
> -7T—---------J------------------------
+Xk+i)(xjt+i+xk+2) 
k=l
/ n - t • Using the substitutions x —
a2 4- b2 4- c2
4(a + b 4- c)(p 4- q + r) h(a 4- p) 4- (b 4- q) 4- (c 4- r)]2. 
o
This inequality is clearly true.
a2 + b2.4- c2 — p2 + q2 4- r2.
The given inequality becomes
2 ap 4- bq 4- cr + a2 4- b2 4- c2. > - (a 4- b 4- c) (p 4- q 4- r),
o
(g 4- p)2 4- (b 4- q)2 4- (c 4- r)2 > ^(a 4- b 4- c)(p 4- q 4- r). 
o
t________________________ 2
ax 4- by 4- cz 4- \/(a2 4- b2 4- c2)(x2 4- y2 4- c2) > - (a 4- b 4- c) (x 4- y + s). 
o
Kvant, 1989
2 hXj
and it is clear. Thus, fc„ = n — 2.
So, we have to prove the inequality
« ~2y-'_________Jfc+i______■ ■
(ijt 4-Xfe+i)(xfc+i 4-Zfe+2)
Using the Cauchy-Schwarz Inequality, we infer that
’ n
and so it suffices to prove that
2 n n n
h 2 XkXk+l I' XkXk+2’ 
k=1 k—l k=l
n -r2V-' __________Xk+1___________
(Xk 4- Xfc+i)(xfc+i 4- Xk+2)
SolutionsHO
3
= 1 +
X1 + I2 + • ■ ‘ + Xi-l + li-j-l + • • •+,Xn
83. [ Walther Janous ] Let n > 2 and let zi,X2,...,xn > 0 add up to 1. Prove 
hat
________ 1__________
" ~^Xil2-.-Xi-lXi+l ... Xn
______ 1.________ _
—yX1X2...Xi-iXi+i ...Xn
Crux Mathematicorum
Second solution: '
Make the classical substitution a = y + z,b = z + x,c = x + y and clear denomi­
nators. The problem reduces to proving that '
X3 + y3 + z3 + 2(i2j/ + y2z + z2x) > 3(xy2 + yz2 + zx2).
We can of course assume that x is the smallest among x, y, z. Then we can write 
y = x + m, z = x + n with nonnegatives m and n. A short computation shows that 
the inequality reduces to 2x(m2 - mn + n2) + m3 + n3 + 2m2n - 3n2ni > 0. All we 
need to prove is that in3 I- n3 f- 2rn2n > 3n2m m)3 - (n - in)m2 -I- m3 > 0
and this follows immediately from theinequality t3 + 1 > 3, true for. t > —1.
(3a-2c)x2+ (x + c - (a-c)2
which follows immediately from 3a > 2c, 4b 4- 2c - 3a = 3(b 4- c - a) 4- b — c > 0.
First solution:
Q "F c
We may assume that c is the smallest among a,b,c. Then let x = b------—. After
some computations, the inequality becomes
>0 « (3a-2c)(2b-a-c)2+(4b+2c-3a)(a-c)2 > 0
First solution:
The most natural idea is to use the fact that 
n- Xj _ , , n - 1
1-Xj
Thus, we have
and we have to prove the inequality
/ fiH) W+
„ f b c a>2 \ a b c
82. [ Vasile Cirtoaje) Prove that the sides a, b, c of a triangle satisfy the inequality
a b c- + - + - - 1 
oca
n-1
obtained from H’
>
n
But this one is not 
inequalities
n numbers. Then we can 
1, let them be a,, such that
n — 1 +12
n2 - 1
n
1 i -
n - 1 + xn
TST 1999, Romania
n + 1 
n - 1 
ri* 
' X3 J
uygens Inequality.
tt 1 I x, 
■I, 1 ~
Old and New Inequalities
very hard, because it follows immediately by multiplying the
84. [ Vasile Cirtoaje, Ghcorghe Eckstein ] Consider positive, real numbers 
ii,X2,...,in such that Xii2—Xn = !• Prove that
n - 1 + ii
First solution:
Suppose the inequality is false for a certain system of 
find a number k > 1 and n numbers which add up to 
------ i------ = kaA. Then we have 
n - 1 T Xi
Second solution:
We will prove even more, that
...V W \ n ) 111-,,'
It is clear that this inequality is stronger than the initial one. First, let us prove that 
n-1 1 \n 
n - 1)
-n(£-+>)•
Solutions82
>1.+
> ,1-i
+ ... 4- xn
>1.+ ...+
USAMO, 2001
85. [ Titu Andreescu ] Prove that for any nonnegatiye real numbers a,b,c such 
that a2 + b2 + c2 + abc = 4 we have 0 3\/a2b2c2 and 
thus it is enough to prove that abc 26c + a2 + abc => (2 - a)(2 + a) > bc(2 + a) => be b + c - be (&— l)(c - 1) > 0, which is true due to 
our choice.
i-J- ’
+ ... + Xn "
.., — respectively, the in- 
in
In~" 
+ l2~"
Remark. . '
Replacing the numbers Xi,i2>■ • • »in with —,—, 
Il 12 
uality becomes as follows
1 1 
1 + (n - 1)^! + 1 + (n - l)x2
Second solution-.
Let us write the inequality in the form
ii , 12 , , in
■ -f- ----------------------- T“ . . . “T -----------------------
n - 1 + ii n -1 + 12 n - 1 + in
This inequality follows by summing the following inequalities 
1“ n
___________ I] ___________ In
X]"" + X2’“ + •••+ ln~" n-l + Xn
The first from these inequalities is equivalent to
xi~" + + ... + In-" > (n - l)i! "
We have used here the fact that a, (n — 1) ”"\/bi... bj-ibj+i... bn.
83Old and New IiicqiiAlltiefi
(l-v/2^7)2 > 0
- tyabc 2(a + b 4- c - y/ab - y/bc - y/ca).
Now, we will prove that a + b 4- c — 3 y/abc b > c and let a = x 4- y, b = x — y. The 
hypothesis becomes x2(2 4- c) 4- y2(2 - c) = 4 — c2 and we have to prove that 
2 | q
(x2 - y2)(l c) z2z b => y > 0 and 6>0=^z>y>0). Now, consider the 
4z2 — (4 ~function f : [0, y/2 - c] -» /?, f(x) = 2(1 - ex)--------- ----------- -(1 - c). We have
1 ““ c
/'(?) = -2c— 8x • - ---- /(v2 — c). So, we
2 “* c .
have to prove that
f(y/2 - c} > 0 2(l-cv/2^e) > (2-n)(l-r) 3 > c4-2\/2 - r. «
clearly true. Thus, the problem is solved.
Solutions84
i
We have two cases
I
Solution (by Anh Cuong):
We have that
3c 
a + 6 + c
3a 
a + 6+ c
tr
2(1 + y/n)
87. [ Kiran Kedlaya ] Let a,b,c be positive real numbers. Prove that 
a + Vab + \/abc 
3 “
3a
a I b I c
36 
a + b + c
a + b + c
3
a + '/ab + x/abc sin ——% , from
i + v/PTT i + " i+VP + l l + x/PTT
7T
where it follows that k k. -
Vietnamese IMO Training Camp, 1995
' a + b a + b + c 
a'~ 3
36
a + 6-i- c 
3
26
%b 3c + a -H 6
V a + 6a + 6 + c“ 3
Now, just add them up and we have the desired inequality. The equality occurs when 
a = 6 — c.
a + 6
2
85Old And New Inequalities
{>/n} > x/n - x/n - 1 =
siTi(n'{v/n}) > sin > 
Let us prove that the last quantity is at least . This comes down to
or 6(\/n + x/n - I)2 + 3(x/n +
i >
>
32it/z. Find the
Vietnam, 2004
6
First solution (by Tran Nam Dung)'.
We may of course assume that x + y + z = 4 and xyz = 2. Thus, we have to find the
But from the Cauchy-Schwarz Inequality we have 2x/k2 + k - x/k2 + 2k - k > 0.
\/k2 I A-) holds, this case is
2x/k2Tk - x/k2 -h 2k - A + 1
1 + v/PTA?
it
x/n- 1 -i- x/n
7T 
2(1 + ZA-2 + k)
1
x/n i/n — 1 ’
x/n - 1 + s/n
2(1 I- x/n)
7T2 / 
t(1 1
n - 1) > ^2(1 + y/n) and it is clear.
i) The first case
ii) The second case
89. ] Dung Tran Nam ] Let x, y, z > 0 such that (i + y + z)3 = 
. . z4 I / I z1minimum and maximum of ------------ rr.(x + y+ z)4
is when {x/n} x —— we infer that,6
x3 and because sinx > x - — we find that
6
____ 2
Because the inequality (1 I k -I x/k2 I 2A-) > 
also solved.
JT2
> “ 2 •
3 (1 + A- + x/k2 t- 2A-)
is when {x/H} > Let x = 1 - {\/n} - => p > k + 1. Then it is easy to see that 
---------- \ „ and so it suffices to prove that
A- + 1 + x/k^ + 2A-
.sm------------ . —
A- + 1 + x/k2 + 2A-
Solutions8G
extremal values of
Clearing paranthesis,we deduce that
a G 5.
But because
expression are
, 1 , which is an easy task.
Crux Mathematicorum
3i -
90. [ George Tsintifas ] Prove that for any a,b,c, d > 0,
(a + b)3(b 4 c)3(c + d)3(d + a)3 > 16a2b2c2d2(a + b 4 c + d)4.
5^/5- 1
2
3- •
2
, we find that the extremal values of the 
l + v/5 l + v/5\ 
2’2/’
i4 4 / 4 z4 
4*
383 - 165x/5 
256 
respectively (2,1,1).
2 8where a = xy+yz + zx. Because y+z = 4 — x and yz = -, we must have (4 -x)2 > x x
which implies that 3 - \/5 0.
of rhe function /(x) = 4x2 - 8x - - defined for 
X
(a - 16)2 - 112
128
9 /
’ PR ’ a*'ta’ne^ r°r triples I 3 - y5,
5
1 . But this reduces to the study
I 2-1 M 
----- . Now, we have
Second solution:
As in the above solution, we must find the extremal values of x2 4 y2 4 z2 when 
x 4 y 4- z = l,xi/z = —, because after that the extremal values of the expression 
x4 + y4 + z4 can be immediately found. Let us make the substitution x = -,y = 
b c , . , , , , a2 4 b2 4 4c2
-, z = -, where abc = 1, a 4 o 4 2c = 4. Then z + f + r =------- -—— and so we4 2 16 2 
must find the extremal values of a2 4 b2 4- 4c2. Now, a2 4 b2 4 4c2 = (4 - 2c)2----- h 4c2
]C 
and the problem reduces to finding the maximum and minimum of 4c2-8c— where
c
there are positive numbers a,b,c such that abc = l,a4 b4 2c = 4. Of course, this 
4
comes down to (4 - 2c)2 > -, or to c G
c
x4 I y4
4'
x4 I y4 | z4 — (x2 I- y2 I- z2)2 - 2 x2y2 -
= (16 - 2^xy)2 - 2(^xy)2 + 4xyz(x 4 y 4- z) = 
= 2a2-64a 4 288,
87Old and New Inequalities
x = abc, y = bed, z = eda, t = dab.
3
>
+ +
*c > anbn + bncn + cnan.
. Bui for
= £ -3.
Solution:
Let us apply Mac-Laurin Inequality for
91. | Titu Andrccscu, Gabriel Dospincscu | Find the maximum value of the ex­
pression
Thus, it is enough to prove the stronger inequality
(a + b)(b + c)(c 4- d)(d 4- a) > (a + b 4- c + d)(abc + bed 4- eda ~ dab).
Now, let us observe that
(a I- b)(b |- c)(c I d)(d I a) •• (ac \ bd \ ad I bc)(ac I bd I ab t cd) 
= (ac 4- bd)2 4- 52 u?(bc 4- bd 4- cd) > 4 abed 4- 52 (‘2(bc + bd + cd) = 
= (a 4- b 4- c 4- d)(abc 4- bed 4- eda 4- dab).
And so the problem is solved.
abc • bed • eda
4
1
1 - ab
a2b2c2d2 a
4
Thus, we have to prove that (ab)n 4- (be)" + (cu)n an(1 - a)n = an(b 4- c)n > anbn 4- ancn 4- nanb'
, (be)’1 . (car
1 — ab 1 - be. 1 - ca 
where a, b, c are nonnegative real numbers which add up to 1 and n is some positive 
integer.
So, we have proved that in this case the maximum is at most -—;—-
I 3 • 4n 1
a = b = -,c = 0 this value is attained and this shows that the maximum value is
* • *1 . .
3~ .n_t for n > 1. Now, suppose that n = 1. In this case we have
We will find that
I 4
Solution:
First, we will treat the case n > 1. We will prove that the maximum value is
- 4n-1. It is clear that ab, be, ca 108(a6c)2 + 23a6c + +a(l I 6) c(l I a)
(1 4- abc) +a(l + b)
i
2(a + b + c) - abc +3^
a(l + 6) 1 + 6 y]abc
And so we are left with the inequality
3
—= + 3v/a6c-3
yabc>
1 + abc
which is in fact an identity!
93. [ Dung Tran Nam ) Prove that for any real numbers a, 6, c such that a2 + 62 + 
c2 = 9,
1 + "----1 - ca
+ l(iT^) + 3 - £
- E
89Old and New Inequalities
= 2a I 2\/18 - 2a2 3.
We consider without loss of generality that x = max{z, y, z). Rom
(I+l)(y+l)(z+l) = afec+-7-+a+fe+c+- + T + - > 2+a+6+c+- + | + - — 5+z+y+c, 
abc a b c ab c-
1-! 
c
First solution:
With the notations x = a + - — l,u = 6 +---- 1, z = c +------1, the inequality
, b c a
becomes
Now, assume that not all three numbers are nonnegative and let c 2xyz — 6z — 2. Because 2xyz-6~—2 /3. Also, we may 
assume of course that a,b, c are non-zero and that a 0. If a 3a - 1 (a + 1 )2(a - 2) 1 we have 6, c > 1 and the inequality is again. Thus, the problem is solved.
First solution (by Gheorghe Eckstein):
Because max{a,6, c} 0 or exactly of the three numbers is negative. First, we will suppose 
that a, 6,c are nonnegative. If abc > 1, then we are done, because
2(a + 6 + c) - abc 
(c-iy
c
Solution:
We will prove that mn =
with equality if and only if abc = 1. Because y+z = -+b+ a
two cases a) x> 0, yz 0,y> 0,z> 0.
a) x> 0, yz 4 we get •
xy + yz 4- zx > 4 > 3.
b) x> 0,y> 0,z> 0. We denote xy + yz + zx — 3d2 with d > 0. From the mean
inequalities we have
xy 4- yz 4- zx > 3 \/x2y2z2,
from which we deduce that xyz 4, we obtain d3 4- 3d2 > 4, (d - l)(d 4- 2)2 > 0, d > 1 so 
xy l- yz I- zx > 3. With this the given inequality is proved. We have equality in the 
case a = b = c = 1.
95. [ Gabriel Dospinescu ] Let n be an integer greater than 2. Find the greatest 
real number mn and the least real number Mn such that for any positive real numbers 
Xl,X2,...,Xn (with Xn = 2>0>Zn-rl = Zl)>
n
1 
2(n - 1)
___________ Xi____________
Xi-i 4- 2(n - l)zj 4-Xi+i
1
2(n-l)
mn
I2
- > 0 we distinguish
> A-fn - First, let us see that the inequality
Second solution:
Let u = i41,i/ = y4 1,w = z 4- 1. Then we have
uvw — u 4- v 4- w 4- abc 4- -J- >u4ti + w+2.
abc
Now, consider the function f(t) = 2t3+t2(u+v+w)-uvw. Because lini fit) — oo and 
l—oc
/(I) 1 such that /(r) = 0. Consider the numbers 
UVWm = -,n = -,p = —. They verify mnp = m4-n4-p4-2we deduce from problem 49 r r r
that mn + np + pm > 2(m 4- n 4- p) => uv 4- vw 4- wu > 2r(u 4- v 4- w) > 2(u 4- v 4- w). 
But because u = if l,v = y 4- l,w = z 4- 1, this last relation is equivalent to 
xy 4- yz 4- zx > 3, which is what we wanted.
n 
__________ Xj_________ ____~ Xi-i 4- 2(n - l)Ti 4- .Ti-rt
xyz 4- xy 4- yz 4- zx > 4,
1
a
91Old and New Inequalities
for all i. This shows
that mn >
+
and thus
77ln —
Of course, it suffices to prove that for any
1 thenTaking
++
x2 + xy + y2 = (x + y + z)'2 - (xy + yz + zx) - (x + y + z)z,
Solution:
Considering the relation
But it is clear that 
n 
£
________ 1________
x + x”"1 +2(n-l)
1
2(n-l)
1 
y2 + yz + z2
1________
y/x\ I ■ -Ti I 1
Xi
positive real numbers, then
__ 1__ > __ 1—?
z2 + zx + z2 (x 4- y + z)2
Gazcta Matematica
n
-£----
»=l n — 1 +
y/Xj-! -1,4.1
Xi
x approaches 0, we find that mn -. 
x,, X2,..., xn > 0 we have
96. [ Vasile Cirtoaje ] If x, y, z are 
. 1 
x2 + xy + y2
H 
___________ Xj_______________ 
x,_i + 2(n - l)x, + Xj-i
n 
a,, we have co prove that if a.
- and because for Xi = X2 = •
Mn — which solves the problem.
2x, 
Xj-i + 2(n - l)x, + x,.
Solutions92
we gel
1 -
or
or
Vietnam TST, 1996
I
d(3d — 1)2 + (1 -4d + 9a6c) > 0.
which is Schur’s Inequality.
97. [ Vasilc Cirloajc ] For any a, b,c,d> 0 prove that
2(a3 + l)(b3 + l)(c3 + l)(d3 + 1) > (1 + abcd)(l + «2)0 + &2)(l + (1 + abed)4,
we notice that it is enough to show that that
24 n^3 + - IB1 + + a2)4>
Of course, it suffices to prove that 2(a3 + l)4 > (a4 + l)(a2 + I)4 for any positive 
real a. But (a2 + l)4 
(a 4- I)2(a4 + !) 2(a2 - a + I)2 > a4 + 1 (a - I)4 > 0, which follows.
98. Prove that for any real numbers a, b, c,
(a + ft)4 + (6 + c)4 + (c + a)4 > y(a4 + b4 + c4).
(j + y + z)2 _________ 1________
x2 + xy + y2 1 - (ab + be + ca) - c ’ 
y %b =----------- , c =----------- . The inequality can be rewritten
x + y + z x + y + z
___1___ 1 1 >
1— d — c^l-d-b^l— d —a ~ ’
where a, b, c are positive reals with a + b + c = 1 and d = ab + be + ca. After making 
some computations the inequality becomes
9da - (id2 - 3d I 1 I- Oabr. > 0
Old and New Inequalities •13
+ 1 + c + a
12 -I- 4x + y 2x + 4y => 2x + 4y + Iz. And we need to minimize x + y + z. But
2zz)2 = 3(52x2)
-yz)2 =3(52x2)2-4(52xy) (52®2) +
,2- (521)4 -2852i4
Solution:
Let us make the substitution a + b = 2z, b + c = 2x. c + a — 2y. The inequality 
becomes 52(1/ + z ~ x)4 3,y > 
3, x2 > 3y, we have . .
^x2y > 5x2, > y2,xy2 > 9x, 5xy > 15x, xy > 3y and x2y > 27.
Summing up these inequalities, the desired inequality follows.
2xy > 7
2x + 4y
2 > ------------
“ 2xy - 7
-— ------- f-1 + b + c
Solutions94
But, it is immediate to prove1 -I-
that 2 >
x + y -F z >
I
101. | Titu Andreescu, Gabriel Dospinescu ] Prove that for any x, y, z,a,b,c> 0 
such (hat xy + yz + zx = 3,
Second solution-.
We use the same substitution and reduce the problem to finding the minimum 
value of x + y + z when 2xyz > 2x + Ay 4- 7z. Applying the weighted the AM-GM 
Inequality we find that
Solution:
We will prove the inequality
for any a, b, c, x, y, z. Because the inequality is homogeneous in x, y, z we can assume 
that x + y + z = 1. But then we can apply the Cauchy-Schwarz Inequality so that
Now, we transform the expression so that after one application of the AM-GM 
2x + Ay 
2ry - 7
7
and so x I y I z > - I x -I — > —. We have equality for 2 x 2 
......................................................... 15 . . .r
2
-^-(y + z) + -~-(z + x) + -^t(x + y) > >/3(xy + yz + zx) 
u "f c c ~r u u t u
a . , b . . c , , „
(y I- z) 4 —— (z i x) f- ——(x I y) > 3. 
o + c c + a o + a
5 
with equality for x = 3, y = -,z = 2.
. ,2 225 15(z + y + z)2> — =>x + y + z>y
And also 2x + 4y -1- 7z > 10* • 12i • 5n • zs • y3 • z^. This means that (z+ . 
225
'•y + z)2(2z + Ay + 7z) > —xyz. Because 2xyz > 2x + Ay + 7z, we will have
11 7
x I’ I- y ~ z- 4' 2.t y 2r
Inequality the numerator 3.ry - 7 should vanish .r + y + z>z + y +
 2.t I — 
____
2ry - 7
3 1 . 3 . 9 15 ... .
---- - ---- iiuu au j, ■! y i 4 c. n > 
ffflTI
a 
b + c
102. Let a,b, c be positive real numbers. Prove that
(b 4- c - a)2 (c + a - b)2 (a + b - c) 
(b + c)2 + a2 + (c + n)2 + b2 + (a + b)2 + c2
ab 
(c + a)(c + b)
I>2 +
2 
+
xy (x/y+z + Vz + x + x/x + y)2-
(c + a - b)2 
(c + a)2 + b2
(o+t+c)f£±i + £+5 + £i£
\ b -F c c F a a -F b
2(x + y + z) = 5^ \/(x + y)(x -F z) - (x + y F z).
A good exercise for readers is to show that
' ^2 /(x I !/)(x + z) x *• y 2 1 \/3(xy I yz F zr).
b c /
—;—y +-----rz + V 3(xy + yz + zx) (n - 1)
~ann - 1
\ /
n
i*
/
E«?>(n-1)
Solution:
Let a, - an = > 0 for i G {1,2,... , n - 1}. Now, let us look at
\"
Sa”-n-na> “f71-1) 
i=l 1=1
X) + I2 4-------1- Xn-l
and so it is enough to prove that 
(t + y 4- z - 3)2 
x2 + y2 + z2 -I- 3 > j «■ (Ex)2-15£x + 3£xi/+18>0.
But from Schur’s Inequality, after some computations,
7 xy > 2 y x. Thus, we have
fy* x)' - 15 y r + 3 y xy + 18 > (y x)' -9 y x + 18 > 0,
the Inst. one being clearly true since y x > 6.
fz 
t=l
n - 1
we deduce that
as a polynomial in a - an. It is in fact
n-1 n-1 z
a" 1 y (a I Ii)n - na JJ (a I- x,) - (n- 1) (
We will prove that the coefficient of ak is nonnegative for all k G {0,1,... ,n-1}, 
because clearly the degree of this polynomial is at most n-1. For k - 0, this follows 
from the convexity of the function /(x) = xn
Second solution:
Of course, wc may take a + b + c = 2. The inequality becomes
V 4(1-ft)2 >3 _ v 1 27
^2 + 2(1-n)2 “ 5 1 + (1 -a)2 ~ 10’
But with the substitution 1— a = x, l-b = j/,l-c = z, the inequality reducesto 
that from problem 47.
Xi
103. ( Vasile Cirtoaje, Gabriel Dospinescu ] Prove that if dj, a2l..., an > 0 then 
fti + a2 + • • • + 0, the coefficient of ak is
104. [ Turkevici ] Prove that for all positive real numbers x, y, z, t, 
x4 -F y4 + z4 + t4 + 2xyzt > x2y2 -I- y2z2 + z2t2 + t2x2 -F x2z2 F y2t2.
Kvant
n 
ji - k
k 
n- 1+ xh
ac I bd.
>d (a+ b + c — 3 tfabc'j .
Because d \/abc f b I c - 3 v/abc^ . 
b c,v = — ,w = - Using problem 74, we find that 
yabc vabc
u2 + v2 + w2 + 3 > u + v + w + uv + vtu + wu
which is exactly a2 -F b2 + c2 - ab - be - ca > \/abc (a -F b I- c - 3x/abcj. Thus, it 
remains to prove that d2 + 2 > 3md d2 + 2 > 3\^bP, which is clear.
Solution:
Clearly, it is enough to prove the inequality if xyzt = 1 and so the problem 
becomes
If a, b, c, d have product 1, then a2 F b2 I c2 -F d2 -I 2 > ab + be F cd I da I 
Let d the minimum among a,b, c, d and let m = Vahc. We will prove that 
a2 + b2 -F c2 4- d2 + 2 - (ab + be + cd + da + ac F bd) > d2 -F 3m2 + 2 - (3m2 F 3md). 
which is in fact
a2 + b2 + c2 — ab — be - ca
z
which is clearly smaller than
polynomial is nonnegative and so this polynomial takes nonnegative values when 
restricted to nonnegative numbers.
-n t.
i
5 Qi
2 ' bi
/ 11,1 / »
..Jo .
n
Now, using the Cauchy-Schwarz Inequality for integrals, we get
2
2 >.
*E
a2
b? +
= [\t+i-2dt.=
ij=i Jn
= f ( E iai ■ | dt =
Jo /
/•i
^,2 for any i and that for all x 6 1 J -,2 we have
Solution:
a. 
The key ideas are that — e 
bi
the inequality x2 + 1 8(a2b2 + b2c2 4- c2a2)2.
Solution:
Let x = -, v = a
z
Recall Schur’s Inequality
52 + abc(a 4- b 4- c) > 52 a‘3(b + c) ** a^c(a + b 4- c) > 52 fl3(b + c — a)- 
Now, using Holder’s Inequality, we find that
1 
b'Z =
___
J 
+ c - a 
Combining the two inequalities, we find that 
I abc 
b + c — a
T — y , 1
y-' 4- c - a
a?
Now, the observation that and the inequality | > 1 4- p- allow us to
5 a3write -a2 > — + atbt and adding up these inequalities yields 
2 bi
t=l i=l ' z i=l i=l
Using (1) and (2) we find that
a? a2 a?i I? , 2 2 2\Vi+b2+'"+b;~W^ +“2 + -'- + “n).
which is the desired inequality.
i. We find the equivalent form if — 4— 4— = 1 then 
c r. y z
(x2 4- y2)(y2 4- z2)(z2 4- x2) > 8(x2 4- y2 4- z2)2.
We will prove the following inequality
\ •*' y 
for any positive numbers x, y,z..
Write x2 4- y2 = 2c, y2 4- z2 = 2a, z2 4- x2 = 2b. Then the inequality becomes 
abc 
b 4- c - a
Solutions100
> 1.+ + +
>
>+
>0.
a2 + b2
a
>0. •
1 
(14-a)2
1
(1 + b)2
1 
(1 + c)2
1
1 + ab'
1
1 + cd
c2
a2 + b2
Equality holds if a = b = c - d = 1.
c2 + a'
a \ ab(a — b) 
" b + c (b + c)(b2 — c2)
1
(1 + d)2
Gazeta Matcmatica
108. ( Vasile Cirloaje ) If a,b, c, d are positive real numbers such that abed = 1, 
then
 ab(a - b)
(a + c)(a2 + c2). 
 = (a2 + b2 + c2 + ab + be + ca) • Y* —---- -—&\v~2-----2\
•. 7 (b + c)(c +a)(b2 + c2)^ + a2)
109. ( Vasile Cirtoaje ] Let a, b, c be positive real numbers. Prove that 
a2 . b2 c2 a b c > —---- -f. ------ 4. -----_
b + c c 4- a a I- b
Gazeta Matcmatica
Solution:
We have the following identities
a nb(a - b) + ac(a - c) 
b + c (b + c)(b2 + c2)
bc(b - c) I ab(b - a) 
(c + a)(c2 + a2)
Thus, we have
^b2 + c2
Solution:
If follows by summing the inequalities 
1 , 1 
(1 + a)2 1 (1 + b)2 
1 . r’
(1+c)2 + (1+d)2 
The first from these inequalities follows from
1 1 1 ab(a2 + b2) - a2b2 - 2ab + 1
(1 4- a)2 + (1 1- b)2 1 4- ab (1 + a)2(l + b)2(l + c)2
‘ ab(a - b)2 + (ab -1)2 „
' • ~ (l + a)2(l + b)2(l + ab) “
a2
f^ + c2
b2 b
c2 + a2 c + a
c ac(c - a) + bc(c - b) 
a + b (b + a)(b2 + a2)
b2 I c2 +
101Old find New Iiicqunlitin*
TST 2004, Romania
bi =
= Clibi + 02^2 + ' ' • + Clnbn —
= Si(6j — 62) + 32(62 — 63) + •■■ + sn_i(6n_j - 6n) + sn6n + s„+i(-6i)
2
l. ■. be real numbers and lei S be a 
non-empty subset of {1,2,... ,n}. Prove that
First solution:
Denote St = Oj + • • • + ajt for k = 1, n and also sn+j = 0. Define now
if i € S 
, otherwise
E ~ s«)2 = 
l)2
n+1 
E*
'n + 1 z>.
Z2s 4 y S2i+is2;+i.+4 y
4=1 0
y siSj. 
l 0, we can write
} SiSj(liXj + 1) iH
(s? + ••• + s’ + 1) + s’li(l2 +13 + ••• + Xn) + ••' + S’zn(ll + 
i2 + • • • + in-i) = -s'ix'i--------s’x’ + n(s? + • • • + s’).
S24S2J “ 2
’ ' n+l n+l ,
)2 
l2 € S, then we apply the inductive step 
with the numbers ai + — — > 2ab and thus we have (aj +’ a2 4-------t- a*)2 4-
(a2 4- a3 4- • • • + a*-i)2 > 2aia* (*). Also, the inductive step for 03,... ,an shows that
Now, let us turn back to the solution of the problem. Let us call a succession 
o a> s a sequence and call a sequence that is missing from S a gap. We group the 
successive sequences from S and thus S will look like this
{nu > at> 11»• • •, 
Solutions1114
-k.
x/n — l(ai + 0-2 4- * • • + — n) >
- (n - 1) n~^a2lal...a2na2 + a2 + • • • + a2 >
k
2004 - k
112. [ Gabriel Dospinescu, Calin Popa | Prove that if n > 2 and aj,a2,...,an 
are real numbers with product 1, then
a2 + a2 + • • ■ + a2 - n >----- - • x/n - 1 (a, + a2 + ■ • • + an - n).
71 — 1
1. f(x + y + 1) + /(-I) > /(x) + f(y) if -1 0 and y 0 then f(x) + f(y) a3+lr3+(l-a-b)3 1 > (a+b)(a2—ab+b2)+l—3(a+b)+3(a+b)2-(a+6)3 
 3(a + b)(l - a)(l - 6) > 0, which follows. The second statement is clear and the 
third one can be easily deduced in the same manner as 1.
From these arguments we deduce that if (ti,t2> • • •, £2004) = t is the point where 
the maximum value of the function g : A = {x 6 [-1, l]2004 + ■ •• + xn = 0} —»
2004
—» R,y(ii,...,x2oo4) = '2-jJ(xk) (this maximum exists because this function is 
defined on a compact) is attained then we have that all positive components of t are 
equal to each other and all negative ones are -1. So, suppose we have k components 
equal to -1 and 2004 - k components equal to a number a. Because ti4-t24----^2001 =
k '
0 we find that a = —- and the value of g in this point is (2004-fc) y
I fa
Thus we have to find the maximum value of (2004 — k) (7 —— — k, when k is in 
the set {0,1 2004). A short analysis with derivatives shows that the maximum is 
attained when k — 223 and so the maximum value is i/223 • x/l 7812 - 223.
d2 + o.2 + • + Q2 — n
Solution:
We will prove the inequality by induction. For n = 2 it is trivial. Now, suppose the 
inequality is true for n -1 numbers and let us prove it for-n. First, it is easy to see that 
it is enough to prove it for aj,...,a„ > 0 (otherwise we replace ai,a2,...,an with 
|ail> lazL - • - > |QhI* which have product 1. Yet, the right hand side increases). Now, 
let an the maximum number among ai,a2,...,an and let G the geometric mean of 
ai,a2,... ,an_j. First, we will prove that 
2n 
n — ’ 1
> a2 + (n - 1)G2 - n - • vTi- l(an + (n - 1)G - n)
n — 1
which is equivalent to
105Old and New Inequalities
>
- (n — 1)G).
\/n — 2 — n 4- 1
• \/n - l(fl|d-------Fan-i —(«—1)^):
which is the same as 1 4- >
1 -F
1 4-1 4-
• tyn - l(an. 4- (n — 1)G - n)
• 1/n^T- n >
We have
/'(z) = 2-
n(n-l)
>2
H1+
(n - l)(in 4- -1) 
x .
1 
n(n — 2)
2(n-l) 
n - 2
2n 
n — 1
. 1
n(n - 2)
2n 
n — 1
e j
n-2 1
1 
n-2
ai + • • • + an-i
G
1 
n-2
1 
n-2
• >/n — 1 • G • (ai + • • ■ + fln-1
-nv5Tl >0
for all x > 1 (we took x =
/(z) = x'
(n — l)n-1 
(n-2)"
1
n(n — 2)
and it follows for n > 4 from
xn — 1 
x2
_n-i x 4-^(n-l) + n — 1
----------- n
x
.2(n-l) + n — 1 
----------- n
x
2n 
n — 1 
x/n - 1 ,
— ----- This becomes
‘~x/n — 2 
\ n(n-l) 
) >
_n-1 X +
and
(n- l)n~r. 1 
(n — 2)n “n-2
For n = 3 and n = 4 it is easy to check. 
Thus, we have proved that
aj 4- 02 4------- F a2 — n —
Q__ .
—-—- • >/n — l(ai 4- 02 4- •—Fan - n) >
n — 1 . . ■ . ,
. ^TTI(Qn + (n _ 1)G - n) 
n — 1 ■
2ti------ - ■ Vn-1 (ai 4- a2 4- • ■ ■ 4- an-i - (n - 1) "-^0102 ... an_i). 
n — 1
Because, •—. an_j 
0, it is enough to prove the inequality
a? + -*- + an-i -(n-l)G2 >
Now, we apply the inductive hypothesis for the numbers 7-,..., % 1 wliich have
G G
product 1 and we infer that
Qi + --+Qn-i
G2
and so it suffices to prove that
^^•"-y^2(a14---Fon_1-(n-l)
> a2 -F (n - 1)G2 - n - 
and it is enough to prove that 
n-1 
x2
i). Let us consider rhe function 
G
n-1
Solutions1U6
because
> nl'j+ ■•■ +x
Gazeta Matematica
First solution:
 /(I) = 0. This proves the inequality.
1 
(n-l)n-i-
2
1 +1/2
We presume that x 1. We have
' ‘ ‘ l-x2y
(1 + x2)(l + y2)
2
1 + y2
+ - = in-1 + 
x
113. [ Vasile Cirtoaje ] If a,b,c are positive real numbers, then 
l~2a~ I 2b I~2c~ .
Va + ^^V^ + c + Vc + u~
2
1 +y2
the problem reduces to proving
2
1 + y2
1 - x2y2
(1 + xy)2
2 
-----r H 0,(v^z - x/z+T)2 > 0,
107Old and New Inequalities
 xy + 1 + (zy + I) 0
and so exactly two of the numbers z, y, z are smaller chan 1, let them be z and y. So, 
we must prove that if z and y are smaller than 1, then 
' 2zy 
zy + 1 
Using the Cauchy-Schwarz Inequality, we get 
r~2~ / 2zy
y + 1 V xy + 1
and so it is enough to prove that this last quantity is at most 3. But this comes down 
to
E 2(zy 4- z 4- y 4- 1) 
z 4- y 4- z z4-y> \/2zy(Ty 4- 1)
which follows from i/2zy(zy 4- 1) 1, the left hand side is at most
But
E— z 4- y 4- z. In this
case we can apply the Cauchy-Schwarz Inequality to get. ;.
3E A-z 4 1 •
_ i =
(x+ I)(y 4-1)
r + 7TT~‘ 
' 1 
z 4-1
1 '
1/4-1 ~ ’ 
1 
z 4- 1
and so we are left with the inequality
1 - zy 1 - zy
(z4- l)(y 4- 1)
Solutions108
(xy + yz + zx) +4-
(y + *)2
Iran, 1996
i
(a - b)2 > 0.
(a-b)2.(b-c)’2>(a - c)2 +
(a - b)2.(b-c)2>
s at most
3
First solution (by lurie Boreico)'.
With the substitution x 4- y = c, y 4- z = a, z + x — b, the inequality becomes after 
some easy computations
which shows that the left hand side is at least 
solution.
1 
b
1
c
2----- b > c. If 2c2 > ab, each term in the above expression is positive and 
we are done. So, let 2c2 ac, 2a2 > be. Suppose that 
2b2 1 +1 > 
ac be
Second solution:
Since the inequality is homogenous, we may assumethat xy + yz + zx - 3. Also, 
we make the substitution x 4- y 4- z = 3a. FYom (x 4- y + z)2 > 3(xy + yz + zx), we 
get a > 1. Now we write the inequality as follows
1 : 1 1 3
(3a - z)2 (3a - y)2 + (3a - z)2 “4’
4[(zy 4- 3az)2 + (yz 4- 3az)2 -I- (zx 4- 3ay)2] > 3(9a - xyz)2, 
4(27a4 - 18a2 4- 3 + 4azi/’) > (9a - zys)2,
3(12a2 - l)(3a2 - 4) 4- zyz(34a - xyz) > 0, (1)
2_ _ _1 
be a2
We can immediately see that the inequality (a - c)2 > (a - b)2 F (b - c)2 holds 
and thus if suffices to prove that
/ 2_ _2 _ _1_ _ T
\ ac \ be a2 b2
114. Prove the following inequality for positive real numbers x,y,z 
1 1 1
(x 4- y)2 + (y 4- z)2 + (z 4- r)2
2 2 _L _1 ac + be a2 b2
Old and New Inequalities 109
(2)
and
r (x) =
r =
115. Prove that for any x,y in the interval [0,1],
\/l + x2 + \/l + y2 + \/(le— x)2 + (1 - y)2 > (1 + x/5)( 1 - xy).
, c = 1 + \/5.
12(3a2 - l)2 + 208a2 > (17a -xyz)2.
We have two cases.
i) Case 3a2 — 4 > 0. Since
34a — xyz = | [34(x + y + z)(xy + yz 4- zx) — 9xyz] > 0, 
the inequality (1) is true. ' •
ii) Case 3a2 — 4 0,
it follows that 3a3 - 4a + xyz > 0. Thus,
12(3a2 - I)2 + 208a2 - (17a - xyz)2 > 12(3a2 - I)2 + 208a2 - a2(3a2 + 13)2 = 0.
Solution (by Faruk F. Abi-Khuzam and Roy Barbara - ”A sharp in­
equality and the iradius conjecture”);
Let the function F : [0,1]2 —♦ R, F(x,y) = \/l + x2 + \/l + y2 + 
\/(l - x)2 + (1 - y)2 — (1 + x/5)(l - xy). It is clear that F is symmetric in x and 
y and also the convexity of the function x —♦ >/l + x2 shows that F(x, 0) > 0 for all 
x. Now, suppose we fix y and consider F as a function in x. It.’s derivatives are
T * 1 — X
. J'(x) = (1 .1 Vb)y I -7=y== - /. .2 .. .2
Vi2 + 1 v(l - x)2 + (1 - y)2
i + (i-v)2
x/d + x2)3 ^((1 _ x)2 + (I - y)2)3
Thus, f is convex and its derivative is increasing. Now, let 
f 6^3 
(1 + x/5)2
The first case we will discuss is y > 7. It is easy to see that in this case we have cy > 
1 • .. (the derivative of the function y —» y2c2(y2 - 2y + 2) - 1 is positive)
vy2-2y + 2
and so f'(0) > 0. Because f' is increasing, we have /'(x) > 0 and so j is increasing 
with /(0) = F(0. y) = F(y, 0) > 0. Thus, in this case the inequality is proved.
Solutions110
,y e [0, r] and x 6
f'(x) 0.
(n - 1 )(aj* + a'2 + • • • + aj() + naia.2.. .an > a" + •■• + y 6Now, let us discuss the most important case, when x e
the points 0(0,0),>1(1,0),B(l, 1),C(0,1),M(l,y),N(x, 1). The triangle OMN has 
oerimeter
116. [ Suranyi ] Prove that for any positive real numbers ai, (a! + a2 +
Due to case 1 and to symmetry, it remains to show that the inequality holds in 
[ r r
the cases x e 0, -
In the first case we deduce immediately that
x
\/l + x2
r i]N’
H 71
+ n.a."|} + nan+i • a, + an+i • a.
H]-Let
\/l + x2 + \/l + y2 + \/(l - x)2 + (1 - y)2 and area 1 - xy
2 ‘
But it is trivial to show that in any triangle with perimeter P and area S we have the
P2 ,______ ,______ ___ ________________
’ inequality S /(l - x)2 + (1 - y)2 > 
___ 12v3
x/eTsyr - xy > (1 + x/5)(1 - xy) due to the fact that xy > r2. The proof is complete.
/ n
— (1 + an+i)
\i= 1
0, we will have F(x,y) = /(x) > 0 for all 
.3/ e [0,r].
Thus, we have f
the first case discussed we have f 
points (x,y) with x € 0, i
Solution:
Again, we will use induction to prove this inequality, but the proof of the inductive 
step will be again highly non-trivial. Indeed, suppose the inequality is true for n 
numbers and let us prove it for n + 1 numbers.
Due to the symmetry and homogeneity of the inequality, it is enough to prove it 
under the conditions ai > a2 > • • • > an+i and aj + r’ 
i=l i = l
But from the inductive hypothesis we have
1 - x
+ (x - I)2
-|-Qn)(a’1’ ‘-l-a” 1 + ---+o" x).
IllOld and New Inequalities
and so
+n — On + l
_n—1
~ an+.l > o.
>0an+l
and
>0.— an+i
Let
’I' On 1 n ■
1
Now, we will break this inequality into 
n '
J] a 0 with product 1,
E (Xi-ij)2 >Exi ~n- 
l a" for all t. Thus, the inductive step
n n
an+IEar-1 - (n- o^+iE*2"-
Using this last inequality, it remains to prove that
E-?+1-E 52
fc=2 k=2
(n \
-2x1 (- (n~ l •
\k=2 /
Solution,'.
Of course, the inequality can be written in the following form
> -(n- 1) (^x'l
\k I
1
G2 (n - 1)G, so it suffices to prove that
. - *=2
(n- 1)52 x2 - ( ^xk 
Jc=2
I n 
>2G £xfc 
\fc=2
We will prove that this inequality holds for all X2,1... ,xn > 0. Because the inequality 
is homogeneous, it is enough to prove it when G = 1. In this case, from the induction 
step we already have
n
ik- 1
.. We will prove this inequality by induction. The case n •- 2 is trivial. Suppose- the • 
inequality is true for n - 1 and let us prove it for n. Let
/ n
/(xi,X2,...,x„) - —(n — 1) xk
n -2 2n - 2
G'^n-2 + n - (Jn-2
and this one is an immediate consequence of the AM-GM Inequality.
clearly true.
Thus, we have proved that J(xi,X2, ■ ■ ■ ,xn) •.. ,x,t}. 
It is easy to see that the inequality /(xi,i'2> • ■ • >^>1) S /(xi, G, G,..., G) is equivalent 
to -
n
Xk 
,k=2
(n~2)^x 2k+n-l> [ 
k=2
and so it suffices to prove that
n
 J2(xfc -1)2 > 0, 
k=2
+ „>{(„-l)C+d^T
Old and New Inequalities 113
. Indeed, using Suranyi’s In­
fl - (n — l)a,).
Now, let us observe that
2
2-
But for n > 3, we can apply Jensen’s Inequality to deduce that
1
n
abc
a =
flirt2 ■■ ■an 
1 — (n - l)a,
E^o - 
119. [ Vasile Cirtoaje ] Let 3. 
For n = 3 it reduces to proving that
^Vb + c-a
which was proved in the solution of the problem 107.
E«r
where ui,a2,... ,an 2 is an integer.
k 1
71
(\/l - (n - l)afc
and so, by an immediate application of Holder Inequality, we have
3
n
=> naia2---an > E"'’
E^
LJ------ >
it
Solutions114
Prove that
> 1 - a2’
for
a2 + a| + , ■. -F a?a =
x -
from
>
From
x2 — a2 —
we obtain
(k - l)(x - a) =
+ (*-1)
+ +
Since x2 - a* =
(o - afc)(x - afc) >0,
(a 4- at)(l + ax)' _ 
(l-x2)(x + a)
I-------- -- "T'
We assume, without loss of generality, that ai > «2 > ... > at, therefore a > ak. 
Using the notation
ka2 - a2 
k-1
1 - x2
a 
1-a2
na
1-a2'
(k — l)x
1 — x2 ’
ka2-«i 
k — 1
a2 + a2 + • • • + a2 ka2 - a2 
k - 1
Q2 ~ «t 
k — 1 ’
x
1 - x2
- Qk [-(lb aafc) 
1-a2
Qk a
1 - a2 1 - a2
- - a2—
A "
rF...+
Uk 
l~ak
v3a > at it follows that x > u > —Thus, by the inductive hypothesis, we have 
o
(k - l)x
1 — x2
4 T^F2 +
and it remains to show that
1 - a2 +
-^- + 1-a?
------ j + • • • + 1 - aj
and
_^ + 
1-a2
-(a - Ofc)(l I- aat) , (k - l)(x - a)(l -F ax) a
(l-a2)(l- a2) + (l -x2)(l - a2) “ F^a2
 (a - aji)(x - afc)[— 1 + x2 + a2 + xat + a(x -F at) + a2 -F axat(x + at) -F a2xafc]
(1 - a2)(l -a?.)(l -x2)(x + a)
-a;.= fc(°-°4(°-m)|ilfollowsthtt
_ A-(a - at)2(a + at)
(k- l)(x + at)
Solution:
We proceed by induction. Clearly, the inequality is trivial for n= 1. Now we 
suppose that the inequality is valid for n = k — l,fc>2 and will prove that 
aj , a2 ak ka
1-a?
ka- r^2-
Old and New Inequalities 115
and hence we have to show that
x2 + a2k =
and
>
1
a >
14-
(therefore a = x
n i n 7TZT + 77ZT
(a 4- b + c)(x 4- y + z) = (a2 4- b2 4- c2)(x2 4- y2 4- z2) = 4.
Prove that
120. | Vasilc Cirtoajc, Mircea Lascu | Let a,b, c, x,y, z be positive real numbers 
such that
Consequently, we have
x2 4- uk + xak 4- a(x +ak) 4- a2 4- axak(x + u*) 4- a2xa* > (x2 4- a2) 4- u(x 4- «
Solution:
Using the AM-GM Inequality, we have
4(a64-6c+ca)(xy-t-yz4-zx) = [(a 4- b -r c)2 - (a2 -t- b2 4- c2)j [(x + y + z)2 - (x2 - y2 4- z2)] = 
= 20 - (a 4- b 4- c)2(x2 4- y2 4- z2) — (a2 4- b2 4- c2)(x 4- y 4- z)2 1.
In order to show this inequality, we notice that 
ka2 + (k - 2)a2 
k — 1
ka2
-
k
A- -1 '*
and the proof is complete. Equality holds when ai = a2 = • • • = an.
Remarks.
1. From the final solution, we can easily sec that the inequality is valid for the 
larger condition
x I ak > y/x2 I a2k > a^/
v32.‘ The special case n = 3 and a = — is a problem from Crux 2003.
U
, 1 
abexyz 3a'2 1,
k - I /
n | u 
n-1 n-1
This is the largest range for a, because for ay = 02 = ••• = an_i = x and an = 0
‘ I
-------), from the given inequality we get a > 
n
Solutions116
therefore
(I)
(ay + yz -I- zz)2 > 3zyz(z + y + z),(ab + be 4- ca)2 > 3abc(a 4- b + c),
it follows that
(ab + be 4- ca)2(zy + yz 4- zz)2 > §abcxyz(a 4- b + c)(z 4- y 4- z),
or
1 36abczyz,
 36abczyz. (2)
From (1) and (2), we conclude that
(ab 4- be 4- ca)(zy 4- yz 4- za) 2, find the minimal value of the 
constant kn, such that if xi,x2 xn > 0 have product 1, then
1 1 ‘ 1 
— -f" ~ 4- • • * 4" — 
Vi 4- I'nZi VI 4- knx2 VI 4- k„xn
1 
2n - 1 
(n - 1)2
xu = 1. Suppose this inequality doesn’t hold for a certain system of n 
numbers with product 1, zi,z2,... ,x„ > 0. Thus, we can find a number M > n - 1 
and some numbers a, > 0 which add up to 1, such that
1____
2n - 1 
(^Ip
2n- 1 
(n-l)2
So, it remains to show that
= SGabexyz.
Old auid New Inequalities
and we haveThus, ak 
,k-l
n
>
1 +
n
P[ (di + a-2 4------F ajt -1 -F a*.- i i 4------F an)2 >
holds.
This strong inequality will be proved by induction. For n = 3, it follows from the 
fact that
n
Now, denote 1 — (n — l)at = 6* > 0 and observe that ^6* — 1. Also, the above 
inequality becomes
2
IPl-M •
I[(l-a.)2>
n
 (2n - l)n 
k=l fc=l
Because from the AM-GM Inequality we have
2n - 1
n
our assumption leads to
n(i-bk)2 
k=l
So, it is enough to prove that for any positive numbers ai, a2,..., an the inequality
(n - l)2n
nn
aiftt ... an(ai Faj -I------F nn)n
(iW’
abc abc 27 k2—J /
Suppose the inequality is true for all systems of n numbers. Let ai,a2,... ,an+i 
be positive real numbers. Because the inequality is symmetric and homogeneous, we 
may assume that ai —+-
Thus, it is enough to prove the stronger inequality
2 n:‘,,+2
~—i-^r;-(nVi)"‘^an+,(1 + an-,),,T
i
n-- 1
f(n-l)2 ,
2n-.l 1
Solutions118
2n
1 I>1 I
/
1 +
1 -(- x, where x is
1 +
1 +
Also, (1 -I z)
>
1
122. | Vasilc Cirtoajc, Gabriel Dospincscu | For a given n > 2, find the maximal 
value of the constant kn such that for any Z;, x2,..., xn > 0 for which z, + z2 4------ 1-
z„ — 1 we have the inequality
(1 -Zi)(l-z2)...(l -rn) > knx\x2...xn.
1 + (n + l)z 
(z + I)""1
tlO>i+1 
n - 1
n«n+i 
n - 1
\ f[(l “*»*)
2n
>
2
>
■a>i+l (1 + 
Now, using Huygens Inequality and the AM-GM Inequality, we find that
/ V"
which is trivial.
So, we have reached a contradiction assuming the inequality doesn’t hold for a 
certain system of n numbers with product 1. which shows that in fact the inequality 
*2ti “ 1is true for k„ = --------— and that this is the value asked by the problem.(n-1)-
Remark.
For 7i ~ 3. we find an inequality stronger than a problem given in China Math­
ematical Olympiad in 2003. Also, the case n - 3 represents a problem proposed by 
Vasile Cartoaje in Gazela Matematica, Seria A.
n3nt2
(n - l)2n ■ (n + l)n+1
r , 1 c "C1 + an+l)if “n-n > maxfaj,n2l...,an) > — • bo, we can put----—-------- =
nonnegative. So, the inequality becomes
z 
n(z + 1)
Using Bernoulli Inequality, we find immediately that
z \2n > 3z + l 
n(z 4-1)/ — x + 1
> 1 I- (n - l)z and so it is enough to prove that 
3z + 1 1 4- (n + l)z
z + 1 ~ 1 + (n - l)z
1 I
1 - «>
119Old and New Inequalities
n
n
 (y/n - l)n • U y/ai ■ JJ(1 + y/ai).
Glossary
Si = G] + (12 + ... + Gj,l = 1,2, n,
then
This inequality is a special
= an. This inequality is a special
121
• . (4) Bernoulli’s Inequality
For any real numbers x > — 1 and a > 1 we have (1 + x)° > 1 + ax.
with equality if and only if oi = 02 = ... = a, 
case of the Power Mean Inequality.
n n-1
52 — b,+i) + Snbn.
(2) AM-GM (Arithmetic Mean-Geometric Mean) Inequality 
If Gi,02,.-,an are nonnegative real numbers, then
1 n- Ea* > (ala2...an)”,
(3) Arithmetic Mean-Harmonic Mean (AM-HM) Inequality
If ai,a2,...,an are positive real numbers, then
i n
■. ■ 
1-1 1
n
1 
n 152-
with equality if and only if ai = a2 =
case of the Power Mean Inequality.
(1) Abel’s Summation Formula
If ai,G2)... ,an, bi,b2,—,bn are real or complex numbers, and
Solutions122
n.
1 / nProve the following inequalities
a) xyz > 27;
b) xy + xz + yz > 27;
c) x + y 4- z > 9;
d) xy +■ xz 4- yz > 2 (1 + y 4- z) + 9.
28. [ D. Olteanu ] Let a,b,c be positive real numbers. Prove that
a + b a b 4- c b c 4- a c
■ • ■ "' -f- ■ ■ • ■' “f- , —■■ - —
b 4- c 2a4-b-l-c c + a 2b4-c4-a a4-b 2c4-a4-b
Gazeta MatematicS
Problems12
also holds for any n.
IMO Shortlist, 1986
Russia, 2002
++
where a,b.c,d arc real numbers whose sum of squares is I.
 2. 
Prove that
34. Given are positive real numbers a, b, c and x, y, z, for which a-l-x — b + y = 
c 4- z = 1. Prove that
35. [ Viorel Vajaitu, Alexandru Zaharcscu ] Let a, b, c be positive real numbers. 
Prove that
36. Find the maximum value of the expression
a '(b 1 c + d) + b'(c 4 d + a) + c'1 (d + a 4- b) + d‘*(a -I- b c)
 0 up to I and n > 2.
Crux Mathematicorum
ciicij + asa^ + ... + anaf > a? a' + ^/ a b c \ 
a b c “ \b4-c c4-a a + bJ
33. Find the maximum value of the constant c such that for any 
Xj,x2,...,x„,• • • > 0 for which Xk+i > Xi 4- X2 4- • • • 4- x«.- for any k, the inequality
a
b + c
b
c+ a
1
4
1
ay bz
Old and New Inequalities Id
, where z = max {x, y, z}.
4-+1 - a2 1 - b2
1
4-
42. [ Manlio Marangelli ] Prove that for any positive real numbers x,y,z, 
3(x2y + y2z 4- z2x)(zy2 + yz2 + zx2} > xyz(x 4- y + z)3.
41. [ Mircea Lascu, Marian Tetiva ] Let x, y, z be positive real numbers which 
satisfy the condition
47. [ Titu Andreescu, Gabriel Dospinescu ] Let x,y,z 3a2b 4- 3b2c I- 3c2 a
1 -b2 
b
3> -
~ 4
“ 10
40. Let ai.as,... ,a„ > 1 be positive integers. Prove that at least one of the 
numbers “{/aL “^03,.... “t/aj is lass than or equal to >/3.
Adapted after a well-known problem
1 -c2
4- -------
1 
14-y2
(2z-l)2 
(z(2z+l))
I-a2 
--------4- a
46. [ Calin Popa ] Let u,b, c be positive real numbers, with a,b, c € (0,1) such 
that ab 4- be 4- ca = 1. Prove that 
a b c 
1-c2
xy 4- xz 4- yz 4- 2xyz - 1.
Prove that the following inequalities hold
a) xyz-~;
c) - 4- - 4- - > 4 (x 4- y 4- z);
x y z
oil1,,d) -4----- 1------ 4 (x 4- y 4- z) >x y z
1 a2 145. Let do — - and a.k~i — o.k 4—Prove that 1---- 6(n 4- b 4- r) - f 7 + - ) . k a b c)
1
1 4- x2 +
Problems14
50. Prove that if x, y, z are real numbers such that x2 + y2 + z2 = 2, then
x + y + z 3 and ai,o.2, ...,an be real numbers such that 
ai -f az 1.. .-Hn > n and a2 + a% I-... I d„ > n2. Prove that max{aila2,..., an} > 2.
/ nn 
St
1
— Xi • Xo(i)
48. | Gabriel Dospinescu ] Prove that if y/x + y/y + yfz = 1, then 
(1 - x)2(1 - t/)2(1 - z)2 > 215xyz(x + y)(y + z)(z + x)
55. If x and y are positive real numbers, show that xv + yz > 1.
France, 1996
49. Let x, y, z be positive real numbers such that xyz = x + y + z + 2. Prove that
(1) xy + yz + zx > 2(x + y + z);
3
(2) y/x + y/y + y/~Z 
- Xi
n
52. Let X].X2,...,xn be positive real numbers such that
i=l
that
Old and New Inequalities 15
MOSP, 2001
n
a3 4- b3 4- c3 4- abed > min
Kvant, 1993
Va
Korea, 2001
56. Prove that if a, b,c > 0 have product 1, then
(a + b)(b + c)(c 4- a) > 4(a 4- b 4- c — 1).
57. Prove that for any a, b, e > 0,
(a2 4- b2 + c2)(a + b - c)(6 + c - a)(c 4- a - b) (14- a2)(l + b2)(l + c2)(a- b)2(b- c)2(c- a)2.
AMM
60. Let a, b, c, d > 0 such that a + b 4- c = 1. Prove that
11 d
4’ 9 + 27
/ n \
(Xlt/2 - X21/1)2 0. Prove that
1 1 1 « b c „3 + a + b + c+ —I- 7 4— + 7 + - + — >3a b c b c a
62. [.Tit.u A.ndrcescu, Mircea Lascu ] Let a, x, y, z be positive real numbers such 
that xyz = 1 and a > 1. Prove that.
xa ya zQ 3
-------- 1- —— 4--------->y + z z + x x + y 2
59. [ Gabriel Dospinescu ] Prove that for any positive real numbers ii,X2,... ,xn 
with product 1 we have the inequality
n” ■ (x? + 1) > 4- 52 .
i=l \i = l i=l /
63. Prove that for any real numbers xi,...,xn,yi,...,yn such that zj4----- i-z2 =
Vi+••• + !/’ = 1>
(a4- l)(b4- l)(c4-1)
1 4- abc
Kvant, 1988
Problems16
(fii 4- aj + • • • + on).
TST Romania j
67. Prove that
APMO, 2004
are true. _■
TST 2001, USA
70. ( Gabriel Dospinescu, Marian Tetiva ] Let x, y, z > 0 such that
x 4- y 4- z — xyz.
Prove that
(x~l)(p-!)(?- !) 9(fl5 + 6c + ca) 
for any positive real numbers a, 6, c.
2n 4- 1
3
68. [ Vasile Cirtoaje ] Prove that if0 0;
b) x2y 3/3
- 4 ‘
by/c cy/a ay/b
a(y/3c 4- y/ab') b(y/3a + y/bc) c(\/3b4- y/ca)
69. [ Ti in Andreescu ] Lcla.b.cbc positive real numbers such Lhata4-b+c> abc. 
Prove that at least two of the inequalities
2 3 6 2 3 6 2 3 6 „
abc b c a cab
64. [ Laurencin Panaitopol ) Let ai, a2,... ,au be pairwise distinct positive inte­
gers. Prove that
o2 4- a2 4- • • • 4- o2 >
Old and New Inequalities 17
 rr 4- 4 4-
a2 4- b2 + c2 + 2abc + 3 > (1 + a)( 1 4- b)(l + c).
 T
I
74. [ Gabriel Dospinescu, Mircea Lascu, Marian Tetiva ] Prove that for any po­
sitive real numbers a,b,c,
. 75- [•Titu Andreescu, Zuming Feng ] Let a,b, c be positive real numbers. Prove 
that.
72. [ Titu Andreescu ] Let a, btc be positive real numbers. Prove that 
(a5 - a2 + 3)(b5 - b2 4- 3)(c5 - c2 •+ 3) > (a -I- b 4- c)3.
2 
n(n - 1)
n3 - b3 
a 4- b
76. Prove that for any positive real numbers x,y and any positive integers m, n, 
,n+n-1z).
71. [ Marian Tetiva ] Prove that for any positive real numbers a,b, c, 
(a - b)2 l (b - c)2 I (c - a)2
73. [ Gabriel Dospinescu ] Let n > 2 and xi,X2>• • •>®n >. 0 such that 
/ n
(2a 4- b 4- c)2 
2a2 4- (b 4- c)2
(2b 4- a 4- c)2 
2b2 4- (a 4- c)2
(2c 4- a 4- b)2 
2c2 4- (a 4- b)2
n 1 \
Y"- =n2 + 1- 
Xk /
n 1 la
k-1
c3 - rr3
4-----------c+ a
+ym+n(n-l)(rn-l)(xm+n )+(m+n-l)(xmyn+xnym) > mn(xm+n-ly+y"
b3 - c3
+ b 4- c
1.4- ab 4- abed + 1 4- be 4- bede +
77. Let a, b,c, d, e be positive real numbers such that abede■= 1. Prove that
•' a + abc b+bed c + cde . d-rdea . e + eab 10
■ -J- ■ ■ ■ -f-' 1 * 1 ------.
14- cd4-cdea 1.4- de 4- deab . 1 4- ea 4- eabc 3
Crux Mathematicorurri\
n \
(9) Convex function
A real-valued function f defined on an interval / of real numbers is convex 
if, for any x,y in / and any nonnegative numbers a,/3 with sum 1, 
f(ar + 0y) b? > ... > bn then ^Ja.b, / f(x)dx ■ / g(x)dx 
J a J a J -
(10) Convexity
A function /(x) is concave up (down) on [a,b] C R if /(t) lies under 
(over) the line connecting (ai,/(aj)) and (bi,/(bi)) for all
a 
> (ojbi + O2b2 + ... + anbn)2,
with equality if and only if a, and b, are proportional, i = 1,2,
123Old and New Inequalities
also called convex and concave, re-
Ai/(xi) -F A2/(z2) I- — + An/(zn) > f(XtXi 1- A2z2 + ... I Anzn)
+ ■■• + a2> •••> an-i).
1 1
i
f[ar + n^«
n
Y,aibi
t-1
n
Si > S2 > . • • > Sn,
(14) Mac Laurin’s Inequality
For any positive real numbers zj-, z2,. -.,in>
(12) Holder’s Inequality
If r,s are positive real numbers such that —I— = 1, then for any positive 
real numbers ai, a2,..., an, 
bi,b2,.. . ,bn,
. (13) Huygens Inequality
If Pi,P2,---,Pn,a\,a2,...,On,b\,b2,...,bn are positive real numbers with
Pi + P2 + ... + pn = 1, then
(11) Cyclic Sum
Let n be a positive integer. Given a function f of n variables, define the 
cyclic sum of variables (zi, x2,xn) as
for any xi,x2,-.-,xn in the interval [a,b]. If the function is concave down, 
the inequality is reversed. This is Jensen’s Inequality.
/ n
1 1 
n
i i 
n
with R.
Concave up and down functions are 
spectively.
If f is concave up on an interval [n,b) and Ai,A2,...,An arc nonnegative 
numbers wit sum equal to 1, then
\!
53/(®i.X3. ••.,»«) = /(xi,i2,...,a:n) + f(x2,x3, ...,zn,zi) 
eye
Solutions124
where
X/, ‘ Xj3 ' • • • ' Xf*
>
3)
E(a* + f,‘)r
n
E^_-
1)
2)
+nIIftfc - (E
*=1 fc=l
sfc= ‘
1 -I- 9xyz
4
Cll + 0-2 + ... + an
V n ~ n 
with equality if and only if ai — «2 - ••• — o.n.
(15) Minkowski’s Inequality 
For any real number r 
Oi,O2,... ,an, bi, 6'2>••• i^n,
1
E
• 1 0, xr(x-y)(x-z) + yr(y- 
i)(y - x) 4- zr(z - x)(z - y) > 0. The most common case is r = 1, which has 
the following equivalent forms:
x3 4- y3 I- z3 -I- 3x1/2; > xy(x 4- y) + yz{y 4- z) 4- zx{z 4- x);
xyz > (x 4- y - z)(y 4- z - z)(z 4- x - y);
(19) Suranyi’s Inequality
For any nonncgalivc real numbers ai,a2> =1
1 and any positive real numbers
(17) Root Mean Square Inequality
If ai,02,-..an are nonnegative real numbers, then
/Qi T a‘2 + - + “n >
’ if x 4- y 4- z = 1 then xy + yz + zx x2y2 + y2z2 + z2t2 + t2x2 +- x2z2 + y2l2.
(21) Weighted AM-GM Inequality
For any nonnegative real numbers a\, a2,.... an, if Wi, w2, ■■■, are nonneg­
ative real numbers (weights) with sum 1, then
Wjo.1 + W2a2 + ••• I H'nan > '’j'1 "2? ■ • • an" >
with equality if and only if ai ~ a2 ~ ••• - On-
■
Further reading
127
i
I
I
1. Andrcescu, T.; Feng, Z., 101 Problems in Algebra from the Training of the USA 
IMO Team, Australian Mathematics Trust, 2001.
2. Andreescu, T.; Feng, Z., USA and International Mathematical Olympiads 2000, 
2001, 2002, 2003, MAA. -
3. Andreescu, T.; Feng, Z., Mathematical Olympiads: Problems and Solutions from 
around the. World, 1998-1999, 1999-2000, MAA.
4. Andreescu, T.; Kcdlaya, K., Mathematical Contests 1995-1996, 1996-1997, 
1997-190S: Olympiad Problems from around the World, with Solutions, AMC.
5. Andreescu, T.; Enescu, B., Mathematical Olympiad Treasures, Birkhauscr, 2003.
6. Andreescu, T.; Gelca, R., Mathematical Olympiad Challenges, Birkhauser, 2000.
7. Andreescu, T.; Andrica, D., 360 Problems for Mathematical Contests, GIL 
Publishing House, 2002.
8. Becheanu, M., Enescu, B., Inegalitali elementare.. .gi mai pulin elementare, GIL 
Publishing House, 2002
9. Bcckenbach, E., Bellman, R. Inequalities, Springer Verlag, Berlin, 1961
10. Drimbe, M.O., Inegalitali idei gi metode, GIL Publishing House, 2003
11. Engel, A., Problem-Solving Strategies, Problem Books in Mathematics, Springer, 
1998. • 1
12..G.H. Littlewood, J.E. Polya, G., Inequalities, Cambridge University Press, 1967.
13. Klamkin, M., International Mathematical Olympiads, 1978-1985, New 
Mathematical Library, Vol. 31, MAA, 1986;
14. Larson, L. C., Problem-Solving Through Problems, Springer-Verlag, 1983.
15. Lascu, M., Inegalitali, GIL Publishing House, 1994
16. Liu, A., Hungarian Problem Book III, New Mathematical Library, Vol. 42,
MAA, 2001. • ' ■
17. Lozansky, E; Rousseau, G„ Winning Solutions, Springer, 1996.
18. Mitrinovic, D.S., Analytic inequalities, Springer Verlag, 1970
19. Panaitopol, L., Bandila, V., Lascu, M., Inegalitali, GIL Publishing House, 1995
20. Savchcv, S.; Andreescu, T., Mathematical Miniatures, Anncli Lax New 
Mathematical Library, Vol. 43, MAA.
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>0.++
79. Prove that if a, b, c are positive real numbers then,
i
KMO Summer Program Test, 2001
Kvant, 1989
3
 2 and let xi,x2,...,xn > 0 add up to 1. Prove 
that
1 
n - 1 + x,,
TST 1999, Romania
Crux Mathematicorum
78. [ Titu Andreescu ] Prove that for any o, b,c, e (o, the following inequality 
holds
sin a ■ sin(a - b) • sin(a - c) sin b • sin(b - c) ■ sin(b - a) sin c • sin(c - a) • sin(c - b) 
sin(bd-c) sin(c.d-a) sin(a d-b)
TST 2003, USA
82. [ Vasile Cirtoaje ] Prove that the sides a, b, c of a triangle satisfy the inequality 
abc, - + - + - - 1 
b c a
 2 
ax d- by + cz + y/(a2 + b2 I- c2)^2 I y2 I- z2) > ~(a + b + c)(x y + z). u
84. [ Vasile Cirtoaje, Gheorghe Eckstein ] Consider positive real numbers 
ij,i2>•••>in such that iiX2—xn = 1. Prove that • ■
1 ; 1
„ „ ( b c a
>2 -+t+-\a b c
80. [ Gabriel Dospinescu, Mircea Lascu ] For a given n > 2 find the smallest 
constant fcn with the property: if ai,..., an >0 have product 1, then
________ 0102__________ O2O3 , ._____________________ QnQi
(a? + a2)(c4 + Qi) (a|+ a3)(a| + a2) (a2 + ai)(a? + a„)
Old and New Inequalities 10
— x/abc 0,
• • (n +-b)3(b + c)3(c + d)3(d + a)3 > 16a2b2c2d2(a + b + c + d)4.
Crux Mathematicorum
88. Find the greatest constant k such that for any positive integer n which is not 
a square, |(1 + y/n) sin(7r>/n)| > k.
93. [ Dung Tran Nam ] Prove that for any real numbers a, b, c such that a2 + b2 4- 
c2 = 9,
mar{(y/o - y/b)2,(y/b- y/c)2,(y/e- y/a)2}.
86. [ Titu Andreescu ] Prove that for any positive real numbers a, b, c the following 
inequality holds 
a + b + c
3
' a + b
91. [ Titu Andreescu, Gabriel Dospinescu ] Find the maximum value of the ex­
pression
89. [ Dung Tran Nam ] Let x,y, z > 0 such that (z + y + z)3 = 32zyz. Find the 
2/^ 1/^ |* 2^
minimum and maximum of (z + y + z)4
87. | Kiran Kcdlaya | Let a,b,c be positive real numbers. Prove that 
a + x/ab + \/abc , 
3 “ V
(ab)n | (fee)" [ (ca)»
1 — ab 1 - be 1 — ca 
where a, b, c are nonnegative real numbers which add up to 1 and n is some positive 
integer.
92. Let a,b,c be positive real numbers. Prove that
1 1.1
a(l + 6) + b(l + c) + —-—>___ 2___
1 + a) x/abc( 1 + v^abc)
20 Problems
>+ z2 + zx + z2
Gazcta Matcmatica
98. Prove that for any real numbers a,b, c,
Vietnam TST, 1996
+ +
97. [ Vasile Cirtoaje ] For any a,b,c,d > 0 prove that
2(o;' + 1)(6' + !)(? + IJIrf1 + !)>(! + M)(l + a’)(l + t>2)( 1 + c!)(l + d!).
e+l-l «+l-i6+1-1 
c
94. [ Vasile Cirtoaje ] Let a,b,c be positive teal numbers. Prove that 
«+!-. b+ 1-i
(a + b)4 + (b + c)4 + (c + a)4 > y(a4 + b4 + c4).
a+|-lj >3.
■■ 0 
such that xy + yz + zx = 3,
T~—(y + -) + + *) + + y) 3-b + c c + a a + b
100;. [ Dung Tran Nam ] Find the minimum value of the expression - 
a 
where a,b, c are positive real numbers such that 21ab + 2bc + 8ca 9
x2 + xy + y2 y2 + yz + z2 z2 + zx + z2 ~ (x + y + z)2'
Gazeta Matematica
99. Prove that if a,b,c are positive real numbers such that abc = 1, then
• 1 ' 1 1 ' 1 1 1
1 +a + b + 1 + b + c + 1 + c + a ~ 2 + a + 2 + b + 2 + c
■ ' Bulgaria, 1997
2 3 
+ T + - b c
95. [ Gabriel Dospincscu ] Let n be an integer greater than 2. Find the greatest 
real number mn and the least real number-Mn such that for any positive real numbers 
Xi, X2,..., x„ (with Xji — xq, — X]),
n .
m (n - 1)
where an is the least, among the numbers ai, a2,..., an.
. 10.5. Prove that for any real numbers , a2,..., an the following inequality holds
ata j.
> I-1 I
4-
106. Prove that if ai, a2,... ,an, bi,... ,bn are real numbers between 1001 and 
2002, inclusively, such that a2 4- a| 4- • • • 4- a2 = b2 4- b% 4- ■ • • + b2, then we have the 
inequality . ••
108. [ Vasile Cirtoaje ] If a,b,c,d arc positive real numbers such that abed = 1, 
then
107. [ Titu Andreescu, Gabriel Dospinescu ] Prove that if a,b,c are positive real 
numbers which add up to 1, then
(a2 4- b2)(b2 4- c2)(c2 4- a2) > 8(a2b2 4- b2c2 4- c2a2)2.
104. [ Turkevici ) Prove that for all positive real numbers x,y,z.t, 
x4 4- y4 4- z4 4- t4 4- 2xyzt > x2y2 4- y2z2 4- z2t2 4- t2x2 4- x2z2 + y2t2.
Kvant
1 
(1 + c)2
c2 
a2 + ti2
n
E" 0 then
fli + «2 I......... I ftn-i
--------------- ;------------- n - 1
2
s E
109. [ Vasile Cirtoaje ] Let a, b, c be positive real numbers. Prove that 
a2 b2 c? a b c 
b2 4- c2 c2 4- a2 a2 4- b2 "b + c"*'c + a^a4-b'
Gazeta MatematicS
2U b2
1 
(1 4- d)2
Gazeta MatematicS
>4(a 4- b — c)2 
(a+ b)2 4- c2
1 , 1 
(1-Fa)2 ' (14-b)2
102. Let a,b, c be positive real numbers. Prove that 
(b + c - a)2 (c 4- a - b)2 
(b 4- c)2 4- a2 + (c 4- a)2 4- b2
■•4- — (1 + \/5)(l - zy).
112. | Gabriel Dospinescu, Calin Pupa ] Prove that if n > 2 and ai,a2,...,an 
are real numbers with product 1, then
2/i z—
+ di + • • • + a2 - n >----- - • x/n - l(ai + a2 + • • • + an - n).14 n - 1
2
- 52 (Qi + ••• + aj)2-
r 2a 
:------7 4"a I- b
111. [ Dung Tran Nam ] Let zi ,z2..., z2ooi be real numbers in the interval [-1,1] 
such that Xj -1- x% +... + zi^n = 0. Find the maximal value of the Zi +z24----- Fz2oo4.
113. | Vasilc Cirtoajc ] If a, b, c arc positive'real numbers, then
— 9
(z 4- y)2 + (y + z)2 + (z + z)2 J ~ 4
117. Prove that for any Z],z2,...,zn > 0 with product 1,
l (m | a2-|----- 1 an)(a"-1-l a-"-1 F-• -I u"-1).
Miklos Schweitzer Competition
Old and New Inequalities 23
a —
Prove that
(a + b 4- c)(x 4- y + z) = (a2 + b2 4- c2)(x2 4- y2 4- z2) = -1.
Prove that
+ ••■ +4-
I
120. [ Vasile Cirtoaje, Mircea Lascu J Let a.b,c,x,y, z be positive real numbers 
such that
118. [ Gabriel Dospinescu ] Find the minimum value of the expression 
n -----------------------
y' / Q ifl2 ••• On
" y 1 - (n - 1 )at
02
1 - Oo
of + 02 + . 4- a2 
n
r-+
119. [ Vasile Cirtoaje ] Let 01,02,... ,on 2 is an integer. n — 1
_±J_ + 
1-a2
> na—.
- 1 - Q2
121. [ Gabriel Dospinescu ] For a given n > 2, find the minimal value of the 
constant kn, such that if Xi ,X2, • • •, xn > 0 have product 1, then
1 1 1 ■ ' . ■ 
■ 4- ■ - 4--------- F ~ 2, find the maximal 
value of the constant fcn such that for any Xi, x2,..., xn > 0 for which x2 4- x2 4-• • • 4- 
x2 = 1 we have the inequality
(1 — X1)(1 -I2)...(l -x„) >knXxX2..:Xn. '
CHAPTER 2
Solutions
25
SullltiOIIH26
1. Prove that the inequality
holds for arbitrary real numbers a, b, c.
Kbmal
(a + b + c)2 + (3 - a - b - c)2 = 2
and the conclusion follows.
> +
and
and
7(1 - a)(l - b)(l - c) 1 for all real numbers x the last quantity is at least -y.
Second solution'.
We have the inequalities
7a2 + (1 - b)2 + 7b2 + (1 - c)2 + 7c2 + (l-a)2 >
|a| + |I-b| |b| + |l-c| |c| + |l-a|---------- —-------- -j. ---------- —------- --------
72 72 72
7a2 + (1 - b)2 + 7b2 + (1 - c)2 + 7c2 + (1 - a)2 > ^y
First solution:
Applying Minkowsky’s Inequality to the left-hand side we have
\/a2 + (1 - b)2+ 7b2 + (1 - c)2+ x/c2 + (1 - a)2 > 7(a + b + c)2 + (3 - a - b - c)2.
Denoting a + b + c = x we get
2 9 9
+ 2 “ 2’
27Ol«l and New Inequalities
= 1,
as desired.
Gazeta MatematicS
>
Tournament of the Towns, 1993
Summing up, we obtain
y/abc 4- \/( 1 — a)( 1 — b)(l - c) 8.
Solution:
From the AM-GM Inequality, we have
2(\/a + '/b + >/c) > x/a + Vb + y/c +'3x/abc = y/a + y/b + y/c + 3.
b + c c + a a + b^ ,-r-
—1=—I---- 7=—F--- 7=- > y/a + v b + y/C + 3.y/a y/b y/c
:3 + 2z2 + bx + 1 — 0 has at least one real root, then
c + a a + b
'b ' y/c
28 Solution!)
1. Prove that
 
Z2(3-2t) > 0 and thus |x34-y3+z3-3xyz| y 4- 
cz 0.
Old and New Inequalities 29
the last one being equivalent to (x — a)2 + (y - b)2 4- (z — c)2 > 0.
>2
(a + b -F c) 2
2
(a 4- b + c) >4-
which is well-known.
22
c
a 4- b
First solution-.
We rewrite the inequality as
4
8. [ Hojoo Lee ] Let a, b,c > 0. Prove that
y/ a4 + d2b'2 + b4 4- \/b4 4- b2c2 4- c4 + x/c4 4- c2a2 4- a4 >
> a \/2d2 + be. + by/2b'2 4- ca + c\/2r2 -F ab.
Gazeta Matematicd
+ (« F b)2
a
(b 4- c)2
It remains to prove that
2 +
a b c
(b 4- c)2 (c + a)2 (a 4- b)2 
Applying the Cauchy-Schwarz Inequality we get 
b
(c I- a)2
a b
---------4- ■ 4-
b + c c + a
Second solution:
Without loss of generality we may assume that a 4- b 4- c = 1. Now, consider
Xthe function f : (0,1) —» (0, oo), /(r) = —------rx. A short computation of derivatives
(1 X)
shows that / is convex. Thus, we may apply Jensen’s Inequality and the conclusion 
follows.
Solution:
We start from (a2—b2)2 > 0. We rewrite it as 'la* 4--la2b24--lb4 > 3a4 ' tudlr F 3b4.
It follows that x/a'1 4- a2b2 1 b2 > ~/a2 I b2).
Using this observation, we find that.
(52 x/a4 4- d2b2 4 b*) > 3 •
But using the Cauchy-Schwarz Inequality we obtain
(^2a\/2a24-bc) - b + c c + a a 4- b ~~ 2
Solutions30
When does equality hold?
JBMO 2002 Shortlist
(1)
and
(2)
o3rb:, + ? >
(3)>
Using the AM-GM Inequality we obtain
a + b >
a + b). (4)
9. If a,b,c are positive real numbers such that abc = 2, then 
a3 + b3 + c3 > a\/b + c 4- i>v/c + a + cVa + b.
10. [ Ioan Tomescu ] Let x,y,z > 0. Prove that
___________ ryz___________
(1 -I- 3x)(x + 8y)(y 4- 9a)(z -I- 6)
First solution:
Applying the Cauchy-Schwarz Inequality gives 
3(a2 4- b2 4- c2) > 3(a + b 4- c)2
'a+ b 3nbr, which is true by the AM-GM 
Inequality. We have equality if a = b = c = s/2.
abc (\/(a 4- b)(b 4- c)(c 4- n))
> abcVSabc = 3 = 6.
Thus
(a\/b 4- c 4- b\/a 4- c 4- c\/a + b)2 > 6(a\/b 4- c 4- b\/a 4- c 4- c 
The desired inequality follows from (3) and (4).
aVb 4- c 4- b\/a 4- c 4- c
Second solution:
We have
aVb + c 4- b\/c 4- a 4- c
(n2 4- b2 4- c2)2 0 with a 4- b 4- a = 1.
which is equivalent to Schur’s Inequality.
, for all
Thus,
a2 - 2aX} 4- x2 Xi 
18abc 4- 1 - 2(ab 4- be 4- co) + 1 > 6(ab 4- be + ca) 
8(ab 4- be + ca) 4(ab + be + ca) 
(1 - 2a)(l - 26)(1 — 2c) 
(b 4- c - a)(c + a - b)(a + b - c) 2 and a > 0 such that xj 4- 
2d 
X2 + ... + xn = a and x? 4- x2 4- ... 4- x2 I-+
>
> a 4- b + c.>
> a -f- b + c>
x3 + y3 + z3 > x2y + y2z + z2x.
13. [ Adrian Zahariuc ] Prove that for any a,b, c € (1,2) the following inequality 
holds
(a + b)(b + c) > 4l>\/ac,
the last one coming from ti + b > 2y/ab and b + c > 2\/bc. Writing the other two 
inequalities and adding them up give the desired result.
First solution:
If ab + be + casolves the 
problem:
b\/a 
4b\/c - C\/a
Solution:
The fact that a,b, r.e (1,2) makes all denominators positive. Then
b^/a
46\/c - Cy/ii
(a + b + c)2
t i r
—F 7 + —a b e
abc
b(a + b + c) > \/a(4bx/c - c>/a) 
Second solution:
Replacing a. b, c by La, lb, Ic with I = - preserves the value of the quantity in 
V abc
the left-hand side of the inequality and increases the value of the right-hand side and 
makes at ■ bl • ct = abctA = 1. Hence we may assume without loss of generality that 
y z ~
abc = 1. Then there exist, positive real numbers x, y,z such that a = -,b = - ,c = -.
The Rearrangement Inequality gives
(ab + be + ca)2 
a i b -|- c 
abc
14. For positive real numbers a, b,csuch that abc a + b + c. b c a
cvb a~Jc----------------------1----------- 1-------- 
4cy/a - ay/b 4a\/b-b\/c
abc a2c + b2a + c2b 
r + - + — — --------- ;---------b c a abc
(here we have used the fact, that abc. 
as desired.
14- >
>
Solution:
We set x = 
equivalent, to
Solution:
We have
ay l bx-ac-xz 
= 5(0-i)2
6
ab 4- ac 4- be'
Junior TST 2003, Romania
3 
a 4- b 4- c
6
x 4- y 4- z
x:* 4- y3 4- z-’ 
xyz
15. [ Vasile Cirtoaje, Mircea Lascu ] Let. a,b,c,x,y,z be positive real numbers 
such that a + x > b 4- y > c I z and a 4- b -I- c = x 4- y 4- z. Prove that ay 4- bx > ac 4 xz.
16. [ Vasile Cirtoaje, Mircea Lascu ] Let a, b, c be positive real numbers so that 
abc = 1. Prove that
1+—’— 
xy 4- yz + zx
Forth solution:
, Jab* J ca* 9/be* ~ , 9,9 9
Let x = y ~^T'V = y ~^~,z ~ \ ^orLsecluen*-*y> a = XV — zx ,c = yz*, 
and also xyz •
1 
a'V =
a(y-c)-\-x(b-z) - a(a I b-x-z)4 x(b-z) - a(a-x) I (a \ x)(b-z)
4- - (a2 - x2) 4- (a 4- x)(b - z) — i(a - x)2 4- |(a 4- x)(a - x 4- 26 - 2z) = 
= |(u ~ *)2 + |(« + x)(b- c + y - z) > 0.
The equality occurs when a = x, b = z, c = y and 2x > y + z.
abc
7 -f- —I--- —b c a
i z = - and observe that xyz = 1. The inequality is 
6 c
2a 6
-r + - > 3 6 c
2& c 2c g
Similarly, — + - > 36 and — 4-- > 3c. Adding these three inequalities, the conclusion 
is immediate.
Solutions34
> 1 +
1 + >
The last inequality is equivalent, to
JBMO 2002 Shortlist
-j-> — Ta-b ab(a + b) « (a - b)2(a + b) > 0,
c2
>
O'
b2
> 1.I........ I
1 + X2 + X2X3
Russia, 2004
9 
(s t y i z)2
9
(x + y + z)2’
6
x + y t z'
2
> 0 and this ends the.proof. • .
18. Prove that if n > 3 and Xj,.r2,...,.rn > 0 have product 1, then
1 1 1 
........... 'I ---------------------- -j- • • • -j- ----------------------
1 4- X| + .T|.X’2 1 + X2 + 2:2^3 1 + Xn + X„X\
From (x + y + z)2 > 3(xy + yz + zx) we get
1+^— 
xy + yz + zx
so it. suffices to prove ( hat.
Second solution:
By the Cauchy-Schwarz Inequality we have
, . (a ' c3(a+t+c)^—+_+—
so we only have to prove that
a2 b2 c2— + — + — >a + b+c.b c a
But this follows immediately from the Cauchy-Schwarz Inequality.
17. Let o,b, c be positive real numbers. Prove that
a3 b3 c3 a2 b2 c2
T7 + ~5 + -2>-7- + — d---- .b2 cz a2 b c a
1--------3—
x + y + z
First solution:
We have 
a3 a2
— > — Ta-ba3 + b3> ob(o + b) (a - b)2(a + b) > 0, b2 b
which is clearly true. Writing the analogous inequalities and adding them up gives 
a3 b3 c3 a2 , b2 L c2 a2 b2 c2
b2 c2 a2 b c a b c a
a2 b2 c2\2 
~r +---- 1---- )
b c a J
Solution:
We use a similar- form of the classical substitution , x2 = xn = —
Qi 0.2 o.n
In this case the inequality becomes
flj O2 Qn.
----------------------- ------ ------------------------- F ■ • ’ -f > 1 
fli + a2 + 03 o2 + 03 + a.|-----------an + Oi + a2
Old and New Inequalities 35
and it is clear, because n > 3 and a, + Oj+i + cq+2 0 o x + y - 2xy 4y/2x3y3z3 => x3y3z3 . xyz 1 - 2 • |
o 4
d) This is more delicate; we first notice that there are always two of the three 
numbers, both greater (or both less) than -. Because of symmetry, we may assume 
that x, y | and then
Prove that
, 1
a) xyz 
and
z 2xy + z2 + 2xyz 
-> 2xy (I I z) 0 verify
x2 + y2 + z2 + 2xyz - 1
•r2 I ?/
because - and - are both greater than 1. x y
xz + yz- 2xyz 0,
\x y J
Old and New Inequalities 37
and
| 0 satisfy the condition
we get from here
xy l xz l yz
Solution'.
It is immediate to prove that
|sin(i4- y)| 1.
Gazcta Matcniaticft
20. [ Marius Olteanu ] Let xi,X2,X3,X4,xr, € R so that xi 4- x3 4- x3 + x4 + xr, = 0.
Prove that
x + y -I- z = xyz 
then xy + xz + yz > 3 -r x/x2 4- 1 + \/y2 + 1 + v^z2 4- 1.
First solution:
We have
xyz = x + y 4- z > ‘2^/xy 4- z => z{^/xy)2 - "2y/xy - z > 0.
Because the positive root of the trinomial zt2 — 2t — z is
1 -I- x/1 -I- z2
( •r'
| —^-+~- » Zy/xy > 1 4- \/l 4- z2.
Of course, we have two other similar inequalities. Then,
> Xy/yz i yx/xz I z^/xy >
> 3 4- \/x2 4- 1 + \/y2 4- 1 4- \/z2 4- 1,
and we. have both a proof of I he given inequality, and a little improvement of it.
3K Solutions
(xy -I- xz 4- yz)2 > 2xyz (x -I- y 4- z) I- x2y2z2 = 3 (x 4- y I- z)2.
so that
xy + xz + yz > 3 + \/3(x2 4- y2 4- z2) 4- 9.
But
>2,4- +
JBMO, 2003
,2
(1)> 1+ +
 
is a consequence of the Cauchy-Schwarz Inequality and we have a second improve­
ment and proof for the desired inequality:
Solution:
Let us observe that y 
> 3 (x + y + z)2 - 6 (xy 4- xz + yz) 4- 9 = 3 (x2 4- y2 4- z2) 4- 9,
xy 4- xz 4- yz > 3 4- \/3(x2 4- y2 4- 22) 4- 9 > 
> 3 4- \/x2 4- 1 4- \/y2 4- 1 4- \/z2 4- 1.
a
2c4- b
. 1 +y:
2
1 -I- x2
1 4- y 4- z2
b
2a + c
c
2b 4- a
, c = 1 4- z2, it is sufficient to
,2
- and 1 4- y 4- z2 > 0, so
Second solution:
Another improvement is as follows. Start from
l + l + l>-l-4-- + - = l=> x2y2 + x2z2 4- y2z2 > x2y2z2, 
x2 y2 z2 xy xz yz
which is equivalent to
1 4- x 4- y2
1 -1- x^___
1 4- y:
2
and the similar relations. Setting a = l4-x2ii=14-y2, 
prove that
22. [ Laurencin Panaitopol ] Prove that
1 4- x2 1 4- y2
1 4- y + z2 1 4- z 4- x2 
for any real numbers x, y, z > -1.
Old and New Inequalities 39
b-
> 15.+ 7= I 4
> I->4- 4-4- +
> 2.a + b
E >
>
E-^ (Chebyshev’s Inequality)
and
E£
3
7-V B
B
C
a
2c I b
E°*
3
A + 4C-2B
9 ’ C~
a2 + b 
b 4 c
C
A ‘
2
>
Solution:
Using the Cauchy-Schwarz Inequality, we find that
(EM 
52 a2(b+c) + 52 “2 + 52ab
And so it is enough to prove that
52q2(^ + c) 4- 52 a2 4- ab
The last inequality can be transformed as follows
1 + (52q2) - 2 52 a2(6+c) 4" 2 52ab
25La2-‘252a:' + 25Lftb ~ (Efl2)2 + 252a:‘^S°2’
and it is true, because
>2 14-(52q2) - 252q2(6+c)+252ab-
~ 1 + (52a2)
c
2b I u
C + 4B- 2A 
9 
and the inequality (1) is rewritten as
jB C Aand---- F — 4- — > 3 and the conclusion follows.
ABC
An alternative solution for (1) is by using the Cauchy-Schwarz Inequality:
b c a2 b2 c2 (a + b + c)2
-------------- -j- — ————— “I"------------------- --- ............ .. ^-> ~~ ■ ' - ' 
2u I c 2b I a 2ac I ab 2ub I cb 2bc I ac 3(ab I be I ca)
(E°’)2
23. Let a, 6, c > 0 with a 4- b 4- c = 1. Show that 
a2 4- b b2 + c c2 4- a - --------- f.-------- 
o 4- c c 4- a
B C A
A 1 B ' C
Because A, B,C > 0, wo have from the AM-CM Inequality that
CAB, 
A + B + C " 3
for any a,b,c> 0. Let A = 2c 4- b, B = 2a 4- c, C = 2b 4- a. Then a = 
B + 4A-2C 
9 
A 
B
B
C
Solutions40
Kvant, 1988
£a4 1998.
Vietnam, 1998
Let
>
> (n-1)
Solution:
The condition
In any of the cases a = b + c, b - c + a, c = a 4- b, the inequality
Ea2-2Za6
is clear. So, suppose a + b / c, b + c / a, c + a / b. Because at most one of the 
numbers b + c-a, c + a-b, a + b~ c is negative and their product is nonnegative, 
all of them are positive. Thus, we may assume that
Solution:
1998
xn be positive real numbers satisfying
1 . . 1 1
i2 + 1998 + "■ + xn + 1998 ' 1998’
tfx}x2...xn
n-1
24. Let «,b,c > 0 such that a4 + b4 4- c4 2 and x>,..
__ 1
Xi +. 1998
= ai- The problem reduces to proving that for any positive real 1998 + X{
numbers ai,fl2,... ,an such that ai + a2 + • • • + a„ = 1 we have the inequality
This inequality can be obtained by multiplying the inequalities
1 _ | _ fli 4------ h Qi-1 + a,4-i + •' • + fln
a 0.
Old 27. Then
27
and
Hence all the three numbers must be greater than 1. Set
a = i- l,b = j/-l, c-z-1.
We then have a>0,b>0, c>0 and
x = a + 1, y = b+1, z — c + 1.
which is the same as
a2 + b2 + c2 + a + b + 27;
b) xy + xz + yz > 27;
c) x + y + z > 9;
d) xy + xz + yz > 2 (z + y + z) + 9.
Replacing these in the given condition we get
• • • (a + I)2 4-(6 + I)2 4-(c 4-I)2 = (a + 1) (b 4-1) (c-t-1)
Solution:
a), b), c) Using well-known inequalities, we have
xyz = x2 4- y2 4- z2 > 3\/x2y2z2 => (xyz)' > 27 (xyz)2 ,
26. [ Marian Tetiva ] Consider positive real numbers x,y, z so that 
x2 + y2 + z2 = xyz.
If we put q = ab + ac + be, we have
q 3\](xyz)2 > 3s/272
x + y + z > 3 itfxyz > 3 \X27 = 9.
d) We notice that z2 x 1 z > 1.
Solutions42
Thus
= abc 4- ab + ac4- be 4 (x + y 4- z) - 9.
Try it!
Russia, 2002
hence
and we arc done.
One can also prove the stronger inequality
Solution-.
Rewrite the inequality in the form
(x-l)(y-l) + (x- l)(z- 1) 4- (y — 1) (z - 1) > 12 => 
=> xy + xz + yz > 2 (x 4- y 4- z) + 9;
 x2 + 2y/x 4- y2 4- 2^/y 4-
Now, from the AM-GM Inequality, we have
q 4- >/3q + 2 6 => q = ab 4- ac 4- be > 12.
Now, recall that a — x - 1, b = y - 1, c = z - 1, so wc get
27. Let x,y,z be positive real numbers with sum 3. Prove that 
\/x + y/y + \/z > xy + yz + zx.
4
x3 9
X 4- 2 0.
z2 4- 2yfz > 9.
y2 4- 2y/y > 3y, z2 4- 2y/z > 3z,
28. [ D. Olteanu ] Let a, b,c be positive real numbers. Prove that 
a + b a b+ c b c4-a. a 
b 4- c 2a 4- b 4- c c + a 2b 4- c 4- a a + b 2c4-a4-b
Gazeta Matematica
Old and New Inequalities 43
India, 2002
£x2>
4- >+ 3(ab 4- be 4- ca)
c2 — ac + a2 "r a2 — ab 4- b2 ~ a + b-j-c
Proposed for the Balkan Mathematical Olympiad
'U»
3
30. Let a,b, c be positive real numbers. Prove that 
a3 . b* ■ '• .c3 ' .
b2 — be + c2 c2 — ac + a2 a2 — ab + b2
4— + ------~ x + y
>-l
2
-
Solution:
Let us take - b
29. For any positive real numbers a,b,c show that the following inequality holds
abc-c4-aa4-bb + c 
v + - + ->----- r +------ + t------ •b c a. c + ba + cb + a •;
y z x y z—I----- 1----------- 1---------- 1---------
x x 4- y y + z z + x
(x 4-y 4-z)2 [ 
xy 4- yz 4- zx
be,.,
— x, - = y, - = z. Observe thatc a
a 4- c 1 + xy 1 - x
b 4- c 1 + y 1 4- y'
Using similar relations, the problem reduces to proving that if xyz = 1, then
x - 1 y - 1 z - 1 „------ + -- ----- + - ------ > 0 «■ 
y+1 z+l x+1 “
O (I2-l)(2+iy+(y2-l)(l + l) + (z2_l)(y+I)>0 
 y x2z+y x2 > y x+3.
But this inequality is very easy. .Indeed, using the AM-GM Inequality wc have 
^x2: > 3 and so it remains to prove that y x2 > which follows from the 
inequalities
Solution: /
We set x = a + b, y = b + c and z = a 4- c and after a few computations we obtain 
the equivalent form 
xyz x y 
- 4- - 4- - 4-----------b —y z x x + y y + z
But using the Cauchy-Schwarz Inequality, 
x.y.z. x .1. y z (x + y + z)2 (x + y + z)2
y z x x + y y + z z + x xy 4- yz 4- zx xy + yz 4- zx 4- x2 4- y2 4- z2 ■:
 2(x 4- y 4- z)4 - > 8(x 4- y 4- z)4
2(xy 4- yz 4- zx)(xy 4- yz 4- zx 4- x2 4- y2 4- z2) “ (xy + yz + zx + (x + y + zy)2 
and the conclusion follows.
Solutions44
, it suffices to prove that:
> a + b + c.
2
>
> a + b + c++
>0.
>
First solution'.
Since a -F b + c >
Third solution:
Let
3(ab + be + ca) 
a + b + c ■ ‘
a'1
b2 - be + c2
which is just. Schur’s Inequality.
Remark.
The inequality 
a3 b3 c3
b2 - be + c2 + c2 - ca + a2 a2 - ab -F b2 
was proposed by Vasile Cartoaje in Gazeta Matematica as a special case (n = 3) 
of the more general inequality 
2an -bn-cn 
b2 - be+ C2
2cn - an - bn 
+ a2 - ab + b2
Second solution:
Rewrite our inequality as
(b + c)a3 
: ■ b3 + c3
But this follows from a more general result:
If a,b,c,x,y,z > 0 then
a . a3
b2 - be+ c2
Thus we have to show that
(a2 + b2 + c2)2 > (a + 6 + c) • ^2 a(^2 “ + c2).
This inequality is equivalent to
a4 + b* -F c* + abc(a + b+c) > ab(a2 + b2) + bc(b2 + c2) + ca(c2 + a2),
2bn - cn - an 
c2 - ca + a2
3(ab F be + ca) 
a -F b -I- c
a3 ' b3 c3
b2 — be + c2 + c2 — ca + a2 a2 — ab + b2
From the Cauchy-Schwartz Inequality, we get
(r-2)
53 a(b2 - be + c2)
b-Fc -3 V •
But this inequality is an immediate (and weaker) consequence of the result from 
problem 101.
Old and New Inequalities 45
and
-E
A + B
)=4(E 2
b + c
Ysab
'fa + a — a =c + a •
a
3
Aa —
/
true. So we only need to prove that
a> a F b + c ^2
/
(fa -F c)(fa2 - be F c2) 
bi2 - be + c2
1
3f
arc
------ \
-F a2 - a
/
\
1 
b2 — be + c2
b* -I- c3
fa2 - be + c2
• a 
a + fa + c 
We have equality if and only if a = fa = c. -
Aa =
+ a2 — a121,11
3
and also the similar inequalities
/
E \
\
We consider the convex function /(t) = 
we finally deduce that ■
y v-+ V 3
31. [ Adrian Zahariuc] Consider the pairwise distinct integers zi,X2,...,in, 
n > 0. Prove that . .
x? + x| + • • • + x2 > Xii2 + X2X3 + • • • + xnii + 2n - 3.
2
>2. .
2^a.
3y«fa
• • (E“)2 
+ a2 + a }
1/^ + t2. Using Jensen’s Inequality 
V o
and the similar relations with Ab and Ac. So A > Aa+Ab 4- Ac. But because a) 
(52 ab) we get
£>r
3
3
So we get ■
(ZL a'3) (12 fa2 - bc + c2 ) =
= 5 (y (6+c)^2 ~bc+c’)) (y 
from the Cauchy-Schwarz Inequality. Hence
a = \/a + fa • Vb + c - 5^ a.■4>i(£V6Tt)2-2£
We denote
Solutions46
and
Si >
and
S2 >
So
>
But
so
and the problem is solved.
32. [ Murray Klamkin ] Find the maximum value of the expression z2z2 + x^x-j + 
• • • + Zn-ixn + t2zi w*’cn xi>x2, • • • ,xn-i,xn > 0 add up to 1 and n > 2.
Crux Mathcmaticorum
Solution'. ■
The inequality can be rewritten as
n
-Xi+1)2 > 2(2n —3).
n
^2(ii -zi+1)2 > 4n-6
k i / k
The inequality > y ( 5~^a«
11 "\l I
equality) implies that
n n
- Zi+1)2 = 52 z, - Zi+1 = 0 mod 2
t=i i=i
S2 = -ZAf + l)2 + ••• + (z„ - Z1)2 + (Z1 -Z2)2 + •■• + (z,n-l ~ ®m)2-
2
(which follows from the Cauchy-Schwarz In-
(x.w - zm)2 
n - (iW - m)'
(zAf - zm)2 
M - m
Z,Z2 + Z2Z:j + • • • + Tn-iZn + X2 Zj (Xm - Z,n)2 1 1 
--------- —— -J- ■
M - m n- (M - m)
> (n—I)2-= 4n -8 + - > 4n - 8. 
n n
Old and New Inequalities 47
2
C1 =
\/x7 + \/X2 + • • • + y/xTi 
for all xi,x2,... ,x„_i,xn > 0 which add up to 1. Let us prove first the inductive 
step. Suppose the inequality is true for n and we will prove it for n + 1. We can of 
course assume that x2 = min{ii,x2,... , xn+i}. But this implies that
X?X2 F XjX3 -»-•■• + X2 Xi b2c, -y > . Because
c 3
4 •we have proved that a2b+ b2c + c2a 0 for which Xfc+i > ii -F x2 + • • • + x* for any k, the inequality
c 
a+ 2 V +
for any n. Taking the limit, we find that c > 1 + \/2. Now, let us prove that 1 -f- \/2 
works. We will prove the inequality
\/Xl + y/xi + • • • + y/x^. 3.
ex)
Russia, 2002
3 4- (xyz 4- abc)
+ 4-
= E£
34. Given are positive real numbers a, b, c and x, y, z, for which a+x=b+y= 
c + z = 1. Prove that
J
by induction. For n = 1 or n = 2 it is clear. Suppose it is true for n and we will 
prove that x/rT + y/l? + • • • + + V^n+T 6.1-c 1-a 1-b a b c
1 1
---------F i-----a 4- c b 4- c
1-a
b
1
4
Old and New Inequalities 40
We have 1 
t(t4-1)
2 
t
3-
4
Solution'.
The idea is to observe that fl'’’(i>rC+d)+63(c+rf+a) + c3(rf+a+6) + d,t((i+6+c) is 
equal to ab(a2 4- b2). Now, because the expression ab(a2 4- b2) appears when writing 
(a — b)4, let us see how (.he initial expression can be written:
£ + O’) = + + _
3 52 «■* + ° 52 “2&2 - 52(a ~6)4
4 ———
— •"+ p so the inequality is equivalent to
a ” a 4- 1 -4abc‘
We will prove now the following intercalation:
a ~ 4 4a.be T a 4- 1 4abc'
The inequality in the right follows from the Cauchy-Schwarz inequality:
and the identity 52(a + 1) = 4- The inequality in the left can be written as 52 —
1 + 9abc---- ----- , which is exactly Schur’s Inequality.
36. Find the maximum value of the expression
a3(b 4- c + d) + b3(c 4- d + a) 4- c3(d 4- a + b) 4- d3(a 4- b + c)
where a,b, c, d are real numbers whose sum of squares is 1.
Second solution:
Because the inequality is homogeneous we can consider without loss of generality 
that a + b 4- c = 1 and so the inequality is equivalent to
v \ 1 1
2—* a(a 4- 1) ~ 4abc'
I4 a2a4 + a^a? + ... + oia3
Iran, 1999
38. Suppose that qj 2.
Prove that
First solution-.
We have (x + y)(x + z) 
Hence
= xy + (x2 + yz) + xz > xy + 2xx/iiz + iz = (x/xy + y/xz)2.
Second solution:
FYom Huygens Inequality we have ^(x + y)(x + z) > x + y/yz and using this 
inequality for the similar ones we get
x_______ _
x+ y(x + y)(x + z)
Solution:
A quick look shows that as soon as we prove the inequality for n — 3, it will be 
proved by induction for larger n. Thus, we must prove that for any a ca(c1 - a3) (c3 - b3)(ac - be) 0, which 
is clear.
39. [ Mircea Lascu ] Let a, b, c be positive real numbers. Prove that 
b + c c + a a + b 
-------+ -T- +--------4 a b c
X
X + y/(x + J/)(l -I- z)
x +/xy + y/xz y/x +/y +/z 
and this solves the problem.
_____ y-----------+ — • 
y + \/(l/+ z)(l/+ *) z + V(z + x)(2 + !/)
x + y , 2 -
2x + y/yz 2j/ + y/zx 2z + y/xy
Now, we denote with a = [, — c = allj the inequality becomes
x y z
1 1 1 3, which follows from the AM-GM 
Inequality.
a b a \ 
L------ h ------ + ----- L I •b + c c + a a + b J
- X + y/xy + y/xz
51Old and New Inequalities
and
(x + y 4- z)3 > 27zyz;
41. [ Mircea Lascu, Marian Tetiva ] Let x,y, z be positive real numbers which 
satisfy the condition
Solution:
Using the inequality
(23-1) 
(3(23+1))
we infer chat
Solution.'.
a) We put t3 = xyz', according to the AM-GM Inequality we have
1 = xy + xz + yz + 2xyz > 3t2 + 243 (24 - 1) (4 + I)2 4well-known
(x + y + z)2 > 3 (xy + xz + yz)
c) - + - 4- - > 4 (z + y + z); 
xyz
d) - + - + -- 4(z + y + z)>
x y 2
Solution:
Suppose we have > 3s for all i. First, we will prove that n* 3 
and let us prove it for n +1. This follows from the fact that
1 + - 3^ = ^3 • 3* > 
n 4 n
x- iThus, using this observation, we find that > 3® > => ai+i > a, for all i,
which means that aj contradiction.
40. Let ai,a2,... ,an > 1 be positive integers. Prove that at least one of the 
numbers a^/a2, • • • > is less than or equal to s/3.
Adapted after a well-known problem
1 
+
. b . b , 4c
1
x + y 
4n a + a 
b + c ~ b c' a + c ” a ' c “““ a + b
Adding up these three inequalities, the conclusion follows.
Solutions52
xy 
can be also rearranged as
((2z + I)2 - 4) =
9
= 4'
if we put q = xy + xz + yz, p = xyz.
Now, one can see the following is also true
2 (2z — 1) (2z 4-3)
22 4-1
(x 4- y)
4
2
22 + 1
.2
Wc also have 1 = (2z 4- l)xy 4- z (x I- y) 0, which shows us that
2
22 4- r
then we have 2s3 > 54xyz = 27 - 27 (xy 4- xz 4- yz) > 27 - 9s2, i. e.
3 q = xy + xz + yz > -.4
c) The three numbers x,y,z cannot be all less than because, in this case we 
get the contradiction
2s3 4- 9s2 - 27 > 0 (2s - 3) (s -I- 3)2 > 0,
3where from 2s-3>0«$>-. '
Or, because p 
\xy /
„ t 2
s2 > 3g = 3(1 - 2p) > 3 (1 - -
The inequality to be proved
111,.- + - + ->4(x4-J/4-z) xyz ,
1__
(2z 4-I)2'
a 3 n 1 .
xy -I- xz 4- yz I 2xyz 
We have 1 = (2z 4- 1) xy 4- z (x 4- y) > (2z 4- I)xi/ 4- 22^/xy, which can also 
be written in the form ((22 4- 1) y/xy - 1) (^/xy + 1) $ 0; ai‘d this one yields the 
inequality
4z2- 1 (2z- 1) (2z 4-1)
2 z
1
z
Old and New Inequalities 53
>
(2z + 1)>
>
>
42. [ Manlio Marangelli ] Prove that for any positive real numbers x, y, z, 
3(r2y 4- y2z + z2x)(xy2 4- yz2 4- zx2) > xyz(x 4- y 4- z)3.
_______________3y^/7yz_______________
^3 ■ (y2z + z2x 4- x2y) ■ (yz2 + zx2 + xy2)
 4z2 4- 6z > 4z2 4- 4z 4- 1;
2
2z 4- 1
4z- - 4- z
(2z - I)2 
2 (2z 4- 1) ’
Remark.
It is easy to see that the given condition implies the existence of positive numbers 
a, b, c such that x = , ° ,y = ------ ,z = —^-r. And now a),i») and c) reduce
b + c J c + a’ a+b
immediately to well-known inequalities! Try to prove using this substitution d).
7 + 7 “ 4 (* + y) = (* + ’/) y
±-4 
xy
2 (2z - 1) (2z 4-3) 
where from we get our last inequality
111,, x (2z- I)2
- 4- - 4- - - 4 z 4- y.4- z) >
x y z z (2z 4-1)
Of course, in the right-hand side z could be replaced by any of the three numbers
which is > - (two such numbers could be, surely there is one).
Solution:
Using the AM-GM Inequality, we find that
1 .i. y2* , xy2
3 y2z 4- z2x + x2y yz2 + zx2 4- xy2
and two other similar relations
1 z2x yz2-4------------------------------ 1------------ “-------------- >
3 y2z I z2x 4- x2y yz2 4 zx2 4- xy2
(the assumption z > allows the multiplication of the inequalities side by side), and 
this means that the problem would be solved if we proved
2(2z4-3) 2z4- 1
2z 4 1 ~ z
but this follows for z > and wc arc done.
d) Of course, if z is the greatest from the numbers x, y, z, then z > -; we saw 
that
3z y.ryz
s/3 • (y2z I z2x I x2y) ■ (yz2 I- zx2 I- xy2)
I [ x2y t zx2 > _______________ Sx^xyz_______________
3 y2z-\-z2x-\-x2y yz2 + zx2 4- xy2 ~ ^3 . (y2z + Z2X h . (y,2 h ZTz b xy2j 
Then, adding up the three relations, we find exactly the desired inequality.
Solution*54
1
++
1 
Ofc+l
43. [ Gabriel Dospinescu ] Prove that if a, b, c are real numbers such that 
max{a,b,c} - min{a,b,c} 3a2b 4- 3b2c 4- 3c2 a
Solution’.
Clearly, we may assume that a = min{a, b, c) and let us write b = a4-x,c = a4-y, 
where x, y e [0,1]. It is easy to see that a3 4- b3 4- c3 - 3abc = 3a(x2 - xy4-y2) 4- x3 + y3 
and a2b 4- b2c + c2a - 3abc ■= a(x2 - xy + y2) 4- x2y. So, the inequality becomes 
1 4- x3 4- y3 > 3x2y. But this follows from the fact that 1 4- x3 + y3 > 3xy > 3x2y, 
because 0 0. But this is true from Schur’s Inequality 
applied for a, b, c and ab, be, ca. ■
_1_
«4
1 
n 4- ak
1-b2 
b
a
1 - a2
1 - a2 ------- +
a
44. [ Gabriel Dospinescu ) Prove that for any positive real numbers a, b, c we have
W? 4- (? 4- > 6(a 4- b 4- c) 4- | 4- -Y
J \ ca) \ ab J \a b c)
 —■
an ’n + ak n 4-1 *r=0
and from this inequality and the previous one we conclude immediately that
1---- 
«•
1
— cn(2 —x)r
to prove the following difficult
II
47. ( Titu Andreescu, Gabriel Dospinescu ] Let x,y, z 0 for any I tan A > 3
 215xyz(x + y)(y + z)(z + x)
' A , we can set a = tan —, b =
Solution:
We put a — y/x, b= y/y and c = y/z. Then 1 - x = 1 - a2 = (a -I- b + c)2 - a2 = 
(b + c)(2a + b + c). Now we have to prove that ((a + b)(b + c)(c + a)(2a + b + c)(a + 
2t» + c)(a + d + 2c))2 > 2u,a2b2c2(a2 4- b2)(b2 + c2)(c2 + a2). But this inequality is true
Solution:
It is known that in every triangle ABC the following identity holds 
tan — tan — = 1 and because tan is bijective on [o, — )
B Ctany, c = tan—. The condition a,b,c 6 (0,1) tells us that the triangle ABC is 
acute-angled. With this substitution, the inequality becomes
A 2 A2 tan — 1 — tan —
2 -> oXp 2.
1 — tan2 — 2 (an —
2 2
tan /I tan B tan C(tan A + tan B-(-tan C) > 3(tan A tan Z?+tan B tan G+tan C tan A) 
 (tan A + tan B + tan G)2 > 3(tan A tan B + tan B tan C + tan C tan A),
clearly true because tan A, tan B, tan C > 0.
5—^ (6 + c — a)* 3
a2 + (6 -(- c)2 “ 5‘
In fact, this problem is equivalent to that difficult one. Try to prove this!
Solutions56
Then
ba
(1) We have
xy + yzj4- zxI
o a3 4- b3 + c3 4- 3abc >> 2
«•
50. Prove that if x, y. z are real numbers such that x1 4- y2 + z2 = 2, then •
x + y + z 0) and (2a 4-b 4-c) (a 4-2b 4-c) (a 4-b 4-2c) > 8(b4-c)(c4-a)(a4-b).
b + c 
y
a
a
(a + b)4
8
x = -—
49. Let x, y, z be positive real numbers such that xyz = x + y + z + 2. Prove that
(1) xy 4- yz + zx > 2(x 4- y 4- z);
3
(2) Vx + \/y + \/z1.
1 + x 1 4- y 1 + z
c + a a + b
’2 = —
a b 
------- --------- t-b + c c + a
This can be proved by adding the inequality
l~~a " b 1
V b 4- c c + a ~ 2
with the analogous ones.
>. 2(l + s + 2)„LL£,S+? + S+5.i^ + l+‘.‘+^ > 
■ - . ■ a b b c c a
b + c c + a a + b 
------- 1-----+---------- a b c
> ab(a + b) 4- bc(b 4- c) + ca(c 4- a) a(a - b)(a - c) > 0,
which is exactly Schur’s Inequality.
(2) Here we have
a b X 
---------h 7----- 1 i a 4- c b 4- c J
1 1 l 1
■------= a, :------- — b, —— = c.14-x 14-y 14-z
3
2
57Old and New Inequalities
2
2 + xyz -x-y-z = (l- z)(l - xy) + (1 - x)(l - y) > 0.
Now, if z > 1 we have
This ends the proof.
x + y + z - xyz — x(l - yz) + y + z n
/\
 0,
___________ ,2 j. „2
 2(yz)3 y2 + z2 > 2yz.
1 ‘ 1 
n
= 1. Prove
that
Vojtech Jarnik
= a,. The inequality becomes
>
-• >>i
n
Inequality for the
Solution 2:
With the same notations, we have to prove that
(n -1) E J — a' — - x~~■“ V al + a2 + • • • + ai-l + ai+l 4- ... 4- On
First solution:
Let ~
n
E
i 1
Oi x-' I a V 1 ~ ai
But the last inequality is a consequence of Chebyshev’s Inequality for the n-tuples 
(ai,a2,...,an) and
which is Chebyshev
( 1 1__ __ 1
\ 1 - xf 1 - x22' ’'' ’ 1 - x2n) ‘
n t \■ ( E 1 _x2 J ’ 
\i=l1 ’ /
systems (xi,x2,... ,xn) and
So, it is enough to prove the inequality
>
>
> >
>
and we are done.
USAMO, 1999
2
!
> 0,
I
n 
£
n 
E
But using the Cauchy-Schwarz Inequality and the AM-GM Inequality, we 
deduce that
Now, using the fact that x, > 0, 
the above inequality yields
n 
(wc have used the fact that 4 and x, 2.
n
4n -I- (n — 4) x.
2
, which combined with
n / n
n2 3 and ai, da. • * a„ be real numbers such that 
Qj + a2 +.. . + an > n and a2 + a| + • • • + °n — n2- Prove that max{ai, a2,... ,an} >2.
Solution:
The most natural idea is to suppose that a* 0. Then wc have ^(2 — x n => x» — n a^so
n 
Xi — Il
(n - 1) x/ n - 1 • y/a~t 
x/a? + x/o2 + ■ • • + +■ yaj-i +-■•• +
i 
-—+ 'aiT?
x/a? + x/d2 + . . . + y/Cli-1 + y/aTr\ + . . . + y/ttn 
y/n - 1 • y/ai 
1 1 —= + -= +-•••+• 
vai Va2
1 1 
-— + ••• + -p=
'Qi+1 VQn
y/Q-i 
y/n — 1
n 
E
SolutionsGO
-4..
Since
1
-4 = 0.+
l
56. Prove that if a, b,c > 0 have product 1, then
(a + b)(b + c)(c + a) > 4(a + b + c - 1).
MOSP, 2001
I
a 
a + b - ab
4
(c + d) + (a. + ft) ’
4(ft I d) 
(c + d) + (a + ft)
4(a l c) 
(ft + c) + (d + a)
55. If x and y are positive real numbers, show that xv + yx > 1.
France, 1996
54. [ Vasile Cirtoaje ] If a, ft, c, d are 
a - b b- c c-d d- ci n
ftlc c I d d I a alb
Conjecture. (Vasile. Cirtoaje.)
If a,b,c,d,c. arc positive real numbers, then
a - ft ft - c c-d d — e. e - a------+-------+-------+------- 4-------- > 0. 
ft + c---c + d d + e e + a--- a + ft- “
Solution.’.
We have
a - b b - c c-d d - a a + c ft + d c + a d+b------4.------- =------4 -|.------- 4-----------4 = 
b + c-- c + d d + a a + ft ft + c---- c-f-d d + a a + ft
, , ( 1 1= (a + c) I —— + 3——\ ft + c d + a
Solution:
We will prove that ab >----- :----- - for any a, b 6 (0,1). Indeed, from Bernoulli
OL + 0 ~ GO *
Inequality it follows that a1-6 = (1 + a - I)1-4 
—^—+ —y—
x + y-xy x + y-xy x + y x + y
1 1 4 1 1 >
ft + c d + a “ (ft + c) + (d + a) ’ c. + d a + ft ~
wc fiCt
a -ft ft - r c-d. d - ri
— - ■ 4- —4" ~ 4* — >
ft+c c + d d + a a + ft
Equality holds for a - c and ft - d.
1
a + ft
positive real numbers, then 
c-d
d I a
c + a
d + a
1
a + ft
b+d
c + d
+ (b + d) f—— + ——7 
\c 4~ d cl 4* 0
Old and New Inequalities til
ab 4- be 4- ca 4- > 4.
ab 4- be 4- ca 4- > 4
And so it is enough to prove that
(ab + be 4- ca)3 > 9(a + b 4- c).
> 3 > 1,
because
(ab 4 be 4- ca)2 > 3abc(a + b+ c) = 3(a 4- b 4- c).
57. Prove that for any a, b, c > 0,
(a2 4- b2 4- c2)(a 4- b — c)(6 4- c - a)(c 4- a - b) +bc l-ca) (a4-b4c)(ab4-bc4-ca)/?2 > abc(a2 Ft24-c2).
0 4-c)
Solution:
Clearly, if one of the factors in the left-hand side is negative, we are done. So, 
we may assume that a,b,c are the side lenghts of a triangle ABC. With the usual 
notations in a triangle, the inequality becomes
16 K2 
a + b + c
But this follows from the fact chat (a 4- b 4- c)(ab 4- be 4- ca) > 9abc and
3
a 4- b 4- c
Now, we can apply the AM-GM Inequality in the following form
1
a 4- b 4- c
3
a 4- b 4- c
J (ab 4- be 4- ca)2 
y 81 (a 4- b 4- c)
But this is easy, because we clearly have ab 4- be 4- ca > 3 and (ab 4- be 4- ca)2 > 
3abc(a 4- b 4- c) = 3(a 4- b 4- c).
_ 1 
a I b I c
2. ,
— (ab -h be l ca)
Second, solution:
8Wc will use the fact Chat (a 4- b)(b 4- c)(c 4- a) > -(a 4- b 4- c)(a‘b + be 4- co).
2So, it is enough to prove that - (ab 4- be 4- ca) 4-
Inequality, we can write
8
9
> 1. Using the AM-GM
Solutions62
or
and>
n n
n
>4- 14-x?1 4- x"
and
>F
Thus, we have
59. [ Gabriel Dospinescu ] Prove that for any positive real numbers Xi,X2, •. . ,x„ 
with product 1 we have the inequality
n
rf[(l+«?) 
i=l
i=l \>=1 t=l */
Solution:
Using the AM-GM Inequality, we deduce that 
x? x? x^_i , 1
l + x^
1+xS
58. [ D.P.Mavlo ) Let a,t,c > 0. Prove that
„ , 1 1 1 a b . c3 + a + 6 + c4----- F t 1-----F 7 4- ■a 0 c b c a3 + a + 6 + cd----- F 7 4-----F 7 4- - 4— >3a b c b
n • xn
: n 2a, b2 a I - > 2b,c2b | 7 > 2c. c " a ~ b
abc£a4-£i + £a2c+52j >2(£a + £ab).
But this follows from the inequalities
n2bc I - > 2ab, b2cn I - > 2hr.,c2nb \ T > 2rn 
cab
nn
? n(l-Fx")>^4-^-.
\ t=i x"
(a+l)(b+l)(c4-l)
1 + abc
Kvant, 1988
Solution:
The inequality is equivalent to the following one 
_ „ 1 _ n \\ib +? a
atc+1
Old and New Inequalities 63
n n n
a3 4- b3 4- c3 4- abed > min
Kvant, 1993
the inequality d
— ~ abc
> 24-
*E
61. Prove that for any real numbers a, b, c we have the inequality
J2(l + a2)2(l 4- b2)2(a - c)2(b - c)2 >(14- a2)(l + b2)(l + c2)(a - b)2(b- c)2(c -