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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) 1 + (n — 2)+ + grt 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 Lj 1 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 bổ 1 " ' =1 » 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

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we get (ety+z)? _ 1 a tayty? ,_ ty tyes ze 2 ? (etytz)? z+u+z or (ety+z)? _ 1 #2 -+z +2 1—(ab+be+ ca)— e` where a = — “ _—, b= —#_—, c=——`—— The inequality can be rewritten EZErytz %-+1 +2 EZErytz as 1 1 1 > 9, 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* + )

~Il

(a+b)*+(b+e)°+(e+a)2>

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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 1 1 1 1 1 1 Ita+b Irb+e— l+c+a = Dba 245 2+¢ Bulgaria, 1997 Solution: Let x =a+6+c and y= ab+ bc+ ca Using brute-force, it is easy to see that 244 3 12+4 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 5 2

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

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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 11 „22+ — 11 7 et—+y-—+——£>274+—+42,/1+-— But, it is immediate to prove 2z 2x 2œ — 7ï 2z x? IS 3 9 _ 15 that 2/1+ => 5 and soz+y+z>5+"+— > > We have equality for x x 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 b+e („+ z)+ b (z+z)+ +) >3 ỹ c+tu +b Solution: We will prove the inequality C a+b (z +9) > V3(z + 9z + z#) a b —(yt2)+——@t2)4 b+e

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to obtain a % b+e cra! Gab - (E5) v2: Eas 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: bac ;(#+) >3, V(+)( + 2) — (+ +2), which may be obtained by applying the Cauchy-Schwarz Inequality, as follows b c ——(U+z “+t a) +o +94 +2)4+ +)= a b+e a+b ytez 2+7 ztrụ 2 beet cha atb (Vytzt+Vvzt+a+Vr+y) he (a+b+2) ( )~3z+u+z)> 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

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and so it is enough to prove that (c+yt+z2—-3) _ 3 3 ?+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 a1 t=1 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

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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 n ) Lj,Vig + Li,_, S 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)

q b c

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

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105 Prove that for any real numbers ay, d2, ,@, the following inequality holds he 2 mr as a) ` đ¿ | < » ———~t¡d¿ (> ijai +71 Solution: Observe that n i ` n ae 1 a » TT T8) = ii -ja; | tt? dt = i,j=l J ij=1 0 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

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3 a; a? by 8 aj 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) = al 1 ¿=1 Using (1) and (2) we find that i= sin

~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: 1 1 1 Let « = -—,y = -,z = —- We find the equivalent form if —- + -—+—- = 1 then b Cc ry 2z (#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)

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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 1 1 1+4}? A +b? —14ab' 1 1 1 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

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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

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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, n+l 2 ntl 2 n+l n+1 (Soo) + (S's) < S s72 +1) + sj(n — z4) = ¡=1 i=1 k=1 k=1 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 k 2 (doen eS (te tay) i=0 1</<7<2k+1

Proof of the lemma

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

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3217-Now, let us turn back to the solution of the problem Let us call a succession of 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

5 = 4i; đá, 11; y Diy thy y Digs Dig t1y +++ y Digthys +++) Qiny+ ++, Qi, tk, f

where 1; +k; < 4j41 — 1 Thus, we write S as a sequence, followed by a gap, followed

by a sequence, then a gap and so on Now, take 8s; = aj, + + +4@i,+4,,52 = @ij4k,41+

-++ + Gign-1, -,S2rp—1 = Gi, H+++ + Qi,+k, Then, 2 x «| = (s1 + øa+ - + Sap—1)? 7ES » (s¡ + - +)” < » (œ + - + a;)Ÿ 1</<j<2r—l1 1<i<j<n lA

the last inequality being clearly true, because the terms in the left hand side are among those from the right hand side

Third solution:

We will prove the inequality using induction For n = 2 and n = 38 it’s easy

Suppose the inequality is true for all k < n and let us prove it for n If 1 ¢ S, then

we just apply the inductive step for the numbers az2, ,@,, because the right hand side doesn’t decrease Now, suppose 1 € S If 2 € S, then we apply the inductive step with the numbers a; + a2,a3, ,@n Thus, we may assume that 2 ¢ S It is easy to

(a + b)?

2

see that (a+b+c)? +c? > > 2ab and thus we have (a; + a2 + + ax)? +

(ag +a3 +++-+ag_1)” > 2a,a, (*) Also, the inductive step for a3, ,an shows that 2 ~ | < XS Wt taj)’ i€S\{1} 3<i<j<n So, it suffices to show that nm aj +90 3” ai¡< (+: tai)” + ieS\{1} i= i Me: (aạ + - + ø¡)” || we

But this is clear from the fact that a? appears in the right hand side and by summing

up the inequalities from (*)

111 [| Dung Tran Nam ] Let 71, 22 , £2004 be real numbers in the interval [—1, 1]

such that x? + x3 + + #3004 = 0 Find the maximal value of the 7, +22 + -+ 22904

Solution: 1

Let us take a; = x? and the function f : [—1,1] > R, f(x) = 23 We will prove

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1 f(@t+y+1)+f(-1l) > f(x) + fly) if -l<a,y <0 2 f is convex on [—1,0] and concave on [0, 1]

3.Ifx >Oandy <Oanda+y<Othen f(x) + f(y) < f(-1)+ f(@+y+1) and if xt+y>Othen f(x) + f(y) < f(x+y) The proofs of these results are easy Indeed, for the first one we make the substitution 2 = —a*,y = —b? and it comes down to 1 > a?4+b?+(1-a—b)? & 1> (a+b)(a?—ab+b?)+1—-3(a+b)+3(a+6)?-(at+b)? s = 3(a+6)(1 —a)(1 — 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 (¢),to, ,te004) = t is the point where

the maximum value of the function g: 4 = {z € [—1,1]2994|zy + - + z„ = 0} —> 2004

—> lĐ,g(4, ,22oo4) = ` ƒ(zz) (this maximum exists because this function is k=1

defined on a compact) is attained then we have that all positive components of ¢ 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 t} +tg+ -+teoo4 =

k 3 k

0 we find that a = 3004 — and the value of g in this point is (2004—k) 3004 — & —k

k

Thus we have to fnd the maximum value of (2004 — k) ¢/ 04k k, when & 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 </223 - V17812 — 223

112 | Gabriel Dospinescu, Calin Popa | Prove that if n > 2 and a1,aa, , đụ, are real numbers with product 1, then

2n

m— Ì - Wn — l(œI + aạ + - + đa — mì)

g7 + gã +::-+a2—Tn >

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 a1, ,@, > 0 (otherwise we replace a1,@2, ,@n with

|a1|, |a@a|,.-.,|@n|, which have product 1 Yet, the right hand side increases) Now,

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„2n

*m=T Wn — 1 (ai +øs + -+aa_—1 — (n — 1) "J/aids 0n—1)

Because, *-1⁄41da đứ„_¡ < l and ai+aa+ -T+dnạ_1—(n—]) *=J/0103 daạT—1 > 0, it is enough to prove the inequality

2

đa; + -+a2 ¡—(n—1)G2>3 —¬ - /n— 1-G - (œi + - +aa_† — Ín — 1)G)

Now, we apply the inductive hypothesis for the numbers a eng or! which have product 1 and we infer that 2 2 ajte-+a7_, 2(n—1) „s; ay tes + 4n-1 and so it suffices to prove that 2(n — Ì 2n, an “Vn — 2(a1 +++ -+an—1—(n-1)G@) > > Vin = Mar +> +an—1—(n—G), Lae 1 Yn — 1 ; which is the same as 1 + > This becomes nín — 2) — *Wn-2 1 n(n—1) 1I+—— 2 ( ~ m5) ° and (n — 1)n~! 1 1 1 \"? e 1 AW (n — 2)" = n—2 a |) 1 +2 < n—2 1 +2 < 2

For n = 3 and n = 4 it is easy to check

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because

n—-1 1 n—-1 1 1 1

ert gh 14 ap x (n — 1)z (n — l)z Sn pf — _ (n — 1)n—!

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Second solution: Clearly, the problem asks to prove that if syz = 1 then 2 2, — We have two cases The first and easy one is when xy + z + zz > z++z In this case we can apply the Cauchy-Schwarz Inequality to get 2 2 <4/3 2.Vz+1ŠV32.z.1 1 3 » <5 - À3 )2(zp+z++1)<32+z++z+zu+z+zz) © S z+0+z<zy+z+zz

and so in this case the inequality is proved

The second case is when # + z + zz < z + + z Thus,

(z — 1)(T— 1)(z— 1) =z++z— z — z2 — zz >0

and so exactly two of the numbers #, , z are smaller than 1, let them be x and y So, we must prove that if x and y are smaller than 1, then 2 2 2z +4/—+4/—— <3 x+t1 +1 zyt+1 Using the Cauchy-Schwarz Inequality, we get 2 2 2U 1 1 2U +4/—-+ <2 + ——— + z+l +1 + 1 zø+l £#+l + 1 and so it is enough to prove that this last quantity is at most 3 But this comes down to 1 1 L_ 2z ttl _— + l 2 1+ = tr zu+l wy Because we have + —— > 1, the left hand side is at most = i 1 1 l— xy

etl + — — |] = — yt+l —_—_ (a + 1)(y +1)

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114 Prove the following inequality for positive real numbers x, y, z

1 1 1 9

te +29 (sụp tyxs? teax) >4

lran, 1996

First solution (by Iurie Boreico):

With the substitution s+y =c,y+z=a,z+2 = b, the inequality becomes after some easy computations

2 1 ›

—— => |(a—b)“ >0 ¥ (4-3) @-02

Let a > 6 > ec If 2c? > ab, each term in the above expression is positive and

we are done So, let 2c? < ab First, we prove that 2b? > ac, 2a? > be Suppose that 2b? < ac Then (64+ c)? < 2(b? +c?) < a(b+c) and sob+c < a, false Clearly, we

can write the inequality like that

We can immediately see that the inequality (a — c)? > (a — b)? + (b — ce)? holds

and thus if suffices to prove that 2 2 1 1 1 1 2 2 —+—-—_— b—c)”>|>+— — — b)* (= 7 be a? =) TH (i 7 cab =) (a — 6) 1 1 2 2 1 1W But it is clear that = +—5-—- — <{|—-—-— J and so the right hand side b c2 ab dc b =€ (a — b)?(b Be — ©)” Also, it is easy to see that c is at most 2,2 _ 11.1 1 (a — b)? ac be a b —ac be b2c2 (a — ð)”(b — c)° b2c2 ? which shows that the left hand side is at least and this ends the solution Second solution:

Since the inequality is homogenous, we may assume that xy + yz + zx = 3 Also,

we make the substitution x + y + z = 3a From (x + + z)Ÿ > 3(zw + 0z + zz), we

get a > 1 Now we write the inequality as follows 1 + 1 + 1

(3a—2z)? (8a-—y)? (3a— z)

4[( + 3az)2 + (uz + 3à)Ÿ + (zz + 3ay)?] > 3(9a — zuz)Ÿ, 4(27a1 — 18a? + 3 + 4azz) > (9a — xyz)”,

3(12a2 — 1)(3a? — 4) + zuz(34a — zuz) > 0, (1)

3

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12(3a” — 1)? + 208a? > (17a— xyz) — (2)

We have two cases

i) Case 3a? — 4 > 0 Since

1

34a — xyz = 9 B4(e +y+z)(ay + yz+ zx) — 9xyz] > 0, the inequality (1) is true

ii) Case 3a? — 4 < 0 From Schur’s Inequality

(z++z)”— 4(z+>~z)(z + 0z + zz) + 9ryz > 0,

it follows that 3a? — 4a + xyz > 0 Thus,

12(3a? — 1)? + 208a? — (17a — xyz)* > 12(3a” — 1)? 4+ 208a? — a? (3a? + 13)? = = 3(4— 1la? + 10a’ — 3a®) = 3(1 — a”)?(4 — 3a’)? > 0

115 Prove that for any x,y in the interval [0, 1],

Vite? + /14y?4+ JA —2)? + (1—y)? > (14+ V5) (1 = ay)

Solution (by Faruk F Abi-Khuzam and Roy Barbara - ”A sharp in-

equality and the tradius conjecture” ):

Let the function F : [0,1 + R,F(z,) = Vite? 4+ J14y? + (1—2x)?+(1—y)? — (14+ V5)(1 — ay) It is clear that F is symmetric in x and y and also the convexity of the function « + 1 + z2 shows that F(z,0) > 0 for all

x Now, suppose we fix y and consider F' as a function in « It’s derivatives are , _ £ _ l—z ƒ@)=0+ v5 + VT =z?*+q—”? and 1 (1 —y)? ƒ"(ø) = + vũ+z?”” (a =2)? += 9) Thus, f is convex and its derivative is increasing Now, let _ | 6V3_ _ r= !~ TT /°=1+ vỗ, The first case we will discuss is y > : It is easy to see that in this case we have cy > 1

and so f’(0) > 0 Because f’ is increasing, we have f’(#) > 0 and so f is increasing with f(0) = F(0,y) = Fy, 0) > 0 Thus, in this case the inequality is proved

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Due to case 1 and to symmetry, it remains to show that the inequality holds in

1 1 1

the cases ø € 0, Al , € |0,r] and z € | Al YE | At

In the first case we deduce immediately that

l—z

vi+z? v1i+(-1)2

1 1

Thus, we have f' (¡) < 0, which shows that ƒ 1s decreasing on 0, Al Because from

the first case discussed we have f @ > 0, we will have F(ax,y) = f(x) > 0 for all

f(z) <r(1+ v5) +

1

points (2, y) with x € 0 đ , € |0,r|

1 1 Now, let us discuss the most important case, when x € | Al „1 € [ i: Let

the points O(0,0), A(1,0), BO, 1), C(0, 1), M1, y), N(x, 1) The triangle OMN has

perimeter

1— VI+z2+ vw1+?+Vv(LI—z)?+(1— ø)2 and area >

But it is trivial to show that in any triangle with perimeter P and area S we have the 2 P inequality S < ——~ Thus, we find that /1+22+,/1+y2+/(1-2)?+(1-y)?> 12/3 V6V3/1 — ry > (1+ V5)(1—2xy) due to the fact that zy > r? The proof is complete

116 [| Suranyi | Prove that for any positive real numbers a, a2, ,@p the fol- lowing inequality holds

(ø—1)(œ†+aý+ -+an)+naiaa dạ„ > (ai +aas+ -+an)(ap | +an t+ -tanty Miklos Schweitzer Competition 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 a, > øa > - > Gn 41 and đi + ag +++: +4, = 1 We have to prove that

nr nr n n

n> apt + nant) + nany1 -J[«: + Qn+1 ‘| [a — (1+ 4@p41) (>: đ + cha) >0

i=1 i=1 i=1 z1

But from the inductive hypothesis we have

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