Tuyển tập đề thi vô địch bất đẳng thức thế giới P1

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Preface

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 already 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, questions 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 are 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 Tanadsescu, 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

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CHAPTER 1

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1 Prove that the inequality JES OF + VPS oP + EF ap > holds for arbitrary real numbers a, b,c Komal 2 | Dinu Serbanescu | If a,b,c € (0,1) prove that Vøabe+ w(1— a)(1— b)(1— e) < 1 Junior TST 2002, Romania 3 | Mircea Lascu | Let a,b,c be positive real numbers such that abc = 1 Prove that b+e Tat et Fe 2 Vat vb+ ve +3 cta atb Gazeta Matematica 4 If the equation x* + ax? + 227 + be + 1 = O has at least one real root, then a? +b? > 8

Tournament of the Towns, 1993

5 Find the maximum value of the expression z° + y? + 2° — 3xyz where x? + y? + z* =1and z,y,z are real numbers

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Old and New Inequalities 9 When does equality hold? JBMO 2002 Shortlist 10 [ loan Tomescu | Let x,y, z > 0 Prove that +z (1 + 32)(x + 8y)(y + 9z)(z + 6) When do we have equality? 1 Sma: Gazeta Matematica

11 [ Mihai Piticari, Dan Popescu | Prove that

5(a? + b2 + e2) < 6(aŸ + bỶ + c3) + 1,

for all a,b,c > O witha+6+c=1

12 [| Mircea Lascu | Let 71, %2, ,%, € R,n > 2 anda > 0 such that x + 2 T Prove that x; € [ *) , for all đa + +0„ = a and #7 +z2 + +2 < ¿€{1,2, ,n} 18 [ Adrian Zahariuc | Prove that for any a,b,c € (1,2) the following inequality holds b/a + cvb + a/c >1 4bf/e—cJa 4eVWa-aVvdb 4aVb-bVc — -

14 For positive real numbers a,b,c such that abc < 1, prove that

a be

+-+—=>a+b+c

b ec oa

15 [ Vasile Cirtoaje, Mircea Lascu | Let a,b,c,2,y,z be positive real numbers

such thata+a>b+y>c+zanda+b+c=2+y+2z Prove that ay+bzr > ac+2z

16 | Vasile Cirtoaje, Mircea Lascu | Let a,b,c be positive real numbers so that

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24 Let a,b,c > 0 such that at + 64 + ct < 2(a?b? + bc? + c?a”) Prove that aŠ + bŠ + e? < 2(ab + be + ca)

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Old and New Inequalities 11 25 Let n > 2 and %1, ,2, be positive real numbers satisfying 1 + 1 Ta 1 oi a, +1998 22+1998 ~“ 2,+1998 1998” —— nme > 1998 Prove that Vietnam, 1998 26 | Marian Tetiva | Consider positive real numbers 2, y, z so that +2 +y? +27 = LYZ Prove the following inequalities a) xyz > 27; b) z + zz + 0z > 2ï; c)z++z>9; đ) z +zz+z >> 2(z++z)+9 27 Let x,y,z be positive real numbers with sum 3 Prove that V + + Vz 3 zụ + 9z + zz Russia, 2002 28 D Olteanu | Let a,b,c be positive real numbers Prove that

a+b a ote b ete Cc 3

b+e 2a+b+c c+a 2b+e+a a+b 2c+a+b—4 Gazeta Matematica 29 For any positive real numbers a, b,c show that the following inequality holds ab c_ c+àa a+b b+c _+-+_> b c¢ a ce+b + ate + b+ta India, 2002 30 Let a,b,c be positive real numbers Prove that a3 b3 c 3(ab + be + ca)

RP—k+e @Ồ -ae+da2 @-adth = a+b+e

Proposed for the Balkan Mathematical Olympiad 31 | Adrian Zahariuc | Consider the pairwise distinct integers 71, %2, ,2n, n > 0 Prove that

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32 [| Murray Klamkin ] Find the maximum value of the expression #7 + #23 +

-+ + 9212, +2722, when 21, %2, ,U%n_—1,£n > 0 add up to 1 and n > 2

Crux Mathematicorum

33 Find the maximum value of the constant c such that for any

L1,02, -,%n,°*- > O for which xp4) > 4%, +4 + -+ 2% for any k, the inequality Vi + fiz to + Vin <cVui Fant Fy

also holds for any n IMO Shortlist, 1986 34 Given are positive real numbers a,b,c and x,y,z, for which a+ a2 =b+y= c+2z=1 Prove that 1 1 1 (abe + xy2)(— + + —) > 3 Russia, 2002 35 [ Viorel Vajaitu, Alexandru Zaharescu | Let a,b,c be positive real numbers Prove that ab + be + ca < Ì(a+b+e) s+b+2c b+ec+2a c+a+92b— 4 ‘ Gazeta Matematica

36 Find the maximum value of the expression

a”(b+ec+đd) +~b?(e+d+a) +c(d+a+b) + dđ”(a+b+ e)

where a, b,c,d are real numbers whose sum of squares is 1

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Old and New Inequalities 13 AO Let @1,@2, ,@, > 1 be positive integers Prove that at least one of the numbers %/a2, %%/a3, , °»-¥/@n, *%/a1 is less than or equal to V3

Adapted after a well-known problem

41 | Mircea Lascu, Marian Tetiva | Let x,y,z be positive real numbers which

satisfy the condition # + #z + 9z + 2zz = 1 Prove that the following inequalities hold 1 a ) oye <5 <=; b)z++z> 1 1 1 c) —=+—=+—->4(z++2); roy 2 1 1.1 (2z — 1)” d) Ty z tự tU†?)2 t0 (0z+1D) 3 bo] Go , where z = max {z, y, z}

42 | Manlio Marangelli | Prove that for any positive real numbers 2, y, z, 3(z2 + y?z + 272) (ay? + yz? + za") > xuz(œ + + z)

43 [| Gabriel Dospinescu |] Prove that if a,b,c are real numbers such that

max{a, b,c} — min{a, b,c} < 1, then

1+a@?4+b% +3 + 6abe > 3a7b + 3bŠc + 3c2a

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48 [ Gabriel Dospinescu ] Prove that if /x + /y+./z = 1, then (—=z)?1—ø)”q~ 2z)” > 2'zwz(z + w)(w + z)(z + #) 49 Let zø, 9, z be positive real numbers such that #z = #++z+2 Prove that (1) zu+z+zz>2(œ++2); (3) V#+ VW+ v#< ŠV2yz 50 Prove that if x,y,z are real numbers such that x? + y? + z? = 2, then z++z<zyz+2 IMO Shortlist, 1987

51 | Titu Andreescu, Gabriel Dospinescu | Prove that for any %1,%2, ,%n € (0,1) and for any permutation o of the set {1,2, ,n}, we have the inequality l—#; nr ny dm n >|l+ i=1 i=1 = 1 Prove nr 52 Let 21, %2, ,£p be positive real numbers such that › m ‘ =1 vj that Tr Tr 1 4; >Ín — Ì — ve 2 ( )2_ vn Vojtech Jarnik

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Old and New Inequalities 15 56 Prove that if a,b,c > 0 have product 1, then

(a+ b)(6b+c)\(e+a) > 4(a+b+c-1)

MOSP, 2001

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

1 161 1)(b + 1)(c + 1

Btatbtesn4 rere Sal fy gle We Mer) b b Mes Dery) 1+ abe

Kvant, 1988 59 [ Gabriel Dospinescu | Prove that for any positive real numbers 21, 2%2, ,2n with product 1 we have the inequality n n n 4 n mn, “+1)> j — | án :]@?+Ð> 5 c2] w=l1 w=1 ¿=1 60 Let a,b,c,d > 0 such that ø + b+ e= 1 Prove that 11 d 3+3 + c bcđ > mìn 4 —,— + ——= è¿ œ”+ 0” +c” + dbc > min {5.5 +3} Kvant, 1993 61 Prove that for any real numbers a, b,c we have the inequality

3 (1+42)?(1+ð2)?(ø— e)?(b— e)® > (1+a?)(1+ð2)(1+ c?)(a — b)”(b — e)?(c— a)?

AMM

62 | Titu Andreescu, Mircea Lascu | Let a, x,y,z be positive real numbers such

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64 | Laurentiu Panaitopol | Let a1,a2, ,a@, be pairwise distinct positive integers Prove that 2n+1 aj +a5+ -+a;,> 23 (a, tag + -+4y) TST Romania 65 [ Calin Popa ] Let a,b,c be positive real numbers such that a+b+c¢= 1 Prove that b/c cự avb 3v3

a(Vầ + Vab) b(Vần+ Và) c(VAb+ ve +7

66 [ Titu Andreescu, Gabriel Dospineseu | 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 67 Prove that (a + 2)(b? + 2)(c? + 2) > 9(ab + be + ca) for any positive real numbers a, b,c APMO, 2004 68 | Vasile Cirtoaje | Prove that ifÍ 0 < #z < < z and z + + z = zz +2, then a) (L— #p)(1— z)(1 — #z) > 0) 2 3,2 b) ay <1,2°y < mm

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Old and New Inequalities 17 71 | Marian Tetiva | Prove that for any positive real numbers a, b,c,

a®—b bì—c3 -a? (a — b)? + (b-c)* + (c—a)? + < a+b b+e c+a |— 4 Moldova TST, 2004 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 73 [ Gabriel Dospinescu ] Let n > 2 and 21, 2%2, ,%p, > 0 such that S2) (>: ¬ =n?+1 %k , k=1 k=1 2) - S2) >n®P+4+— rat k=1 Ly n(n — 1) Prove that

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

a2 + b2 + e° + 2abe+ 3 > (1+ ø)(1+b)(1+ e)

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

that

(Qa+b+c)? (2b+a+c)? (2c+a+b)? 3

2a2 + (b+c)? 26?+(at+c)? 2c? + (a+b)? ~~

USAMO, 2003

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

(n—1)(m—1)(a™*"ty™*") + (mtn-1 (ary +ary™) > mnlartr ty pyr" ag),

Austrian-Polish Competition, 1995

77 Let a,b,c, d,e be positive real numbers such that abcde = 1 Prove that

a+abc b + bcd c+ cde d+ dea e+eab

ltab+abed 1+6bc+bcde 1+cd+cdea 1+de+deab l+ca+ cabe

> 10 — 3

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

holds

sỉn ø - sin(œ — Ư) - sin(œ —e) sinb-sin(6 — e) - sin(b — ø)_ sinc- sin(e — a) - sin(e — b) >0 sin( + c) sin(e + ø}) sin(a + b) —

TST 2003, USA

79 Prove that if a,b,c are positive real numbers then,

Va4 + b1 + c+ + Veh? + Pet 2a? > Va3b+ Bet Gat Vab? + be? + ca

KMO Summer Program Test, 2001

80 | Gabriel Dospinescu, Mircea Lascu | For a given n > 2 find the smallest constant k, with the property: if a1, ,@, > 0 have product 1, then

a1 Q2 + a203 fives $ — AanG1 F< ky

(a? +a2)(az +a) (a3 +.a3)(a? + a2) (a2 + a1)(a? +an) — 81 [ Vasile Cirtoaje | For any real numbers a, b,c, x,y, z prove that the inequality holds Ww] 2 az + bụ + cz + v/(a3 + b2 + c2)(2 + 2+ z?) > s(œ+b+e)(z+ +2) Kvant, 1989 82 [ Vasile Cirtoaje | Prove that the sides a, b, c of a triangle satisfy the inequality 3(0 4242-1) 50(2 4622) b ¢ oa a b oe 83 | Walther Janous | Let n > 2 and let 71, 2%2, ,%, > 0 add up to 1 Prove that Crux Mathematicorum

84 | Vasile Cirtoaje, Gheorghe Eckstein | Consider positive real numbers #1,Z2, ,„ such that #+zs #„ = l Prove that 1 + + +®———— <1 m-lt+a, n-1+22 nm—-14+ 2p TST 1999, Romania

85 | Titu Andreescu | Prove that for any nonnegative real numbers a, b,c such

that a? + 6? + c? + abc = 4 we have 0 < ab+ be + ca — abc < 2

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Old and New Inequalities 19

86 | Titu Andreescu | Prove that for any positive real numbers a, b, c the following

inequality holds

OFFS | Save < maz{ (Va — v5)?, (vb Ve)”, (Ve - Va}

TST 2000, USA

87 | Kiran Kedlaya | Let a,b,c be positive real numbers Prove that

a + Vab+ Ÿabc _ af a+b atbte

3 — 2 3 ‘

88 Find the greatest constant & such that for any positive integer n which is not

a square, |(1 + /n) sin(zn)| > È

Vietnamese IMO Training Camp, 1995

89 | Dung Tran Nam ] Let #,,z > 0 such that (x + y+ 2)? = 32zyz Find the ei ty* +24

minimum and maximum of ——————_ (ct+y+z)

Vietnam, 2004

90 | George Tsintifas ] Prove that for any a,b,c,d > 0,

(a+ 6)?(b+ e)3(e+ đ)®(d+ a)? > 16a?b?c2d?(œ + b+ e+ đì!

Crux Mathematicorum

91 [ Titu Andreescu, Gabriel Dospinescu | Find the maxinum value oÊ the ex-

pression

(ab)” | bc)” | (ca)”

l—ab l1-be l—ca

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

(of) (OE i)e(oe£a) (ced ra(oed 1) (ba) ea

95 [ Gabriel Dospinescu |] Let n be an integer greater than 2 Find the greatest real number m, and the least real number Ä⁄„ such that for any positive real numbers £1,02, -,%n (with ty, = %,%n41 = #1), Mn < : < My "Sd +92(nT— 1)#¿ +.” 96 | Vasile Cirtoaje | If x,y, z are positive real numbers, then 1 1 1 9 2 3T 2 zt 2 52 2° #2 + #U +9 2 tụz+z z2 + zz + z (z++z) Gazeta Matematica

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 Matematica

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

(z+b)*+(b+e)*+(e+a)°®> sa) (a1 + b* + €) Vietnam TST, 1996 99 Prove that if a,b,c are positive real numbers such that abc = 1, then 1 1 + 1 < 1 + 1 + 1 l+a+6 lI+ư+c ltet+a724+a 2406 2+c Bulgaria, 1997 1.2 8

100 | Dung Tran Nam | Find the minimum value of the expression ¬ + 3 + :

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Old and New Inequalities 102 Let a,b,c be positive real numbers Prove that (b+e— a)? + (c+ø— b)? (b+ @)2 + a? (c+ a)2 + b2 (a+b)?+c? ~ 5 (a+b—c)? >< 3 Japan, 1997 103 [ Vasile Cirtoaje, Gabriel Dospinescu | Prove that if a1, ae, .,đ„ > 0 then aj + g5 † ++- TÐ da — ngida dạ > (n — Ì) (5m ?ì n—1 )

where a,, is the least among the numbers a1, a2, ,@n

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

ai+yttzt+tt+2eyzt> oe ytyest PP 4Prr testy’ Kvant 105 Prove that for any real numbers a1, a2, ,@,, the following inequality holds he 2 mr oa 1) ` đ¿ | » ———f¡đ/ (> ijai +7—1

106 Prove that 1Í aI,ds, , an, ÐỊ, ., 6, are real numbers between 1001 and

2002, inclusively, such that øŸ + gã + - + a2 = b‡ + bã + - + b2, then we have the

inequality

Tạ (đŸ + đổ + + + đã)

TST Singapore 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”)

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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;)? ¿c8 1<i<j<n

'TST 2004, Romania 111 [| Dung Tran Nam ] Let 71, 22 , Z2004 be real numbers in the interval [—1, 1]

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

112 | Gabriel Dospinescu, Calin Popa | Prove that if n > 2 and a1,a2, ,Gn are real numbers with product 1, then 2n nm—1 - n — l(œ + đa + - + đạ — mì g7 + gã +-:- a2 —Tn > 113 [ Vasile Cirtoaje | If a,b,c are positive real numbers, then 2a + 2b + 2c <3 a+b b+€c cta 114 Prove the following inequality for positive real numbers x, y, z 1 1 1 9 (oy +ye+ en ( To Gea mì >1 Gazeta Matematica Iran, 1996 115 Prove that for any x,y in the interval [0, 1], V1+z2+~+v1+z?+V(-—z)?+(1—g)? > (1+ v5)(1— zp)

116 [ Suranyi | Prove that for any positive real numbers ø,ds, , đ„ the following inequality holds

(n—1)(a? +a? + -+a")+nayaz Gn > (a, +ag+-+-+an)(ap | tant +-+-+ar-t),

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118 | Gabriel Dospinescu | Find the minimum value of the expression » 4142 - Aan - —m wh©r© đ1,đa, ,đ„ < 1 add up to 1 and n > 2 is an integer Th — 119 | Vasile Cirtoaje | Let a1,a2, ,G@, < 1 be nonnegative real numbers such that ai t+az+ +a2 _ V3 a= 4, om “>_— n — 3 Prove that, ay 4 ag 4 4 an > na

l-da la IlTa2 — l1—a2

120 | Vasile Cirtoaje, Mircea Lascu | Let ø, , e, #,,z be positive real numbers such that (ø+b+)œ ++z) = (a2 +? + c?)(g” +2 + z7) =4 Prove that beryz < 3 A0CLY Z —— Ue S 36°

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

constant k,, such that if 71,272, ,2%, > 0 have product 1, then 1 1 1 + + - + — VI+kzạzi V1tkn2ze V1 + kn#„ <n-1 Mathlinks Contest

122 [ Vasile Cirtoaje, Gabriel Dospinescu | For a given n > 2, find the maximal

value of the constant k„ such that for any 21, 22, ,2%n > 0 for which z?+23+ -+

x? = 1 we have the inequality

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CHAPTER 2

Solutions

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1 Prove that the inequality

VEO OP + VRE oP + JPET ae > SX

holds for arbitrary real numbers a, Ư, e

a

Komal

First solution:

Applying Minkowsky’s Inequality to the left-hand side we have

V4 +(1—ð)?+02 +(1— e?+V@+(1—a)® > Jlatb+o24 (3—-a—b_-o

Denoting a+b+c=<2 we get

8\" 9 _ 9 (a+b+c)?+(3-a-—b-c) =2(x-3) +5 > 3

and the conclusion follows

Second solution:

We have the inequalities

Ja? + (1-6)? + fb? + (1-0)? + Ve? + (1-a)? > 5 lal += 8) Wl + ite, lel + [tal — V2 v2 v3 32 and because |x| + |1 — | > 1 for all real numbers z the last quantity is at least > 2 | Dinu Serbanescu | If a,b,c € (0,1) prove that Vøbe+ w(1— a)(1— b)(1— e) < 1 Junior TST 2002, Romania First solution: Observe that x? < #3 for x € (0,1) Thus Vabe < Vabe, and

(—ad—b1—-o < 7% —-a (i _b(1_o

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Old and New Inequalities 27 Summing up, we obtain Vøbe+ w(1— a)(1 — b)(1— e) < as desired a+b+c+l-arl-b+tl-c_¡ 3 —— 3 Second solution: We have

Vabe + V{1— a)(1T— b)(1— e) < Vb-We+ W1—b-WT1—e< 1,

by the Cauchy-Schwarz Inequality Third solution:

Let a = sin?z,b = sin? y,c = sin? z, where z,y,z € (0, 3): The inequality becomes

sinz:siny:sinz+cosx:-cosy:cosz < Ì and it follows from the inequalities

sinz-siny:sinz+cosx-cosy-cosz < sinx-siny +cosz-cosy = cos(x —y) <1

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Solution: Let « be the real root of the equation Using the Cauchy-Schwarz Inequality we infer that (z4 + 2z? + 1)” "eae = because the last inequality is equivalent to (a? — 1)* > 0 a? +b? >

5 Find the maximum value of the expression z° + y? + 2° — 3xyz where x? + y? + z* =1and z,y,z are real numbers

Solution:

Let t= xy + yz + zx Let us observe that

(2? +y? + 23 — 32yz)? = (@ ty +2)? — xy — yz — za)? = (L+2#)(1-— t)Ẻ

1

We also have —5 <t< 1 Thus, we must find the maximum value of the expression

1

(1+ 2t)(1—t)?, where =5 < ‡< 1.In this case we clearly have (1+ 2#) -— 1) <1 t?(3—2t) > 0 and thus |x? +y?+2?—32yz| < 1 We have equality for z = l1, =z= 0

and thus the maximum value is 1

6 Let a,b,c, x,y,z be positive real numbers such that «+ y+ z= 1 Prove that

aœ + bụ + cz +2\/(xy + 0z + z#)(ab + be + ca) <a+b+e

Ukraine, 2001 First solution:

We will use the Cauchy-Schwarz Inequality twice First, we can write ax+by+ cz < Va2 + b2 + cÈ - +2 + 2 + z2 and then we apply again the Cauchy-Schwarz

Inequality to obtain:

ax+by+cz + 2/(ry + yz + z#)(ab + be + ca) <

Vie Sia? + V2ab- V2” zụ < 13 .v.» v.vẻ lA lA Second solution:

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the last one being equivalent to (a — a)? + (y— 6)? +(z—c)? > 0

7 | Darij Grinberg | If a, 6, c are three positive real numbers, then a + b + Cc ` 9 (b+c)? (c+ø)Ÿ (a+b)? 4(atb+e) First solution:

We rewrite the inequality as

ø+5+9{ “ )> (b+eŸ (e+a)Š (a+b)

Applying the Cauchy-Schwarz Inequality we get

(at+b+c) a Cc — , +— > ( —+ ” +— )

(b+ce)? (c+ø) (at+b)?)}/~ \b+e cta artb

It remains to prove that * d>[ a b Cc - - b+e cta a+b 3 > — —9 which is well-known Second solution: Without loss of generality we may assume that a+ b+c = 1 Now, consider x

the functi e function f= (0,1) + (0,00), F(®) = {+ Tmp : (0,1 0 = —-

shows that f is convex Thus, we may apply Jensen’s Inequality and the conclusion A short computation of derivatives

follows

8 | Hojoo Lee | Let a,b,c > 0 Prove that

Jat + a2b2 + bt + M + b?c2 + ct—+ V/e1 + c2a2 + a4 > > av 2a? + be + by 2b? + ca + eV 2c? + ab Gazeta Matematica Solution: We start from (a* — 6”)? > 0 We rewrite it as 4a*+4a7b? +.4b* > 3a*+6a7b? +30" v3

It follows that Vat + a2b2 + b2 > ae + 6°)

Using this observation, we find that

(> a* + a?b? +o) >3 » a?)

But using the Cauchy-Schwarz Inequality we obtain

(> av 2a? + be) < (> a”) (> (2a? + be) ) <3 (> 23)

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9 If a,b,c are positive real numbers such that abc = 2, then a +0 4+ >avb+ce+bVetaticvatb When does equality hold? JBMO 2002 Shortlist First solution: Applying the Cauchy-Schwarz Inequality gives 3(a? +b? +c”) > 3(a + b+)? (1) and (2 +b? +c?) <(a+b+oj(a®+b +c’) (2) These two inequalities combined yield (a? +b? +c?)(at+b+c) 3 (a* +b? +e)[(b+0) + (a+) +(a+8)] 6 (aVb+e+bVa+c+cva +Ù)Ÿ 6 a”-+b”+c7 > (3) Using the AM-GM Inequality we obtain aVb+e+bVa+e+ecVa+b > 3 [abe ( (a+ dO + e)(e + a)) > 3V/abeV8abe = 38 = 6 Thus (aVb+e+b/ate+cevat+b) > 6(avb+e+bVate+evatb) (4) The desired inequality follows from (3) and (4) Second solution: We have

avb +e+bVetateVvatb < /3(43 + b2 + c2)(œ + b+ e)

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