Tuyển tập đề thi vô địch bất đẳng thức thế giới P4

But this has already been proved in the problem 84... which is Schur’s Inequality... Therefore, in the initial problem the answer is 21 achieved for Second solution: We use the same subs

is trivial, because #¿_+ + 2(n — l)#¿ + 2441 < 2(n-1)- » x, for all i This shows

k=1 that mp, > Taking x; = x’, the expression becomes

2(n — 1)

ert+ar-!4+2(n—-1) 142(n-l)et+a? 14+2(n—-1)e™-!+xr-?

1 and taking the limit when 2 approaches 0, we find that m, < ———~ and thus

2(n — 1)

1

win 2(n — 1)

Now, we will prove that M, > ¬ Of course, it suffices to prove that for any L1,%2,- ,Ln > O we have

n

2 =1 —Z +; 1 + 2(n — l)z¿ +¡‡A¡ — 2 <5:

But it is clear that

n

of 5 + 2(n — 1); + 2j41 - 1 2,/€j-1 °° fig + 2(n — 1);

=1

nr

1

=>, ¿=1 †„— [+ V?ii Đan

Lj

Taking V1" T— a;, we have to prove that if [[@ = 1 then

» ng < 1 But this has already been proved in the problem 84 Thus,

nm — a;

i=1

1

M, > 5 and because for 71 = %2 = - = &%, we have equality, we deduce that

1

M, = 3 which solves the problem

96 | Vasile Cirtoaje | If x,y, z are positive real numbers, then

Gazeta Matematica

Solution:

Considering the relation

#2 + øụ + 2 =(œ++ 2)? T— (xụ + uz + z#) —(e@+y42)z,

we get

a tayty? ,_ ty tyes ze 2 ?

or

(ety+z)? _ 1

#2 -+z +2 1—(ab+be+ ca)— e`

where a = — “ _—, b= —#_—, c=——`—— The inequality can be rewritten

as

l-d-ec l1-d-b l1-d-a7 where a,b,c are positive reals with a+ b+c=1 and d=ab+ bc+ca After making some computations the inequality becomes

9đ” — 6d? — 3d + 1 + 9abe > 0

or

d(3d — 1)? + (1 — 4d + 9abe) > 0

which is Schur’s Inequality

97 | Vasile Cirtoaje | For any a, b,c,d > 0 prove that

2(aŠ + 1)(b3 + 1)(eŸ + 1)(đ + 1) > (1+ abeđ)(1 + a”)(1 + 02)(1+ e2)(1+ d?)

Gazeta Matematică Solution:

Using Huygens Inequality

Hú +a*) > (1+ abed)*,

we notice that it is enough to show that that

22T + Ð? > [[(+ a9)(1+ a2)

Of course, it suffices to prove that 2(a? + 1)* > (a4 + 1)(a? +1)‘ for any positive real a But (a?+1)* < (a4+1)?(a? +1)? and we are left with the inequality 2(a?+1)? > (a+1)?(at+1) & 2(a?—ø+1)2>a?+1 6 (a—1)* 50, which follows

98 Prove that for any real numbers a, b,c,

(a* + b* + )

(a+b)*+(b+e)°+(e+a)2>

Vietnam TST, 1996

Solution:

Let us make the substitution a + b = 2z,b+ c= 2%,c+a = 2y The inequality becomes Soy +z—z)t<28 » x’ Now, we have the following chain of identities

Svyte-2 =>) (Sox? + 2z — 2a — 2z) =3 (Soa?) +4 (>2)

(S)0z~ su 2)) +43216w+zz =2 =3 (9z?) =4(S)) (S2) +

+16 `z2° — 4 SOI =4 6z} +16 `z?? — ®Sz} <28 9z!

">>"

99 Prove that if a,b,c are positive real numbers such that abc = 1, then

Bulgaria, 1997

Solution:

Let x =a+6+c and y= ab+ bc+ ca Using brute-force, it is easy to see that

the left hand side is 7 4g ẻ, while the right hand side 1s aor tery ow,

#2 + 2z + t + # 9 + 4z + 2 the inequality becomes

wi t4e+y+3 1, 1214z+ _- 2z + 3— zụ < 3-Yy

z2+ 2z +-+zxụ — 9+ 4z~+2 z2 + 2z -+t+zụ — 9+4z~+2u'

For the last inequality, we clear denominators hen using the Inequalitles # > 3,1 > 3,27 > 3y, we have

s7 > 5z”, = > y?, vy? > 92, 52y > 15z,z > 3 and z?u > 27

Summing up these inequalities, the desired inequality follows

1 3

100 [ Dung Tran Nam ] Find the minimum value of the expression — + atc

where a,b,c are positive real numbers such that 21ab + 2be + 8ca < 12

Vietnam, 2001

First solution (by Dung Tran Nam):

1 2

Let - = 2, 5 y, — = z Then it is easy to check that the condition of the problem

C

a,

becomes 2xyz > 2a” + 4 + 7z And we need to minimize x + + z But

2z >7 x> 2x + 4y

— 2z— 7

z(2z — 7) > 2+ + 4ụ >

Now, we transform the expression so that after one application of the AM-GM

2z+4 Inequality the numerator 3z — 7 should vanish z +# -†+ z > #++ = + : —

xy —

14

et—+y-—+——£>274+—+42,/1+-— But, it is immediate to prove

2z 2x 2œ — 7ï 2z x?

that 2/1+ => 5 and soz+y+z>5+"+— > > We have equality for

15 Therefore, in the initial problem the answer is 21 achieved for

Second solution:

We use the same substitution and reduce the problem to finding the minimum value of a+ y+ 2 when 2xyz > 2a + 4y + 7z Applying the weighted the AM-GM Inequality we find that

And also 2a + 4y + 7z > 105 - 125 515 -zð5 -y3 zi 215 This means that (a+

225

+ + z)?2(2z + 4ụ + 7z) > > ty: Because 2ayz > 2” + 4y + 7z, we will have

225 15

(œ++z)> + =z++z2 >

5 with equality for ¢ = 3,y = 2i2= 2

101 [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x,y, z,a,b,c > 0 such that xy + z + zz = 3,

a

(„+ z)+ b (z+z)+ +) >3

Solution:

We will prove the inequality

C

a+b (z +9) > V3(z + 9z + z#)

—(yt2)+——@t2)4 b+e

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

to obtain

a

Thus, we are left with the inequality 4 ` ( ; ) +4 +5< re

c

is equivalent to » craeznSä 3 , which is trivial

c+ø)(c

z + V3(z + z + z#) <

Remark

A stronger inequality is the following:

which may be obtained by applying the Cauchy-Schwarz Inequality, as follows

——(U+z “+t a) +o

+94 +2)4+ +)= a

2(z++z)=ˆ V(œ +9)(œ + z) — (@++2)

A good exercise for readers is to show that

So Vet y(at2z) >+z++z+ V3(aw + 0z + z2)

102 Let a,b,c be positive real numbers Prove that

(b+e— a)? (cta-—b)? | (a+b-c)?

(b+c)2+a? (c+a)?+2 (a+b)+c? —

Japan, 1997

First solution:

Let 2 = + y= cre ,2— a= The inequality can be written

c

b

— 1\2

gà z2+1 75 UV Ù > 2

Using the Cauchy-Schwarz Inequality, we find that

z2 +1 — z?+?+z?+3

and so it is enough to prove that

?+uaz2ass € Q/2”~—15),z+33 zy+18>0

But from Schur’s Inequality, after some computations, we deduce that

So ay > 2S a Thus, we have

(Sx) - 5 et 3 ay + 18> (Siz) -9Éz+ls>0,

the last one being clearly true since ` z > 6

Second solution:

Of course, we may take a+ 6+ c= 2 The inequality becomes

4(1 — ø)? 3 1 27 awa TS ————a<—_-

2⁄22+20-a 25 S >1+-aŠ10

But with the substitution 1-a = 2,1-—6=y,1—c = z, the inequality reduces to that from problem 47

103 [ Vasile Cirtoaje, Gabriel Dospinescu |] Prove that if a1, @2, ,@, > 0 then

a1 +@2 +++ +an-1 }

—_ đụ,

a + đỂ + + + độ — nữa > Ín = 1) [ —

where ứ„ 1s the least among the numDerS đ1, đạ, ; đạ,

Solution:

Let a; — ap, = 2; > 0 for7 € {1,2, ,n —1} Now, let us look at

n—-l1

n n n > a i=1

) a? —n |[& — (n— 1 —— — Gy

nr

as a polynomial in a = ay It is in fact

ay tag t+++4+2n-1\"

„ — a+ (at a)" na ][(6 +) = (x= Ð (

We will prove that the coefficient of a* is nonnegative for all k € {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 f(x) = +”

n—-1 n

3i i=1

n—1

n—1

3z >(n—1)

i=1

For k > 0, the coefficient of a* is

n—-1

n n—k

(2) › x; — Tì › i14: - (212 _—p¬

¿=1 1<?1< -<?„_Šn—1

Let us prove that this is nonnegative From the AM-GM Inequality we have

1<i1<?2< -<?„_—SŠn— Ì

+]

h —k n—k -kE\_ _# tì n—k

ED 1<t4 <ig< -<in_p<n-l (ert teirte tent) = 8 (Ea i= 1

n—-1

n which is clearly smaller than @ » Tp, This shows that each coefficient of the

¿1 polynomial is nonnegative and so this polynomial takes nonnegative values when restricted to nonnegative numbers

104 [ Turkevici | Prove that for all positive real numbers z, y, z, t,

z1 +ựt +zt +it + 2xgzt > 2U) + 2z? + 22t? 12x? + x?z” ty? Ph

Kvant Solution:

Clearly, it is enough to prove the inequality if zyzt = 1 and so the problem becomes

If a,b, c,d have product 1, then a2 + ð2 + e2 + đ2+ 2 > ab~ be + cả + da + ae + bd

Let d the minimum among a, ,c,d and let m = Wabc We will prove that

a+b? +c 4+d?+2-(ab+be+cd+da+tac+ bd) > đỀ + 3m2 + 2 — (3m? +3md),

which is in fact

a? +B +e —ab—be-~ca>d(atb+e~3Vabe)

Because d < abc, proving this first inequality comes down to the inequality

a2 + bŠ + e2 — ab — be — ca > Vabe (a+b +¢-3Vabe)

Yabo ~ Sabo ~ 3 abe

u2 + 02 +? 3 >u +0 0 + tru + 09) + tu

which is exactly a2 + b2 + œ2 — ab— bc — ca > Wabe (a +b+e— 3Vabe), Thus, it

remains to prove that d?+2>3md d2 +2 > 32, which is clear

105 Prove that for any real numbers ay, d2, ,@, the following inequality holds

a)

` đ¿ | < » ———~t¡d¿

(> ijai +71

Solution:

Observe that

» TT T8) = ii -ja; | tt? dt =

1 n

~ I Oo \Gj=1 do tai jay I | dt =

1 fn 2

= / Sria;- th)

0 „=1 Now, using the Cauchy-Schwarz Inequality for integrals, we get

[ (Sia) as Uf me] a) = (S20)

which ends the proof

106 Prove that if a,,a2, ,@n,61, ,6, are real numbers between 1001 and

2002, inclusively, such that af + a3 + -+a? = b?+b3+ -+62, then we have the

inequality

a "+ a + -+-#< -(a+a2~+ -+d2) a 17

TST Singapore

The key ideas are that ~ € 5 | for any 2 and that for all x € 5 | we have

a

5 the inequality #7 +1 < 2# Consequently, we have

and also

+

3 a;

Now, the observation that = = and the inequality = -— > 1+ allow us to

3

a:

write gửi > bs + a,b; and adding up these inequalities yields

i

5 mr Tì g3

see h(E i=1 Tì Qa: 3 nr

) D; + asbs] = › bị + › ab; (2)

Using (1) and (2) we find that

i=

~J (aŸ + aÿ + -:- + aŸ), nr which is the desired inequality

107 [ Titu Andreescu, Gabriel Dospinescu | Prove that if a, b,c are positive real numbers which add up to 1, then

(a7 + b”)(b2 + c?)(c? + a2) > 8(a?bŸ + b?e? + ca”)

Solution:

Let « = -—,y = -,z = —- We find the equivalent form if —- + -—+—- = 1 then

(#2 +ø2)(0° + z”)(z? + g2) > 8(a* ty? + 2°)’

We will prove the following inequality

1 1 1)\°

(x? +y°)(y? + 27)(z? + 2”) (- tot -) > 8(z” +ˆ + z”)”

Write 2? + y? = 2c, 0? + z? = 2a, z? + +? = 2b Then the inequality becomes

abe

DL Vipera 2 out

Recall Schur’s Inequality

Sat + abc(œ + b + c) > À a3(b+c) © abc(a+b+ c) > S a?(b+e— a)

Now, using Holder’s Inequality, we find that

for any positive numbers 2, y, 2

qŠ

Ề +€— b Cc a (x vb+ec— ) | b e a

Combining the two inequalities, we find that

b

HS

and so the inequality is proved

108 | Vasile Cirtoaje | If a,b, c,d are positive real numbers such that abcd = 1,

then

1 + 1 + 1 + 1 >1 (I+a)? (1+2 (1+)? (l4d? 7 ~

Gazeta Matematica

Solution:

If follows by summing the inequalities

1+4}? A +b? —14ab'

1+? G+d? —14ed

The first from these inequalities follows from

1 + Âm _ ab(a? + 6") — ab? — 2ab+1 _

(l+a)? (146)? 1+ab (1+ a)?(14+0)?(1+c)?

ab(a — 6)? + (ab — 1)?

(1+ a)?(1 4 6)?(1 + ab)

> 0

Equality holds if@=b=c=d=1

109 | Vasile Cirtoaje | Let a,b,c be positive real numbers Prove that

Bic @+a® att b+e cta at+db

Gazeta Matematica

Solution:

We have the following identities

a2 a ab(a— 6) +ac(a—c)

b? b _ bc(b—c) + ab(b — a)

c ce — ae(œ—d)+ be(c— b)

a2+b2 a+b (b+a)(b?+a°)

Thus, we have

q2 Qo ab(a — b) ab(a — b)

URES > b+e =» lam —c?) (a+e)(a3+ c3)

ab(a — b)?

(b+ c)\(e+ a)(b? + c2)(c2 + a2) >

0

= (a®+02+c°+ab+ be+ ca) cÀ”

110 | Gabriel Dospinescu | Let a1, a2, ,@,, be real numbers and let S be a non-empty subset of {1,2, ,2} Prove that

» < » (a; + +4a;)?

¡cS 1<i<j<n

TST 2004, Romania

First solution:

Denote sy = a, +-:-+ a, for k =1,n and also s,,1 = 0 Define now

„_ [1 ifies

0 , otherwise Using Abel’s summation we find that

So ai = a,b, + dobp + + +4nbyn =

¿CS

= 81 (by — bạ) + sa(ba — bs) + -'+ 8n—1(8n—1 — bn) + Snbn + 8n41(—b1)

Now, put by — bo = 1, be — ba — #22;, sÖn_—1 — bn = Ln—1,0n = In yEn41 = —b, So

we have

n+l

So ai = » 4:5:

¿c8 ¿=1 n+l and also x; € {—1,0,1} Clearly, »- = 0 On the other hand, using Lagrange

w=1 identity we find that

» (a; + : + a;)”

1<i<j<n

n

ae

w=1 1<i<j<n

n+l n+1 NỔ

1<i<j<n+1

So we need to prove that

n+1 n+l 2 n+l 2 (n+ 1) Ss” s2 > sai] + «)

But it is clear that

S si) + 5 “| = So si(1 + 27)

+ 2 » 8,8; (a2; + 1)

1<i<j<n+t+l1

Now, using the fact that 28,8; < s;° + s7, Ì + #¿#; > 0, we can write

2N ø8/(0aj+Ðl)< Yo (s2+37)+ziz;)=

1<i<jn+1 1<i<j<n+

nm (sp ++++ +8541) + 8fai(zo +23 4-6 an) Tình + 82a (#1 t+ + ao tes+4+an-1) = sia} — +++ — 82 @? tn(si + - +52)

So,

(Soo) + (S's) < S s72 +1) + sj(n — z4) =

n+l

=(n + 1) `” sz, and we are done

k+1

Second solution (by Andrei Negut):

First, let us prove a lemma

Lemma

For any @1,@2, ,@an41 © R we have the inequality

(doen eS (te tay)

i=0 1</<7<2k+1

Proof of the lemma

Let us take s, = a, + -+a, We have

A241 = 81 +83 — 8g ++++4+ Sons — Sax

i=O0

and so the left hand side in the lemma is

2k+1

> s2 +2 ) $2441$2;41 + 2 ) $2482; — 2 ` 8244182;

O<i<j<ck 1<i<j<k O<i<k

1<j<k

and the right hand side is just

2k+1 (2k + 1) » s2 — 2 » 858)

i=1 1<i<j<2k+1 Thus, we need to prove that

2k+1 2k dX si >4 » $2i4182j41 +4 » $2582;

O<i<j<k 1<¡<j<k and it comes by adding up the inequalities

28254182441 < 52/11 + 594415 282582; < 89; +