Tuyển tập đề thi vô địch bất đẳng thức thế giới P5

21 Now, let us prove the second inequality.. Solution: Of course, the inequality can be written in the following form n>-—n-1 S2 + so.. We will prove this inequality by induction.. Beca

and so

n—1 —_

nai TJ > toi 3 a; mm a,

Using this last inequality, it remains to | prove that

ny a; -S đ | — @n41 nS ap —S a, +

—1 +đ„ +1 a; + (n— lay, - | > 0

Now, we will break this inequality into

Gn+1 (Tre +(n—-l)a )đxy 11 — đại J)>0

and

ben — Sa) —An41 s34 — Yer) > 0

Let us justify these two inequalities The first one is pretty obvious

n

Io (n — 1)djp.1 — đu = II — „+1 + đn+1) + (nm — Yang, — any, =

21

+?

> Ong tangy: So (ai — Angi) +(n- Yan, - đa} 1 =0

21 Now, let us prove the second inequality It can be written as

na; — À ad? > dua nề at — À dị

1

Because n › ae — › a~' > 0 (using Chebyshev’s Inequality) and a,41 < —, it n

1=1 i=1

is enough to prove that

1

but this one follows, because na??? + -a; '> a; for all 7 Thus, the inductive step

n

is proved

117 Prove that for any %1,%2, ,%n > 0 with product 1,

n

SY) (a — 27)? > Soa? -

w=1 1<i<j<n

A generalization of Turkevici’s inequality

Solution:

Of course, the inequality can be written in the following form

n>-—(n-1) S2) + (so)

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

ƒ(#1,#a2, ,#p) = —(n — l) (2) + (so)

and let G = ”-⁄2a#3 #ạ, where we have already chosen x, = min{x,22, , 2p}

It is easy to see that the inequality f(r1,%2, ,%n) < f(r1,G,G, ,G) is equivalent

to

(n — LS ax; — S2) > 2z [Som —Ín — va)

n Clearly, we have x; < G and » x, > (n —1)G, so it suffices to prove that

k=2

We will prove that this inequality holds for all zs, ,„ > 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

(ø—=2)3_z? +m—1> (yo)

and so it suffices to prove that

Soap tn-1>S 2x - (z¿T— 1)” >0,

clearly true

Thus, we have proved that ƒ(#1,#as, ,#„) < ƒ(zi,G,G, ,G) Now, to com- plete the inductive step, we will prove that ƒ(zi,G,G, ,ŒG) < 0 Because clearly r= Gr- ;> the last assertion reduces to proving that

(n — 1) ((n-6? + aos) +n > ((m-)6+ a)

which comes down to

n—2 2n — 2

Gonz 7" = "Gn=3

and this one is an immediate consequence of the AM-GM Inequality

118 | Gabriel Dospinescu | Find the minimum value of the expression

» 4142 - Aan - ay m

1

1 add up to 1 and n > 2 is an integer

where @1,@2, ,@n, <

Solution:

We will prove that the minimal value is =: Indeed, using Suranyi’s In-

nh—

equality, we find that

(m — 1) Soa? +NQ1a2 4n > Soap => NA, 42 4n > S a7 (1 — (n — 1)aj)

=1

Now, let us observe that

¬ 3

az (1— (n= 1)ax) = _ _ (i)

and so, by an immediate application of Holder Inequality, we have

n 4 3

»

n—-1 ; k=1 ;

fol 1— (n— 1)ag But for n > 3, we can apply Jensen’s Inequality to deduce that

nooo n 3

Thus, combining all these inequalities, we have proved the inequality for n > 3 For n = 3 it reduces to proving that

b

Viner 2"

which was proved in the solution of the problem 107

119 | Vasile Cirtoaje | Let a1,a2, ,a@, < 1 be nonnegative real numbers such that

ajtaz+ +a2 v3

Prove that

ay 4 ag 4 4 an > na

l-da la IlTa2 — l1—a2

Solution:

We proceed by induction Clearly, the inequality is trivial for n= 1 Now we suppose that the inequality is valid for n = k —1, k > 2 and will prove that

Lae > —_—

for

aj+as+ +a; IV Is

We assume, without loss of generality, that aj > a2 > > ax, therefore a > az Using the notation

ap +a? + -+ap_, ka? — a3

3 from a > a, it follows that « > a> ¬ Thus, by the inductive hypothesis, we have

ay a2 đpy —1 (k — 1)z

I-aP 1d TH Tạp — 1—z2 `

Tay 1—z2 — 1—a2

„3 _ ga - #8 mm a7

we obtain

(k — 1)(z — a) (a — ay)(a + ax)

z+

and

—(a—ag)(1+aaz) (k-1)(a —a)(1 + az) _ a= 4 je eee) (1 — az)(1 - a?) (1 — x?)(1 - a?) 1— a2 1-a? (1 — x?)(x +a) _ (a — ag)(œ — a¿)[—1 + #Ÿ + d; + ray + a(2 + ay) + 07 + ara, (x + ag) + a 2d]

(1— a3)(1— a¿)(1— #?)(œ + a) k(a — ag)(a + ag)

k-1

k(a — ag)? (a + ag) (k-1)(a@+a,z) — `

ka” — d2 k—1 Since 2? — az = — dị —= , it follows that

(a — ag) (@ — ag) =

and hence we have to show that

x? + az + rap + a(# + ag) + aŸ + aag(œ + ag) + a zag > 1

In order to show this inequality, we notice that

ka? + (k — 2)a2 ka?

wet dy = k— 1 “k—1

z-+døyg >\x/z2+a})>a _ — b VWk_1

z + q2 + zay + a(# + ag) + 8Ÿ + azag(% + ag) + a2zag > (z2 + a2) + a(# + ag) + a2 >

k | k

>(sẫ: atte > 3a 1, and the proof is complete Equality holds when a, = ag = = Gn

and

Consequently, we have

Remarks

1 From the final solution, we can easily see that the inequality is valid for the larger condition

1

14+, /2+4 m+—1 This is the largest range for a, because Íor đị = đa = - = đ„ạ + = # and ø„ = 0

— Ì

(therefore a = #\/ * n ), from the given inequality we get a >

3

2 The special case n = 3 and a = v is a problem from Crux 2003

120 | Vasile Cirtoaje, Mircea Lascu | Let a,b,c, x,y,z be positive real numbers such that

(atb+e\(e@tyt+z=(@P@4P4 ee)? +y42)=4

Prove that

beryz < 3

A0CLY Z —

J<ng

Solution:

Using the AM-GM Inequality, we have

4(ab+be+ca)(xyt+yz+zx) = [(a+bt+c)? - (a +0? +c’)] [(ctyt+2) -(2? +y?+2)] =

=20— (a+b+e)*(z? +2 + z?)T— (a?+2+c°)(x+u+z)°<

<20—2W(a+b+e)2(z? +2 + z2)(a2 + b2 +c?)(z+u+z)?=4,

therefore

(ab + bc + ca)(xy + yz + za) <1 (1)

By multiplying the well-known inequalities

(ab + be + ca)? > 3abc(at+b+c), (xu +ụz+ zz)) > 3z0z(w +ụ +2),

it follows that

(ab + be + ca)? (xy + yz + zx)* > 9abczz(a + b+ e)(œ + + 2),

or

(ab + bc + ca)(xy + yz 4+ 2x) > 36abcryz (2) From (1) and (2), we conclude that

1 < (ab + be + ca)(xy + yz + 2x) > 36abcryz,

therefore 1 < 36abcryz

To have 1 = 36abcryz, the equalities (ab + bc + ca)? = 3abe(a + b + c) and (cytyz+zx)* = 3xyz(x+y+z) are necessary But these conditions imply a = b = ¢

and x = y = 2, which contradict the relations (a +6+c¢)(x2x+y+2) = (a@+6%4

c2)(z? + 2 + z?) = 4 Thus, it follows that 1 > 36abcryz

121 | Gabriel Dospinescu | For a given n > 2, find the minimal value of the

constant k,, such that if 71, %2, ,%, > 0 have product 1, then

VI+kzạzi VJV1tknze VI + kn#„

Mathlinks Contest

We will prove that k, = ao Taking #1 = #a = - = #„ = 1, we find that

TL —

2n — 1

mà an So, it remains to show that

(n — 1)?

n

1

ra | te 2n — Ì -= ai

(m—1)2 `

if 1 - 2: s- Zn = 1 Suppose this inequality doesn’t hold for a certain system of n numbers with product 1, 71, 22, ,%, > 0 Thus, we can find a number M >n-1 and some numbers a; > 0 which add up to 1, such that

1 2n — | „

(ø—1)?"

= Mag

1+

Thus, az <>

HSS r (areag~')] > (wea) <1 (w=arg ~')-

i and we have

Now, denote 1 — (n — l)ag = by > O and observe that Shy = 1 Also, the above

k=1 inequality becomes

¬ TIe-%> > (2n — 1)” (ITs -»]

Because from the AM-GM Inequality we have

II < (TH)

k=1

our assumption leads to

IIc — bự)” < PO be bp

k=1

So, it is enough to prove that for any positive numbers a1, @2, ,@,, the inequality

[[ (ai tae t+ +++ aK 1 tony t+ tan)” > Ty 2 «+ An (Qt Fda +++ +n)"

k=1

holds

This strong inequality will be proved by induction For n = 3, it follows from the fact that

He)’ E99) oss

Suppose the inequality is true for all systems of n numbers Let @1,a@2, ,@n41

abe

be positive real numbers Because the inequality is symmetric and homogeneous, we may assume that a, < a2 < +- <a@,4, and also that aj +a.+ -+a, = 1 Applying the inductive hypothesis we get the inequality

m

n—1 [[c- a;)’ > > way" Q1QQ An-

To prove the inductive step, we must prove that

]l(a : +1- aj) > (n+1)m+114 Lae đx>@„+1(1 + On+1)

21

n+l

Thus, it is enough to prove the stronger inequality

1 An+1 ° 3m+2 nr "

i=1

Now, using Huygens Inequality and the AM-GM Inequality, we find that

2n

1+ An+1 > fat Gn+1 > (14 NAn+1

Ila — aj)

Tì

wl

and so we are left with the inequality

pq NOt > n” an+i(L+an+i) 1

n—1 ~ (n—1)??-(n+1)rt!

n( + @n+41) = l+z, where # 1S

n+1

1

if Gn4, > max{aj,a2, ,an} > — So, we can put

n nonnegative So, the inequality becomes

14—2_) ylttrt de

n(x + 1) (x + 1)r7

Using Bernoulli Inequality, we find immediately that

2n

n(œ + 1) e+1 Also, (1+ 2)"~! > 1+(n-—1)z and so it is enough to prove that

32 +1 ` l1+(n+ 1)z z-+l — I+(n-— l)z

which 1s 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

2n — 1

is true for k, = — and that this is the value asked by the problem

(n — 1)

Remark

For n = 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 Gazeta Matematica, Seria A

122 [ Vasile Cirtoaje, Gabriel Dospinescu | For a given n > 2, find the maximal value of the constant k„ such that for any %1,22, ,2%n > 0 for which 27 +23 + -+

x? = 1 we have the inequality

(I—zi)(1— #s2) (ÍT— #z„) > ku#1#2 n

Solution:

We will prove that this constant is (,/n — 1)” Indeed, let a; = 2? We must find the minimal value of the expression

nm

[[a- va

i=l

nm

[va

i=l

when a; +a@2+ :+a@,, = 1 Let us observe that proving that this minimum is (.,/n—1)”

reduces to proving that

[Ïd =s0 >(#= 0"-TJv-]]g + v8

But from the result proved in the solution of the problem 121, we find that

nr — 1)? m nm

Hú —aj)* > == [[«

So, it is enough to prove that

nm

lu + va < (+=)

7=1 But this is an easy task, because from the AM-GM Inequality we get

14+ Ja;))< }1+ =] <{1+—],

form <r] (og

the last one being a simple consequence of the Cauchy-Schwarz Inequality

nr

Remark

The case n = 4 was proposed by Vasile Cartoaje in Gazeta Matematica Annual Contest, 2001

Glossary

(1) Abel’s Summation Formula

If a1, @9, ,@n, 61, b9, ,b, are real or complex numbers, and

S;=a,+aot+ +a;,i = 1,2, ,n,

then

` a;Ö¡ = » Si(bi — bi41) + Sdn

AM-GM (Arithmetic Mean-Geometric Mean) Inequality

If a 1, @2, ,@, are nonnegative real numbers, then

— So ai > (41 02 dn)”, nial

with equality if and only if a4, = ag = = ay This inequality is a special case of the Power Mean Inequality

Arithmetic Mean-Harmonic Mean (AM-HM) Inequality

If a1, @, ,@n, are positive real numbers, then

— ai ———_—

hi " L - _ 1

i=1 ay with equality if and only if a, = a2 = = ay This inequality is a special case of the Power Mean Inequality

(4) Bernoulli’s Inequality

For any real numbers « > —1 and a > 1 we have (1+ 2)* > 1+ az

121

(5) Cauchy-Schwarz’s Inequality

For any real numbers a1, @2, ,@p, and bj, be, , bn

(a‡ + øã + + ø2)(bị + bộ + + b2) >

> (a,b; + asÖa + + a„,b„)Ÿ,

with equality if and only if a; and 6; are proportional, 7 = 1, 2, ,n

(6) Cauchy-Schwarz’s Inequality for integrals

If a,b are real numbers, a < b, and f,g : [a,b] > R are integrable functions,

then

( / “Heltei) < ( / Pee) ( / rey) ,

(7) Chebyshev’s Inequality

If ay < ag < < ay and 01, bo, ,b, are real numbers, then

I)If b: < by < < by then Said; > — (Soa; } bị]:

2)If bị > bạ > > bạ then Ð ` a¡b¡ <

(3) Chebyshev?s Inequality for integrals

If a,b are real numbers, a < 6, and f,g: [a,b] > R are integrable functions and having the same monotonicity, then

b

(b—a) [ Ƒ(z)g(z)dz > [ sou | g(z)dz

and if one is increasing, while the other is decreasing the reversed inequality

is true

(9) Convex function

A real-valued function f defined on an interval J of real numbers is convex

if, for any x,y in J and any nonnegative numbers a, with sum 1,

flax + By) < af(x) + Bf(y)

(10) Convexity

A function f(x) is concave up (down) on [a,b] C R if f(x) lies under (over) the line connecting (a1, f(a1)) and (6), f(61)) for all

ax<a<2<b, <b

A function g(x) is concave up (down) on the Euclidean plane if it is concave

up (down) on each line in the plane, where we identify the line naturally

(14)

with R

Concave up and down functions are also called convex and concave, re- spectively

If f is concave up on an interval [a,b] and Ai, A2, ,An are nonnegative numbers wit sum equal to 1, then

Ai f (#1) + Aof (42) tot Ànƒ(#n) > ƒOlzi + Às#¿ + + Àn#n)

for any %1,2%2, ,%p in the interval [a, b| TẾ the function is concave down, the inequality is reversed This is Jensen’s Inequality

Cyclic Sum

Let n be a positive integer Given a function f of n variables, define the cyclic sum of variables (21, #a, , #„) aS

Sf (1,3, «+5 2n) = f(x1,%2, ,n) + (Z3, 3, ,®n; #1)

cực

+ + ƒ(@„, đ1, đa, , Gp—1)-

Holder’s Inequality

1

If r,s are positive real numbers such that — + — = 1, then for any positive

rs real numbers a1, G2, -,@n,

b,,bo, ,bn,

1

Ss

3 3 4

7=1 < 7=1 = - i=l

nr a Tì Tì

Huygens Inequality

lÍ p1,7Ø», ,Ðn,01,05, ,0n,Ðt,ba, ,bạ are positive real numbers with

pit pot + pn = 1, then

Tú: +ủ¿)?: > II + II:

=1

Mac Laurin’s Inequality

For any positive real numbers 21, %2, -.,%n,

Sy, > S22 2 Sn,