Tuyển tập đề thi vô địch bất đẳng thức thế giới P3

Old

and

New

Inequalities

61

First solution:

Using the identity (a + 6)(6+ c)(c +a) = (a+6+4+c)(ab+ be + ca) — 1 we reduce
the problem to the following one
ab + be + ca + — > —
atb+e
Now, we can apply the AM-GM
b+6

> 4.

Inequality in the following form

,/ (ab + be + ca)?

—————
an.

———.

9(a+b+)

And so it is enough to prove that

(ab + be+ ca)? > 9(at+b+c).
But this is easy, because we clearly have ab + be + ca > 3 and

(ab+ be + ca)?

>

3abc(a+ b + e) = 3(a+b+ ©).
Second solution:
We will use the fact that (a + b)(b+ c)(e +a)

3
> gia + b+e)(ab + be + ca).

2

So, it is enough to prove that 9 (0b + be + ca) +
;
Inequality, we can write

a+b+c

> 1. Using the AM-GM

„| (œb + be + ca)?
>1
8l(a+b+ec)

—

2

g (ab + be+ ca) + —————
a+b+c7—

?

because

(ab + be + ca)2 > 3abe(a
+ b + e) = 3(a+ b+ e).
57. Prove that for any a,b,c > 0,

(a7 + b` + c?)(a+b— e)(b+e— a)(e+a— b) < abe(ab
+ be + ca).

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

(a? +b? +c’).

2

atb+e

< abc(ab+be+ca)

&

(at+b+c)(ab+bc+ca)R? > abc(a*
+b? +c’).

But this follows from the fact that (a + 6+ c)(ab + bc + ca) > 9abc and
0< OH? =9R?-— a2 — b — cẺ.


62

Solutions

58. | D.P.Mavlo | Let a,b,c > 0. Prove that

161 Sal fy gle We
1)(b Mes+ Mer)
1)(c Dery)+ 1
Btatbtesn41 rere
b
b
1 + abe
Kvant,

1988

Solution:
The inequality is equivalent to the following one

La

—

eatery
or

1

abe Ya+

—

2

— +)

Dato

_>939=————“^=

a c+

abe + 1

5

a

> 2(Soa+

am) .

But this follows from the inequalities
2
2
2 + ọa > 2ca
œ“bc
+ Pb > 2ab,“ca
+ —C > 2bc,c“ab
a
and

5
1
2
1
9
1
a-c+ — > 2a,b°a+
— > 2b,c°b+
= > 2c.
e
a
b

59. [| Gabriel Dospinescu | Prove that for any positive real numbers 21, 2%2,...,2n

with product 1 we have the inequality
n

n”-][(@P+l>
al
i

n

n

i=l

i=1

2)

1

3]

n

.

?

Solution:

Using the AM-GM
xt1

1+z7

and

1

l+z‡

Thus, we have

4

Inequality, we deduce that

12

l+zỹ

+

1

——

l+zỹ

fon.

Len—I1

1+z„¡

4

1

l+z?

>

n


Old

and

New

Inequalities

63

Of course, this is true for any other variable, so we can add all these inequalities to
obtain that

which is the desired inequality.

60. Let a,b,c,d > 0 such that ø + b+ e= 1. Prove that
a® +6°
+

+ abed > min

1

1

{ 5, stat:

Kvant,

1993

Solution:

1
Suppose the inequality is false. Then we have, taking into account that abc < 2m
1

1

the inequality d (5 — abe) >@+P43

1

- 3 We may assume that abe < 7

Now,

1

we will reach a contradiction proving that a® + 6? + c? + abed > T It is sufficient to
prove that

a+h+e——

———T——————
7
abc

1

But this inequality is equivalent to
»

a3 = 3abe+1—3

1

aửe +a3+b”+c7>

.

+ l5abc

>

+

1. We

use now the identity
1+
9ab
` ab and reduce the problem to proving that » ab< _

which is Schur’s Inequality.
61. Prove that for any real numbers a, b,c we have the inequality

3

+22)? +b2)?(ø— e)?(b— ø)” > (1+a”)(1+ð2)(1+c”)(a— b)”(b— e)”(e— a)°.
AMM
Solution:
The inequality can be also written as »

5
52
(1
1
ren?

> 1 (of course, we may

assume that a,b,c are distinct). Now, adding the inequalities

(1+ a?)(1 + 07)

(+6)d+e)

J,

(1+c?)(a—b)2 —

(1+a@?)(6-c)?

~

(which can be found using the AM-GM

Inequality) we deduce that

(1+ a7)(1+ 0?)

2>axzza-

14+ 06?
|a— b|le— Ù|

>>

14+ 6?

I6-s0-a


64

Solutions

and so it is enough to prove that the last quantity is at least 1. But it follows from

»

1+?

jy

|l(b— a)(b—e)|—

SS
1 + -

= 1

—c)

and the problem is solved.
62. | Titu Andreescu, Mircea Lascu | Let a, 2, y,z be positive real numbers such
that xyz = 1 and a > 1. Prove that
re

ụ2

Uu+zZ

z+z

+

z2

3

ety

2

First solution:
We may of course assume that x > y > z. Then we have
x
y
Z
+2Z_—

and 2%!

z1+#z

E+Y

> y?-! > 2-1, Using Chobystiew's

E23 (Le)

Inequality we infer that

ee

ì

.

(LS)%

.

Now, all we have to do is to observe that this follows from the inequalities » re1 >3
(from the AM-GM
Second

Inequality)

„3

and s~—

„-+z—2

solution:

According to the Cauchy-Schwarz

(lu+2)+u(e+2)+2(e+y)] (

Inequality, we have:

a

+

+5)

>

(ý 2+2

+z®)

Thus it remains to show that

(0

+y

+2)"

> 3(ay + yz + 20).

Since (x + y + z)* > 3(zy + yz + zz), it is enough to prove that
lta
ite
ita
+?

+27

+22

>r+ytz.

From Bernoulli’s Inequality, we get

of =[L+(z—-U|9° >1++#Z¿-n=
Ita

1—

c4,

1

14

and, similarly,

1+a

1+a

La

3(1-a)

lta

(c+y+z)-(@+y4+z)=

`


Old

=

—

Equality holds for # =

and

New

Inequalities

65

—

(e+w+z~3)

> “—(3ÿwz—

3) =0.

= z = 1.

Remark:

Using the substitution 6 = a+1(6>

2) andz=

-, y= Đ z= - (abc = 1) the

inequality becomes as follows
1

a? (b

1

1

3

+c) 7 b2(e + a) 7 đ*{(a+ b) 2 2

For 6 = 3, we obtain one of the problems of IMO 1995 (proposed by Russia).
63. Prove that for any real numbers 21,...,2%n,Y1,---,Yn Such that 2?+---+22 =

yee dyad,

?ì

(Z1a — #s1)” < 3 ( — > een
k=1

Korea,

2001

Solution:

We clearly have the inequality

(amram

YS wan? = (Set) (Sout)- (Sam) =
- (1- Soa) mm)

1
+?

hr

n

i=1

i=1

i=1

i=1

Because we also have
i=1

mới

2

i=1

<1, we find immediately that

( — 3 su)

ụ + year]

i=1

=1

<2 ( — ` ea]
z—=1

and the problem 1s solved.
64. [ Laurentiu Panaitopol | Let ai, ø›,..., „

be pairwise distinct positive inte-

gers. Prove that

aj tag to +a,

2n+1

2S

(ai + as + --- + an).
TST

Romania


66

Solutions

Solution:
Without

loss of generality, we may assume that a, < ag < --- < a, and hence

a; > 17 for all 7. Thus, we may take b; = a; —i > 0 and the inequality becomes
nr

`

9

nr

bj< +25

i=1

Now,

ib; ; +

mín + 1)(2n
+ 1) _

6

Ù¡ ; +

—————————,
6

+1

using the fact that aj,

> a; we infer that 6) < ¿

< --- < ð„ and from

Inequality we deduce that

w=1

+1

and the conclusion is immediate.
diately that

n(n + 1)(2n
4+ 1)

n

›

:

i=1

Chebyshev’s

numbers

2n +1

>
>—a

7=1

Also, from the above relations we can see imme-

we have equality if and

only if a,,a2,...,a@,

is a permutation

of the

1,2,...,n.

65. [ Calin Popa ] Let a,b,c be positive real numbers such that a+b+c¢=
Prove that
b/c

a(V3c+vab)

+

ca

+

b(V3a+vbe)

avb

1.

` 3v3

c(V3b+ ca) —

4 -

Solution:

Rewrite the inequality in the form
bc

>

3v3

ca +
+a
b

.

With

ơ

the substitution ô =

be

4

ca

ab

.

4/,y =
'đ== the conditlon a + b-+e=
a
c
becomes xy + yz + zx = 1 and the inequality turns into
#
2. V3y

But, by applying the Cauchy-Schwarz

»

a
V3ay + ayz

>

>

3v

“4

+ yz

Inequality we obtain

» 7)

2

V34 3xryz

>

3

ty

V3 + i

v3

where we used the inequalities
(Siz)

>3 (So zy)

3v3

=—,4

and xyz < wa

1

Old

and

New

Inequalities

67

66. | Titu Andreescu, Gabriel Dospinescu | Let a, 6, c,d be real numbers such that

(1 + a?)(1 + 67)(1 +c?)(1 + d?) = 16. Prove that
—3 < ab+bce+cd+da+ac+
bd — abcd < 5.

Solution:

Let us write the condition in the form 16 = H¿ +a)- Ta —i). Using symmetric
sums, we can write this as follows
16=

(1-10

a- Sab +i >

abe + abcd)

(1+%

0 a- Sabi

D abe + abcd).

So, we have the identity 16 = (1 — Ð` ab + abcd)? + (SS a— >> abc)”. This means that
|1 — S> ab+ abcd| < 4 and from here the conclusion follows.
67. Prove that

(a? + 2)(b + 2)(c? + 2) > 9(ab + be + ca)
for any positive real numbers a, 6, c.
APMO,

2004

First solution:

We will prove even more: (a? + 2)(Ù2 + 2)(c? +2) > 3(œ + b + c)?. Because
(x+b+e) < (la| + |b| + |c|)”, we may assume that a,b,c are nonnegative. We will
use the fact that if x and y have the same sign then (1+ 2)(1+y) >1+2+y. So,
we write the inequality in the form

(+2)

> othe

and we have three cases

i) If a,b,c
are at least 1, then TT

(“

2_

4

3

+1)

>14505

2_

5

4

(

a)

2

» 2)

ii) If two of the three numbers are at least 1, let them be a and 6, then we have
a-1.,
3

peal
3

IV

Il

Pal
3

+2
3

(a? +b? +1) (1?+1% +c’) ` (a+b+c)
9
by the Cauchy-Schwarz

9

iii) If all three numbers are at most 1, then by Bernoulli

H(G
and the proof 1s complete.

—

9

Inequality.

+)

c2

Inequality we have


68

Solutions

Second

solution:

Expanding everything, we reduce the problem to proving that

(abc)? + 23 `a?0? +45 0a? +8>
Because
inequality

3300?

>

(abc)? + »

3 `ab

and

a?+2>

23 `a?0? +6>
2À ` ab.

9S — ab.
4À ` ab,

we are left with the

Of course, we can assume

that a,b,c are

non-negative and we can write a = 27,b = y’,c = z”. In this case

2S

ab-S oa? =(z++z)(z+—
z)(u+z— z)(z+z— 9).

It is clear that if x,y,z
is trivial. Otherwise,

are not side lengths of a triangle, then the inequality

we can take x =u+v,y

=v+w,z

= w-+u

and reduce the

inequality to

((u+v)(v+w)(w+u))*+2>

16( + 0 + u)uU.

We have ((u+v)(v + w)(w +u))* +141>3%/(ut vs

(vtw)t(utw)!

it remains to prove that the last quantity is at least 16(u+v+w)uvw.
down to

16

(utv)i(v+w)*(w+u)*

and

This comes

3

> sp (wap) (u ++u).

But this follows from the known inequalities

(utv)(vt+w)(w+u)

8

> gute +w)(uv + vw + wu),

8
(uv + vw + wu)* > 34(uow)3,
Third

wu +

+~+ > 3u).

solution:

In the same manner as in the Second solution, we reduce the problem to proving
that

(abe)? +2>

23 `ab— ».

Now, using Schur’s Inequality, we infer that
9ab
23 2ab—À `a?and as an immediate consequence of the AM-GM
9ab

a+b+c

———

Inequality we have

< 3V (abc)?.

This shows that as soon as we prove that

(abc)? + 2 > 3/(abe),
the problem is solved. But the last assertion follows from the AM-GM

Inequality.


Old

and

New

Inequalities

68.
| Vasile Cirtoaje
| Prove that IÍ 0 < z <
a) (1— zø)(1— z)(1— zz) 3 0;

69

< z and z+

+ z = zz +2, then

b) Jays
ay < 1a’y?
lay" < =.
55
Solution:

a) We have

(l—ay)(1—yz) =1l-—ay—yz+ay?2z =1—a2y—yz+y(@t+y+2—-2) =(y-1* >0
and similarly

(1—øz)—zz) =(1—z)“>0,
So the expressions 1 — xy, 1 — z

(1—zz)(1— zu) = (L— ø) >0.

and 1 — zz have the same sign.

b) We rewrite the relation x+y+z = xyz+2 as (1—x)(1-y)+(1-z)(1—-zy) = 0.
Ifz > 1 then z > # > z > 1 and so (I — z)(1 — ø) + (1— z)(1— zg) > 0, impossible.
So we have x < 1. Next we distinguish two cases 1) xy < 1; 2) zy > I1.
32
1) ty <1. We have ay 2) xy > 1. From y > ,/ry we get y > 1. Next we rewrite the relation x + y+2z=

0z + 2 as # +
T— 2 = (z0
— 1)z. Becausez > # gives
# +
— 2 > (z0
— l)U,
(@T— 1)(2—#— z) > 0 so 2> z(1+ 0). Using the AM-GM Inequality, we have

1+

> 2Vÿ and 1+

2>3z

2

=

= 1+ 548 5301-32. Thus we have 2 > 2e,/y and
32

which means that xy < 1 and xy? < 7

32
The equality x?y = 1 takes place when x = y = 1 and the equality x?y”? = 2z
2
takes place when x = 3° y= r=2.
69. | Titu Andreescu | Let a, b,c be positive real numbers such that a+b+c > abc.
Prove that at least two of the inequalities
2

2

a

Ob

eg

Cc

2

78

b

¢c¢

4a

2

8g

28

ec

a

Log

ob

are true.

TST

2001,

USA

Solution:
1
The most natural idea is to male the substitution -— = x, ; = Y, : = z. Thus, we
a
have #,,z > Ö and z# + z + zz > 1 and we have to prove that at least two of the
inequalities 2z+3-+6z

> 6, 2u+3z-+6z > 6, 2z+3z~+6y > 6 are true. Suppose this is

not the case. Then we may assume that 2z-+ 3„-+6z < 6 and 2z+3z-+ổy < 6. Adding,
1
5-5
we find that 5a+9y+8z < 12. But we have x >
+ ễ, Thus, 12 >
sẽ +9 +8z
Ù

#

+

z


70

Solutions

which is the same as 12(y+z) > 5+9y?+82?+ 1l2yz @ (2z—-1)?4+ (8y+2z—-2)? < 0,
which is clearly impossible. Thus, the conclusion follows.
70. | Gabriel Dospinescu, Marian Tetiva ] Let x,y,z > 0 such that
z +
Prove that

+Z—+xUz.

(— 1)(w— 1)(z— 1) < 63
- 10.

First solution:

Because of x < xyz > yz > 1 (and the similar relations xz > 1, zy > 1) at most
one of the three numbers

can be less than

1. In any of these cases (2 < 1, y > 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
x-l=a,y-—1=6,z-l=c.
Then a,b,c are nonnegative real numbers and, because
zœ=a+l,=b+l,z=c+],
they satisfy

a+1+b6+1+c+1=(a4+1)(b4+1) (c+),
which means
abe + ab+act+
be =

2.

Now let « = Wabc; we have
ab + ac +

be > 3V abache = 32’,

that’s why we get

a
+ 34? <2

(x41) (x?

4+ 22-2)

< 06

& (x +1) (ec +1+ V3) (ec +1- V3) <0.

For x > 0, this yields
Vabe = 2x < V3—-1,
or, equivalently
abe <

which is exactly
The proof 1s complete.

(v3 — 1)

3

;

( — 1)(wT— 1)(z— 1) < 6V3
- 10.


Old

Second

and

New

Inequalities

71

solution:

Like in the first solution (and due to the symmetry) we may suppose that x > 1,
y > 1; we can even assume that « > 1, y > 1 (for x = 1 the inequality is plain). Then
we get xy > 1 and from the given condition we have
+%+1

— 1`
The relation to prove is

(z — 1)(T—
1) — 1) < 6v3-
10 âđ

â 2#z (s + z + yz) < 6V3
9,
or, with this expression of z,

day"
— ay
— (ex +y) xy—**—1 <6V3-96
zy—-1
& (xy —x —y)* + (6V3 — 10) zụ < 6/3 -— 9,

after some

calculations.

Now, we put x =a+1,y=0+41

and transform this into

a2b? + (6v3 10) (a +b-+ab) —2ab>0.
But
a+b>

2Vab

and 6/3 — 10> 0, so it suffices to show that

ab? + (6V3 - 10) (2Vab + ab) — 2ab > 0.
The substitution t = Vab > 0 reduces this inequality to

tt 4 (sv3 — 12) # +2 (6v3-10)t> 0,

“
The

f+ (6v3 - 12) t+2(6V3—10) > 0.
derivative of the function

f(t) =0 + (6V8—12)++ 2 (6V8— 10), + >0
is

f#@=3 ( — (v3 — ))

and has only one positive zero. It is \/3 — 1 and it’s easy to see that this is a minimum

point for f in the interval [0, co). Consequently

f(t) > f(V3-1) =0,
and we are done.
A final observation:

in fact we have

f(t) = (1 v3+1) (t+2v3- 2).


72

Solutions

which shows that ƒ (£) > 0 for £ > 0.
71. | Marian Tetiva | Prove that for any positive real numbers a, b,c,
a®—b
a+b

bì—c
b+c

ce —a*| _ (a—b)* + (b—0)* + (c— a)"
c+a
—

4

Moldova
First

*

TST,

2004

solution:

First of all, we observe that right hand side can be transformed into

2,

a3 — bŸ

a+b

—

3

x8

1

1

(a -)(5-—)

3

+0

«3

1

1

-' (TT)

7

(a —b)(e— b)\(a—0) (> ab)
(œ + b)(ø + c)(b + c)
and so we have to prove the inequality

|l(a — ð)(b — e)(ce— a)|(ab+
be+ ca) _ 1

<5 (Ye

It is also easy to prove that (ø + ð)(b + c)(e+ a)
so we are left with

oa
Using the AM-GM

(X2(4— 9?)

2

_

80).

IV

(a+b)(b + e)(e+ 4)

(œ+b+
e)(ab + be + ca) and

> [«

-%|.

Inequality, we reduce this inequality to the following one

(3) >|H«=9|.

This one is easy. Just observe that we can assume that a > 6 > ¢ and in this case it
becomes

3
(œ — b)(a — e)(b — c) < 2 (a+b+ ø)Ÿ

and it follows from the AM-GM

Inequality.

Second solution (by Marian

Tetiva):

It is easy to see that the inequality is not only cyclic, but symmetric. That is why
we may assume that a > b > c > 0. The idea is to use the inequality
2
2
y.
winyty
x
4+ 5

eros

> ———

——

x+y

<9

>†/+†+—

5

which is true if « > 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_

arg

—>

a2+ab+t?
a+b

and of course
act

a

>b+—,b
>
FOr

Cc
2

>

b2 + be + c?

b+e

>zor

b

sg—


Old

and

New

Inequalities

73

That is why we can write
a? — b?
a+b

+ ab + b2
= (apea’? teeth
a+b
IV

»

b + be -L c2
gether

gŠ -L ae - c2
Và j2Taote
`

+ (c+5} —&=9 (ø+3) =
$) +00)
(a-b) (0

3692

4
In the same manner we can prove that
ng

< 5 ”(ø — b)?

œ+b

—

4

and the conclusion follows.

72. | Titu Andreescu | Let a,b,c be positive real numbers. Prove that

(a° — a? +3)(0? —b7 4+3)(2 —c +3) > (at+b+e)®.

USAMO, 2004
Solution:

We start with the inequality a° —a2+3>a?+2

6

it remains to show that

3
II)

Using the AM-GM

>

(2+)

.

1

>=

#/[(a?

Inequality, one has
a3

+

3
ajt+2

We

(a?—1)(a?—1) > 0. Thus,

1

bb) 3 4+2

+

3 42

3a
+ 2)

write two similar inequalities and then add up all these relations. We

that

[Ie +2> (Sa),

which

is what

3

we wanted.

73. | Gabriel Dospinescu | Let n > 2 and #1,#s,...,#„ > 0 such that

Prove that

DÓI

ao

=
(Sot): - (a)
owe 4+——..
mCn
2

2

will find


74

Solutions

Solution:
In this problem,

a combination

between

identities

and

the Cauchy-Schwarz

Inequality is the way to proceed. So, let us start with the expression

(=Li. 4th4; 2) ° |

1
t

wi

œ
—_—

l|

+

|

Hs

|S

|

S|

1
+

l|

SI

- (E82) Es) E3)~

Li

Z

*j

SIS.

> (r3)

1
mm.

We can immediately see that

Thus we could find from the inequality

`

—
1
(+3 -›)

>0

2

#

(9) a)ae~

that

Unfortunately, this is not enough. So, let us try to minimize

> (r?).
Ly

x

1

%¿

J

This could be done using the Cauchy-Schwarz

(Ee)
3%;

x;

.

Because

»
‡
Li
(=
j

2;
a

Xj

2

c2,
—

2

Inequality:

(eens
Lj

%;

n
2

2) = 1, we deduce that

Sa) (Ea) eras
vi

which is what we wanted to prove. Of course, we should prove that we cannot have
equality. But equality would imply that x, = v2 = --- = &,, which contradicts the
assumption

É⁄)82)-e


Old

and

New

Inequalities

75

74. | Gabriel Dospinescu, Mircea Lascu, Marian Tetiva ] Prove that for any positive real numbers a, Ö, c,

a2 + b2 + e? + 2abe+
3 > (1+ ø)(1+b)(1+
e).
First solution:

Let ƒ(ø,b,e) = a? +? + c2 + abe+ 2— ø—b— e— ab— be — ca. We have to prove
that all values of f are nonnegative. If a,b,c > 3, then we clearly have - + : + : < 1,
which means that ƒ(ø,b,e)

> a? + b + cẰ+2—a—b—c

b

that a < 3 and let m = —

(3 — a)(b—c)*
1

>0.

So, we may assume

Easy computations show that f(a, b,c) — f(a,m,m)

> 0 and so it remains to prove that f(a,m,m)

=

> 0, which is the same

as

(a+1)m? —-2(a4+1)m4+a?—a4+2>0.
This is clearly true, because the discriminant

of the quadratic

equation is —4(@ +

1)(a — 1)2 <0.
Second

solution:

Recall Turkevici’s

Inequality

z* +? + z2 +! + 2xuzt > z2? + 2z? + z2) + 12x? + z2 + 2£
for all positive real numbers x, y,z,t. Taking t = 1,ø = #2,b = 9Ÿ,z = z

and using

the fact that 2/Vabc < abc + 1, we find the desired inequality.

75. | Titu Andreescu, Zuming Feng | Let a,b,c be positive real numbers. Prove
that

(2a+b+c)—

2a2 + (b+c)?

First

(2b+a+c)

(2c+a+b)?

26?+(at+c)?

2c3+(a+b)3

<8

— —

USAMO, 2003

solution:

Because the inequality is homogeneous, we can assume that a+ b+c=

(2a+b+c)?

2a2+(b+c)?

‘Thus

_ a2 + 6a+9

_ M42.

3a2—-6a+9

3


4a+ 3

2a+b+c)

1

32.)

4a + 4

3

4a+3

2+(a—1)?/

) <
—

3. Then


76

Solutions

Second

solution:

Denote x =

b

re
a

+

y=

2

b

eS

(x + 2)? WT

c

. We have to prove that

2z +1 <=e©ŠS
5

But, from the Cauchy-Schwarz

(2-1)?

` ` `"——>-.
an

1

Inequality, we have

ye

` (z++z—3)?

z?+2

—z?+?2+z?+6.

It remains to prove that

Q(x? + y? + 2? 4 Qay + 2z + 2z+ — 6 — 6y — 6z+9) >z?+?+z”+6©

©z?+ˆ°+z”+4(zụ
+ 0z + zz) — 12(œ + + z) + 123 0.
Now # + z + zz > 3/+22z2 > 12 (because #z

> 8), so we still have to prove that

(xtytz)?+24—12(2+y+z) +12 > 0, which is equivalent to (x +y+z-—6)? > 0,
clearly true.

76. Prove that for any positive real numbers x,y and any positive integers m,n,

(n—1)(m—1) (a? ty™*") + (mtn—-l)(ary"+ary™) > mnlartr—ly tytn" 7),
Austrian-Polish

Competition,

1995

Solution:

We transform the inequality as follows:

mx —y)(a™ Fr" — y™*P—1) > (mtn —W(a™—y™)(a" —y")

.m+n—]

_

ụ

n1

ứn +? — 1)

gm

—

m

nr

coy

&

n

— U) — m(œ—W)_ n(z
— 0)

(we have assumed that x > y). The last relation can also be written
#

ey) |

x

min tars |

¥

¥

and this follows from Chebyshev’s

x

ear. f

t” "dt

Ụ

Inequality

for integrals.

77. Let a,b,c,d,e be positive real numbers such that abcde = 1. Prove that
a+ abc
+
l+øaồ+abcd

b + bcd

+
c+ cde
+
d + dea
+
e+eab
1+bc+bcdc
1+ cd+cdea
TI+de+decab_ ` 1+ ca
+ cabc
Crux

Solution:

a=

substitution
x
U

b-2

z

=

,d=

Silo

consider the standard

«Là

We

tu
x

> 10
—

3

Mathematicorum


Old

and

New

Inequalities

77

with x,y, 2,t,u > 0. It is clear that
1 + 1

atabe
—
y
tt
1
tsi
1+ ab + abcd ~ Ta
T+
+

Z

U

tạ.
.
.
1
1
1
Writing the other relations as well, and denoting — = a,, — = aa, T— = đa, TT
x
y
Z
as, we have to prove that if a; > 0, then

` _ 0a 1d.
a,

Using the Cauchy-Schwarz

+agt+a5

1
a4,-=
u

„1.1
-

3

Inequality, we minor the left-hand side with

4S?
2S% — (ag + a4)? — (a, + a4)? — (a3 + a5)? — (ag + a5)? — (a, + a3)?’

where S =

5

So ai.

By applying the Cauchy-Schwarz

Inequality

again for the

denominator Of the fraction, we obtain the conclusion.

78. | Titu Andreescu | Prove that for any a,b,c, € (0, 3) the following inequality
holds

sỉn ø - sin(œ — Ö) - sin(œ —e)
sin(b + c)

sinb-sin(6 — e) - sin(b — ø)_
sin(c + a)

sinc- sin(e — a) - sin(e — b) >0
sin(a + b)
—

TST 2003, USA

Solution:
Let x = sina, y = sinb, z = sinc. Then we have x,y,z > 0. It is easy to see that
the following relations are true:

sina - sin(a — b)- sin(a — c) - sin(a + 6) « sin(a +c) = x(a? — y?) (a? — z”)
Using similar relations for the other terms, we have to prove that:

> 2(2? — y°)(a? —y*) > 0.
With the substitution 2 = /u, y = Vv, z = Vw the inequality becomes » Vu(u —
v)(u — w) > 0. But this follows from Schur’s Inequality.
79. Prove that if a,b,c are positive real numbers then,

Va4 + b1 + c+ + Veh? + Pet 2a? > Va3b+ Bet Gat
KMO

Summer

Program

Solution:

It is clear that it suffices to prove the following inequalities
So at +S oa? b? > S > ab +

áp?

Vab? + be? + ca.
Test, 2001


78

Solutions

and

(Soot) (Shas?) = (Sad) (2á).

The first one follows from Schur’s Inequality
So at+abeS

a > S

a?b+

áp

and the fact that

Sab?

> abe S~ a.

The second one is a simple consequence of the Cauchy-Schwarz

Inequality:

(ab + Betcha) < (ab? + b?c? +? a7)(at + bt 4+ &*)
(abŸ + beŠ + ca®)? < (a2bŠ + b°e? + e?a?)(a* + b + €).
80. | Gabriel Dospinescu, Mircea Lascu | Eor a given ø% > 2 find the smallest
constant k, with the property: if a1,...,@, > 0 have product 1, then
a1 Q2

4

4203

(a? +a2)(az +a)

4

4

Andy,

(a3 +.a3)(a? + a2)


(a2 + øi)(aˆ + an)

Solution:

”

1

Let us take first aj = đứa = --- =ữạ_ 1 —=#,0ạ = —-

We infer that

en

bk

>

Qe2r-1

"= amETTTj@TT+T)

for all z > 0. Clearly, this implies

k,

+

m„ — 2

”(+z)”

`

7+ — 2

”1+ø

> ?— 2. Let us prove that ø — 2 1s a good

constant and the problem will be solved.

First, we will prove that (2? + y)(y? + ©) > zw(1 + z)(1 + g). Indeed, this is the
same as (x + y)(x — y)” > 0. So, it suffices to prove that
1

(1+a1)(L+az)

+

1

1

(1+ a2)(1 +43)

Now, we take a, = „HỘ c.«a đựy — “
#2

»“
k=1

2-1 R+1ER+2

(Z + #+1)(#k+1 + #&+2)

) >9,

which can be also written in the following from
n

Lk

»

hol Lk + Le41

+

#11
(tp + #g+1)(#e+1

+ #g+3)

>

Clearly,
nm

2

nm

xem

< n-2.

(1+ an)(1+ ai) ~

and the above inequality becomes

#1

(1 _

fess + ——______

———————_|.

nee

"+ 2n


Old

and

New

Inequalities

79

So, we have to prove the inequality
n

S-

Chee

ha

>1,

Mek + €p41) (p41

Using the Cauchy-Schwarz

+ Le+2)

Inequality, we infer that
n

n

2
+1

2„
k=1 (tp + #g+1)(#g+q

de

k+1

>

+ #eg+2)

2

—

xẽ

(tp + Le41)(Le+1

+ Le42)

k=1

and so it suffices to prove that
n

2

»

Th

«|

> Sox;

k=1

nr

Tr

+23

k=1

7i

+ 0a.

k=1

k=1

But this one is equivalent to
n

2

›

Liv; > 2 ›

n

%g#k+1 + )

3g'k--3

1
and it is clear. Thus,

k, = n — 2.

81. [ Vasile Cirtoaje | For any real numbers a, b,c, x,y, z prove that the inequality

az + bụ + cz +

(a3 + b2 + c2)(z2 + 2+ z3) >

wl bo

holds

s(œ+b+e)(z++z).

Kvant,

Solution:

far ty? +22

Let us denote # = \/ —————:
z = tr, which imply

a3 -+ b2 + c2

v.

Using

Hung

the substitutions « = tp, y == tq tq and
Poy

Š

C4+PtP=apPigtr.

The given inequality becomes
2

A

(at+p)?+(b+q)?4+(et+r) >

os

> 5 at+b+c)(p+q+r),
Wl

aptbgter+a° +b +e

=(atb+e\(ptqtr).

Since

4(a+b+e)(p+qg+r)<[(a+b+e)+(p+aqg+r)}È.

This inequality is clearly true.

Wl] —¬

it suffices to prove that

(a+ p)? + (b+q)? +(ce+r) >

1989

=[(at+p)
+ (b+ q) + (ctr).


S0

Solutions

82. [ Vasile Cirtoaje | Prove that the sides a, b, c of a triangle satisfy the inequality

3(0 4242-1)
b
¢
oa
First

50(2 4622).
a
b
oe

solution:

We may assume that c is the smallest among a, b,c. Then let x = b— at c After
some computations, the inequality becomes

(3a—2e)z°+ (ø +e- 1) (a—c)? > 0 ©

(3ø—2e)(2b—a—e)?+(4b+2e—3a)(a—e)Ÿ > 0

which follows immediately from 3ø > 2c, 4b + 2e — 3ø = 3(b+e— ø)+b—ec>0.
Second

solution:

Make the classical substitution a =y+2,b=z++2,c=2x+y

and clear denomi-

nators. The problem reduces to proving that

#2 + 9 + z3 + 2(z?u + 0 )z + z +) > 3(z2 + uz + zz”).
We

can of course assume

y=xt+m,z=2+Nn

that x is the smallest

among

x,y,z.

Then

we can write

with nonnegatives m and n. A short computation shows that

the inequality reduces to 2x(m? — mn +n?) + m3 +n? + 2m?n — 3n?m > 0. All we
need to prove is that m? +n? + 2m?n > 3n?m | (n—m)> —(n—m)m? +m? > 0
and this follows immediately from the inequality t? + 1 > 3, true for t > —1.
83. | Walther Janous | Let n > 2 and let 71, 22,...,2%, > 0 add up to 1. Prove
that

Crux
First

Mathematicorum

solution:

The most

natural idea is to use the fact that
n—

x;

4

1-2;

n—l

Byte

te

+ R-pt@U@yi
t+ + fy

Thus, we have

(i=) <0

7 (n-

?—1

l1—

%

—

Ly

?—1

1
x12

-««4—1+1‹-‹--Ÿn

and we have to prove the inequality

lI (: + ~) >
i=l

Ly

=1

2E

(: +—
„

#122

s.-2—19¿L1

---Jn

) .