Bsics of Olympid Inequlities Smin Rist
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Introduction The im of this note is to cquint students, who wnt to prticipte in mthemticl Olympids, to Olympid level inequlities from the sics Inequlities re used in ll fields of mthemtics They hve some very interesting properties nd numerous pplictions Inequlities re often hrd to solve, nd it is not lwys possile to find nice solution But it is worth pproching n inequlity rther thn solving it Most inequlities need to e trnsformed into suitle form y lgeric mens efore pplying some theorem This is wht mkes the prolem rther difficult Throughout this little note you will find different wys nd pproches to solve n inequlity Most of the prolems re recent nd thus need fruitful comintion of wisely pplied techniques It took me round two yers to complete this; lthough I didn t work on it for some months during this period I hve tried to demonstrte how one cn use the clssicl inequlities through different exmples tht show different wys of pplying them After lmost ech section there re some exercise prolems for the reder to get his/her hnds dirty! And t the end of ech chpter some hrder prolems re given for those looking for chllenges Some dditionl exercises re given t the end of the ook for the reder to prctice his/her skills Solutions to some selected prolems re given in the lst chpter to present different strtegies nd techniques of solving inequlity prolems In conclusion, I hve tried to explin tht inequlities cn e overcome through prctice nd more prctice Finlly, though this note is imed for students prticipting in the Bngldesh Mthemticl Olympid who will e hoping to e in the Bngldesh IMO tem I hope it will e useful for everyone I m relly grteful to the MthLinks forum for supplying me with the huge collection of prolems Smin Rist 8 Novemer, 008 iii
iv INTRODUCTION
Contents Introduction iii 1 The AM-GM Inequlity 1 11 Generl AM-GM Inequlity 1 1 Weighted AM-GM Inequlity 5 1 More Chllenging Prolems 7 Cuchy-Schwrz nd Hölder s Inequlities 9 1 Cuchy-Schwrz Inequlity 9 Hölder s Inequlity 14 More Chllenging Prolems 17 Rerrngement nd Cheyshev s Inequlities 19 1 Rerrngement Inequlity 19 Cheyshev s Inequlity More Chellenging Prolems 5 4 Other Useful Strtegies 7 41 Schur s Inequlity 7 4 Jensen s Inequlity 7 4 Minkowski s Inequlity 8 44 Rvi Trnsformtion 8 45 Normliztion 9 46 Homogeniztion 9 5 Supplementry Prolems 1 6 Hints nd Solutions to Selected Prolems References 9 v
vi CONTENTS
Chpter 1 The AM-GM Inequlity 11 Generl AM-GM Inequlity The most well-known nd frequently used inequlity is the Arithmetic men-geometric men inequlity or widely known s the AM-GM inequlity The term AM-GM is the comintion of the two terms Arithmetic Men nd Geometric Men The rithmetic men of two numers nd is defined y + Similrly is the geometric men of nd The simplest form of the AM-GM inequlity is the following: Bsic AM-GM Inequlity For positive rel numers, The proof is simple Squring, this ecomes + ( + ) 4, which is equivlent to ( ) 0 This is oviously true Equlity holds if nd only if = Exmple 111 For rel numers,, c prove tht + + c + c + c First Solution By AM-GM inequlity, we hve +, + c c, c + c Adding the three inequlities nd then dividing y we get the desired result Equlity holds if nd only if = = c Second Solution The inequlity is equivlent to ( ) + ( c) + (c ) 0, 1
CHAPTER 1 THE AM-GM INEQUALITY which is oviously true However, the generl AM-GM inequlity is lso true for ny n positive numers Generl AM-GM Inequlity For positive rel numers 1,,, n the following inequlity holds 1 + + + n n with equlity if nd only if 1 = = = n n 1 n, Proof Here we present the well known Cuchy s proof y induction This specil kind of induction is done y performing the following steps: i Bse cse ii P n = P n iii P n = P n 1 Here P n is the sttement tht the AM-GM is true for n vriles Step 1: We lredy proved the inequlity for n = For n = we get the following inequlity: + + c Letting = x, = y, c = z we equivlently get This is true y Exmple 111 nd the identity c x + y + z xyz 0 x + y + z xyz = (x + y + z)(x + y + z xy yz zx) Equlity holds for x = y = z, tht is, = = c Step : Assuming tht P n is true, we hve Now it s not difficult to notice tht 1 + + + n n n 1 n implying P n is true 1 + + + n n n 1 n + n n n+1 n+ n n n 1 n Step : First we ssume tht P n is true ie 1 + + + n n n 1 n As this is true for ll positive i s, we let n = n 1 1 n 1 So now we hve 1 + + + n n n 1 n 1 n 1 1 n 1 = n ( 1 n 1 ) n n 1 = n 1 1 n 1 = n,
11 GENERAL AM-GM INEQUALITY which in turn is equivlent to 1 + + + n 1 n 1 n = n 1 1 n 1 The proof is thus complete It lso follows y the induction tht equlity holds for 1 = = = n Try to understnd yourself why this induction works It cn e useful sometimes Exmple 11 Let 1,,, n e positive rel numers such tht 1 n = 1 Prove tht (1 + 1 )(1 + ) (1 + n ) n Solution By AM-GM, 1 + 1 1, 1 +, 1 + n n Multiplying the ove inequlities nd using the fct 1 n =1 we get our desired result Equlity holds for i = 1, i = 1,,, n Exmple 11 Let,, c e nonnegtive rel numers Prove tht ( + )( + c)(c + ) 8c Solution The inequlity is equivlent to ( + ) ( + c c ) ( c + c ), true y AM-GM Equlity holds if nd only if = = c Exmple 114 Let,, c > 0 Prove tht c + c + c + + c Solution By AM-GM we deduce tht c + + c c c =, c + c + c c =, c c + + = c
4 CHAPTER 1 THE AM-GM INEQUALITY Adding the three inequlities we get which ws wht we wnted c + c + c + ( + + c) ( + + c), Exmple 115 (Smin Rist) Let,, c e positive rel numers Prove tht ( + ) + c( + c) + c(c + ) ( + c)(c + ) Solution By AM-GM, ( + ) + c( + c) + c( + c) = ( + ) + c( + c) + c( + c) + ( + ) + c( + c) + c( + c) = ( + c) + ( + c) + c ( + ) + ( + c + c) + ( + c + c) ( + c) + ( + c) + c ( + ) + ( + c + c) + c = ( + c) + ( + c) + c ( + ) + ( + c) + c( + ) + c( + c) = ( ( + c) + ( + c) ) + ( ( + c) + c( + ) ) + ( c ( + ) + c( + c) ) ( + c)( + c) + c( + c)( + ) + c ( + )( + c) c = ( + c)( + c) + c ( + c)( + ) + c ( + )( + c) c Equlity holds if nd only if = = c Exercise 111 Let, > 0 Prove tht + Exercise 11 For ll rel numers,, c prove the following chin inequlity ( + + c ) ( + + c) ( + c + c) Exercise 11 Let,, c e positive rel numers Prove tht + + c + c + c Exercise 114 Let,, c e positive rel numers Prove tht + + c + + c + c ( + c + c ) Exercise 115 Let,, c e positive rel numers such tht c = 1 Prove tht + + c + + c
1 WEIGHTED AM-GM INEQUALITY 5 Exercise 116 () Let,, c > 0 Show tht ( + + c) ( 1 + 1 + 1 ) 9 c () For positive rel numers 1,,, n prove tht ( 1 ( 1 + + + n ) + 1 + + 1 ) n 1 n Exercise 117 Let,, c e nonnegtive rel numers such tht + + c = Prove tht + + c + + c + c 6 Exercise 118 Let,, c, d > 0 Prove tht + c + c d + d + + c + d 1 Weighted AM-GM Inequlity The weighted version of the AM-GM inequlity follows from the originl AM-GM inequlity Suppose tht 1,,, n re positive rel numers nd let m 1, m,, m n e positive integers Then we hve y AM-GM, 1 + 1 + + 1 }{{} This cn e written s + + + + }{{} m 1 m m 1 + m + + m n 1 1 }{{} 1 }{{} n n n }{{} m 1 m n m + + n + n + + n }{{} m n 1 m 1 +m + +mn m 1 1 + m + + m n n m 1 + m + + m n ( m 1 1 m mn 1 n ) m 1 +m + +mn Or equivlently in symols mi i ( ) 1 m i mi i mi Letting i k = m k mj = m k m 1 + m + + m n for k = 1,,, n we cn rewrite this s follows: Weighted AM-GM Inequlity For positive rel numers 1,,, n nd n weights i 1, i,, i n n such tht i k = 1, we hve k=1 1 i 1 + i + + n i n i 1 1 i in n
6 CHAPTER 1 THE AM-GM INEQUALITY Although we hve proof if i 1, i,, i n re rtionl, this inequlity is lso true if they re positive rel numers The proof, however, is eyond the scope of this note Exmple 11 Let,, c e positive rel numers such tht + + c = Show tht c c 1 Solution Notice tht 1 = + + c + c + c + + c ( c c ) 1 ++c, which implies c c 1 Exmple 1 (Nguyen Mnh Dung) Let,, c > 0 such tht + + c = 1 Prove tht c c + c c + c c 1 Solution From weighted AM-GM, we hve + + c + + c 1 ( c c ) ++c = + + c c c, + c + c + + c 1 ( c c ) ++c = + c + c c c, c + + c + + c 1 ( c c ) ++c = + c + c c c Summing up the three inequlities we get ( + + c) c c + c c + c c Tht is, c c + c c + c c 1 Very few inequlities cn e solved using only the weighted AM-GM inequlity So no exercise in this section
1 MORE CHALLENGING PROBLEMS 7 1 More Chllenging Prolems Exercise 11 Let,, c e positive rel numers such tht c = 1 Prove tht + c + c + + c Exercise 1 (Michel Rozenerg) Let,, c nd d e non-negtive numers such tht ++c+d = 4 Prove tht 4 cd + c + c d + d Exercise 1 (Smin Rist) Let,, c e positive rel numers Prove tht + + c + + c + + c + c + c Exercise 14() (Phm Kim Hung) Let,, c e positive rel numers Prove tht + c + c + c + + c 4 () (Smin Rist) For rel numers,, c > 0 nd n prove tht ( + c + c + n ) c + n + + c Exercise 15 (Smin Rist) Let,, c e positive rel numers such tht + + c = + c + c nd n Prove tht + c + c + n + + c + n
8 CHAPTER 1 THE AM-GM INEQUALITY
Chpter Cuchy-Schwrz nd Hölder s Inequlities 1 Cuchy-Schwrz Inequlity The Cuchy-Schwrz inequlity is very powerful inequlity It is very useful in proving oth lic nd symmetric inequlities The specil equlity cse lso mkes it exceptionl The inequlity sttes: Cuchy-Schwrz Inequlity For ny rel numers 1,,, n nd 1,,, n the following inequlity holds ( 1 + + + ) ( n 1 + + + ) n (1 1 + + + n n ), with equlity if the sequences re proportionl Tht is if 1 1 = = = n n First proof This is the clssicl proof of Cuchy-Schwrz inequlity Consider the qudrtic f(x) = n n n n ( i x i ) = x i x i i + i = Ax + Bx + C i=1 i=1 i=1 i=1 Clerly f(x) 0 for ll rel x Hence if D is the discriminnt of f, we must hve D 0 This implies ( n ) ( n B 4AC i i 4 ) ( n i i i=1 i=1 i=1 ), which is equivlent to ( n ) ( n ) ( n ) i i i i i=1 i=1 i=1 Equlity holds when f(x) = 0 for some x, which hppens if x = 1 1 = = = n n 9
10 CHAPTER CAUCHY-SCHWARZ AND HÖLDER S INEQUALITIES Second Proof By AM-GM, we hve 1 i + 1 i 1 1 ( i ) ( i ), i + i ( i ) ( i ), n + n i i n n ( i ) ( i ) Summing up the ove inequlities, we hve i i ( i ) ( i ), which is equivlent to ( n ) ( n ) ( n ) i i i i i=1 i=1 i=1 1 + + +, which in turn is equiv- n Equlity holds if for ech i {1,,, n}, 1 + + + n lent to 1 = = = n 1 n We could rewrite the ove solution s follows i = i = i 1 + + + + i n 1 + + + n i i ( 1 + + + ) ( n 1 + + + n) Here the sigm nottion denotes lic sum nd it will e used everywhere throughout this note It is recommended tht you get used to the summtion symol Once you get used to it, it mkes your life esier nd sves your time Cuchy-Schwrz in Engel Form For rel numers i,,, n nd 1,,, n > 0 the following inequlity holds: 1 + + + n ( 1 + + + n ), (1) 1 n 1 + + + n with equlity if nd only if 1 1 = = = n n Although this is direct consequence of the Cuchy-Schwrz inequlity, let us prove it in different wy For n = this ecomes x + y Clering out the denomintors, this is equivlent to ( + ) x + y () (y x) 0,
1 CAUCHY-SCHWARZ INEQUALITY 11 which is clerly true For n = we hve from () x + y + c z A similr inductive process shows tht And the cse of equlity esily follows too ( + ) x + y + c z ( + + c) x + y + z 1 1 + + + n n ( 1 + + + n ) 1 + + + n From (1) we deduce nother proof of the Cuchy-Schwrz inequlity Third Proof We wnt to show tht ( i ) ( c i ) ( i c i ) Let i e rel numers such tht i = i c i Then the ove inequlity is equivlent to This is just (1) 1 1 + + + n n ( 1 + + + n ) 1 + + + n Exmple 11 Let,, c e rel numers Show tht ( + + c ) ( + + c) Solution By Cuchy-Schwrz inequlity, (1 + 1 + 1 )( + + c ) (1 + 1 + 1 c) Exmple 1 (Nesitt s Inequlity) For positive rel numers,, c prove tht + c + First Solution Our inequlity is equivlent to or This cn e written s + c + 1 + c + + c + + 1 + c + c + + 1 9, ( 1 ( + + c) + c + 1 c + + 1 ) 9 + ( 1 (x + y + z ) x + 1 y + 1 ) z (1 + 1 + 1), where x = + c, y = c +, z = + Then this is true y Cuchy-Schwrz Second Solution As in the previous solution we need to show tht ( 1 ( + + c) + c + 1 c + + 1 ) 9 +,
1 CHAPTER CAUCHY-SCHWARZ AND HÖLDER S INEQUALITIES which cn e written s + c + c + + + which is true y AM-GM Third Solution We hve So it remins to show tht 1 +c + 1 c+ + 1 + 1 ( + c)(c + )( + ) ( + c)(c + )( + ), + c = ( + + c) + c ( + c + c) ( + + c) ( + c + c) ( ) + ( c) + (c ) 0 Exmple 1 For nonnegtive rel numers x, y, z prove tht x + xy + y + yz + z + zx (x + y + z) Solution By Cuchy-Schwrz inequlity, x(x + y) ( x ) ( (x + y) ) = 4(x + y + z) = (x + y + z) Exmple 14 (IMO 1995) Let,, c e positive rel numers such tht c = 1 Prove tht 1 ( + c) + 1 (c + ) + 1 c ( + ) Solution Let x = 1, y = 1, z = 1 Then y the given condition we otin xyz = 1 Note tht c Now y Cuchy-Schwrz inequlity x 1 ( + c) = 1 ( 1 1 x y + 1 ) = x y + z z y + z where the lst inequlity follows from AM-GM (x + y + z) (x + y + z) = x + y + z Exmple 15 For positive rel numers,, c prove tht + + + c + c c + 1 xyz =, Solution We hve + 1 ( + 1 ) 1 1 + 1 + 1
1 CAUCHY-SCHWARZ INEQUALITY 1 This follows from Cuchy-Schwrz inequlity + = + + c c + c + c + ( + + c) ( + c + c) + + c + = 1 Exmple 16 (Vsile Cirtoje, Smin Rist) Let x, y, z e positive rel numers Prove tht x y z x + y + y + z + z + x Solution Verify tht LHS = x(y + z)(z + x) + y(z + x)(x + y) + z(x + y)(y + z) (x + y)(y + z)(z + x) (x(y + z) + y(z + x) + z(x + y)) (z + x + x + y + y + z)) (x + y)(y + z)(z + x) (xy + yz + zx)(x + y + z) = 4 (x + y)(y + z)(z + x) (x + y)(y + z)(z + x) + xyz = (x + y)(y + z)(z + x) xyz = 1 + (x + y)(y + z)(z + x) 1 + 1 8 =, where the lst inequlity follows from Exmple 1 Here Cuchy-Schwrz ws used in the following form: x + y + cz ( + + c)(x + y + z) Exercise 11 Prove Exmple 111 nd Exercise 116 using Cuchy-Scwrz inequlity Exercise 1 Let,, c, d e positive rel numers Prove tht + c + c + d + c d + + d + Exercise 1 Let 1,,, n e positive rel numers Prove tht 1 + + + n 1 1 + + + n Exercise 14 (Michel Rozenerg) Let,, c, d e positive rel numers such tht + +c +d = 4 Show tht + c + c d + d 4
14 CHAPTER CAUCHY-SCHWARZ AND HÖLDER S INEQUALITIES Exercise 15 Let,, c e positive rel numers Prove tht ( ) ( ) ( ) c + + 1 + + c c + Exercise 16 (Zhutykov Olympid 008) Let,, c e positive rel numers such tht c = 1 Show tht 1 ( + ) + 1 c( + c) + 1 (c + ) Exercise 17 If,, c nd d re positive rel numers such tht + + c + d = 4 prove tht 1 + c + 1 + c d + c 1 + d + d 1 + Exercise 18 Let 1,,, n nd 1,,, n e rel numers Prove tht 1 + 1 + + + + n + n ( 1 + + + n ) + ( 1 + + + n ) Exercise 19 (Smin Rist) Let,, c e the side lengths of tringle Prove tht + c + c + + c c + 1 Exercise 110 (Phm Kim Hung) Let,, c e positive rel numers such tht + + c = 1 Prove tht c + + < + + c c + Exercise 111 Let,, c > 0 Prove tht ( c + c + c + + + + c + c ) Hölder s Inequlity Hölder s inequlity is generliztion of the Cuchy-Schwrz inequlity This inequlity sttes: Hölder s Inequlity Let ij, 1 i m, 1 j n e positive rel numers Then the following inequlity holds m n i=1 j=1 ij n m m j=1 i=1 ij m
HÖLDER S INEQUALITY 15 It looks kind of difficult to understnd So for revity specil cse is the following: for positive rel numers,, c, p, q, r, x, y, z, ( + + c )(p + q + r )(x + y + z ) (qx + qy + crz) Not only Hölder s inequlity is generliztion of Cuchy-Schwrz inequlity, it is lso direct consequence of the AM-GM inequlity, which is demonstrted in the following proof of the specil cse: y AM-GM, = + + c + p p + q + r + x x + y + z px ( + + c )(p + q + r )(x + y + z ), which is equivlent to ( + + c )(p + q + r )(x + y + z ) px + qy + crz Verify tht this proof lso generlizes to the generl inequlity, nd is similr to the one of the Cuchy- Schwrz inequlity Here re some pplictions: Exmple 1 (IMO 001) Let,, c e positive rel numers Prove tht + 8c + + 8c + c c + 8 1 Solution By Hölder s inequlity, ( ) ( ( ) ( + 8c + 8c) + 8c) ( + + c) Thus we need only show tht ( + + c) + + c + 4c, which is equivlent to This is just Exmple 11 ( + )( + c)(c + ) 8c Exmple (Vsile Cirtoje) For,, c > 0 prove tht + + c + + Solution For the left prt, we hve from Hölder s inequlity, ( ) ( ) ( ) ( + ) ( + + c) + + Thus ( + ) + + c
16 CHAPTER CAUCHY-SCHWARZ AND HÖLDER S INEQUALITIES Now for the right prt, y Cuchy-Schwrz inequlity we hve ( ) ( + + c) + + So it remins to show tht which is Exmple 15 + 1, Exmple (Smin Rist) Let,, c e the side lengths of tringle Prove tht 1 8c + ( + c) + 1 8c + ( + c ) + 1 8c + (c + ) 1 c Solution We hve By Hölder s inequlity we otin 1 8c + ( + c) 1 c ( ) 1 8c 1 8c + ( + c) ( + c) 8c + ( + c) 1 8c 1 c ( + c) 8c + ( + c) ( + + c) (4c + ( + c) + ( + c ) + ( + c ) ) = 1 In this solution, the following inequlity ws used: for ll positive rel numers,, c, x, y, z, x + y + c z The proof of this is left to the reder s n exercise ( + + c) (x + y + z) Exmple 4 (IMO Shortlist 004) If,, c re three positive rel numers such tht +c+c = 1, prove tht 1 1 1 + 6 + + 6c + c + 6 1 c Solution Note tht 1 From Hölder s inequlity we hve 7 + c + c + 6 = Hence our inequlity ecomes 1 c(7 + c + c) (c) ( ) ( c(7 + c + c) 9 ) c
MORE CHALLENGING PROBLEMS 17 Hence it remins to show tht 9( + + c) ( + c + c) 1 (c) [c( + + c)] ( + c + c) 4, which is oviously true since ( + c + c) c( + + c) ( c) 0 Another formultion of Hölder s inequlity is the following: for positive rel numers i, i, p, q (1 i n) such tht 1 p + 1 q = 1, 1 1 + + + n n ( p 1 + p + + p n) 1 p ( q 1 + q + + q n) 1 q Exercise 1 Prove Exercise 1 using Hölder s inequlity Exercise Let,, x nd y e positive numers such tht 1 11 + 11 nd 1 x 11 + y 11 Prove tht 1 5 x 6 + 5 y 6 Exercise Prove tht for ll positive rel numers,, c, x, y, z, x + y + c z ( + + c) (x + y + z) Exercise 4 Let,, nd c e positive rel numers Prove the inequlity 6 + c + 6 + c + c6 + c( + + c) Exercise 5 (Kyiv 006) Let x, y nd z e positive numers such tht xy + xz + yz = 1 Prove tht Exercise 6 Let,, c > 0 Prove tht x 1 + 9y xz + y 1 + 9z yx + z 1 + 9x yz + c + c c + + (x + y + z) 18 c c + + c + c More Chllenging Prolems Exercise 1 Let,, c > 0 nd k Prove tht k + + k + c + c kc + k + 1
18 CHAPTER CAUCHY-SCHWARZ AND HÖLDER S INEQUALITIES Exercise (Smin Rist) Let,, c, m, n e positive rel numers Prove tht (m + n) + c(m + nc) + c (mc + n) m + n Another formultion: Let,, c, m, n e positive rel numers such tht c = 1 Prove tht 1 (m + n) + 1 c(m + nc) + 1 (mc + n) m + n Exercise (Michel Rozenerg, Smin Rist) Let x, y, z > 0 Prove tht x + xy + y x + xy Exercise 4 (Vsile Cirtoje nd Smin Rist) Let,, c, k > 0 Prove tht k + + k + c + c k + 1 < kc + k ( + + c) Exercise 5 (Michel Rozenerg nd Smin Rist) Let x, y, z e positive rel numers such tht xy + yz + zx Prove tht x 4x + 5y + y 4y + 5z + z 4z + 5x 1 Exercise 6 Let,, c > 0 such tht + + c = 1 Prove tht + c + + c c + + c c + + c + c 1 c *Exercise 7 (Ji Chen, Phm Vn Thun nd Smin Rist) Let x, y, z e positive rel numers Prove tht 4 7(x + y + z ) 4 x y z 7(yz + zx + xy) + + 4 x + y y + z z + x 4
Chpter Rerrngement nd Cheyshev s Inequlities 1 Rerrngement Inequlity A wonderful inequlity is tht clled the Rerrngement inequlity The sttement of the inequlity is s follows: Rerrngement Inequlity Let ( i ) n i=1 nd ( i) n i=1 e sequences of positive numers incresing or decresing in the sme direction Tht is, either 1 n nd 1 n or 1 n nd 1 n Then for ny permuttion (c n ) of the numers ( n ) we hve the following inequlities n n n i i i c i i n i+1 i=1 i=1 Tht is, the mximum of the sum occurs when the two sequences re similrly sorted, nd the minimum occurs when they re oppositely sorted Proof Let S denote the sum 1 1 + + + n n nd S denote the sum 1 1 + + + x y + + y x + + n n Then S S = x x + y y x y y x = ( x y )( x y ) 0, since oth of x y nd x y re either positive, or negtive, s the sequences re similrly sorted Hence the sum gets smller whenever ny two of the terms lter This implies tht the mximum must occur when the sequences re sorted similrly The other prt of the inequlity follows in quite similr mnner nd is left to the reder A useful technique Let f( 1,,, n ) e symmetric expression in 1,,, n Tht is, for ny permuttion 1,,, n we hve f( 1,,, n ) = f( 1,,, n) Then in order to prove f( 1,,, n ) 0 we my ssume, without loss of generlity, tht 1 n The reson we cn do so is ecuse f remins invrint under ny permuttion of the i s This ssumption is quite useful sometimes; check out the following exmples: Exmple 11 Let,, c e rel numers Prove tht i=1 + + c + c + c 19
0 CHAPTER REARRANGEMENT AND CHEBYSHEV S INEQUALITIES Solution We my ssume, WLOG 1, tht c 0, since the signs of,, c does not ffect the left side of the inequlity Applying the Rerrngement inequlity for the sequences (,, c) nd (,, c) we conclude tht + + c c + c + c = + + c + c + c Exmple 1 For positive rel numers,, c prove tht + + c + c + c Solution WLOG let c Applying Rerrngement inequlity for (,, c ) nd (,, c) we conclude tht + + c c + c + c Exmple 1 (Nesitt s inequlity) For positive rel numers,, c prove tht + c + c + + c + Solution Since the inequlity is symmetric in,, c we my WLOG ssume tht c Then verify 1 tht + c c + + ie +c 1 c+ 1 + Now pplying the Rerrngement inequlity for the ( ) sequences (,, c) nd 1 +c, 1 c+, 1 + we conclude tht nd + c + + c + Adding the ove inequlities we get ( + c + which ws wht we wnted c + + c + + c + + c + c + + c + c + c + c c + + c + + +, + c ) + c + + c + c + c + + + + = Exmple 14 (IMO 1975) We consider two sequences of rel numers x 1 x x n nd y 1 y y n Let z 1, z,, z n e permuttion of the numers y 1, y,, y n Prove tht n (x i y i ) i=1 n (x i z i ) i=1 Solution Expnding nd using the fct tht yi = zi, we re left to prove tht n x i y i i=1 n x i z i, i=1 1 WLOG=Without loss of generlity
1 REARRANGEMENT INEQUALITY 1 which is the Rerrngement inequlity Exmple 15 (Rerrngement inequlity in Exponentil form) Let,, c 1 Prove tht c c c c First Solution First ssume tht c 1 Then which is true, since c nd c c c Now let 1 c Then c c c c c c c c, c c c c c c c c c, which is lso true Hence the inequlity holds in ll cses Second Solution Here is nother useful rgument: tke ln on oth sides ln + ln + c ln c ln + ln c + c ln It is now cler tht this holds y Rerrngement since the sequences (,, c) nd (ln, ln, ln c) re sorted similrly Note tht the inequlity holds even if,, c > 0, nd in this cse the first solution works ut the second solution needs some tretment which is left to the reder to fix Exmple 16 Let,, c 0 Prove tht + + c c Solution Applying the Rerrngement inequlity for (,, c), (,, c), (,, c) we conclude tht + + c c c c + c + c In the sme wy s ove, we cn prove the AM-GM inequlity for n vriles for ny n This demonstrtes how strong the Rerrngement inequlity is Also check out the following exmple, illustrting the strength of Rerrngement inequlity: Exmple 17 (Smin Rist) Let,, c e positive rel numers Prove tht ( ) ( ) ( ) + c + c c + ( Solution Note tht the sequences +c, c+, sorted Therefore y Rerrngement inequlity we get + c = + c c + 1 c + ) ( 1 nd c+, 1 +c, 1 + c + c 1 c+c ) re oppositely
CHAPTER REARRANGEMENT AND CHEBYSHEV S INEQUALITIES Now from Cuchy-Schwrz inequlity 1 + c + c = ( + c) which ws wht we wnted Cn you generlize the ove inequlity? c + ( Exercise 11 Prove Exmple 114 using Rerrngement inequlity Exercise 1 For,, c > 0 prove tht c + c + c + + c ( + c) ) ( ), c + Exercise 1 Let,, c > 0 Prove tht 8 + 8 + c 8 c 1 + 1 + 1 c Exercise 14 Prove Exercise 1 using Rerrngement inequlity Exercise 15 Let,, c e positive rel numers Prove tht ( + c) + c(c + ) + c ( + ) 1 + c + 1 c + + 1 + Exercise 16 (Yroslvle 006) Let > 0, > 0 nd = 1 Prove tht + + + 1 Exercise 17 Let,, c e positive rel numers stisfying c = 1 Prove tht + c + c + + c Exercise 18 Let,, c e positive rel numers such tht + + c = 1 Prove tht + c + + c + + c + c + + c Exercise 19 (Novosiirsk 007) Let nd e positive numers, nd n N Prove tht (n + 1)( n+1 + n+1 ) ( + )( n + n 1 + + n )
CHEBYSHEV S INEQUALITY Cheyshev s Inequlity Cheyshev s inequlity is direct consequence of the Rerngement inqulity The sttement is s follows: Cheyshev s Inequlity Let ( i ) n i=1 nd ( i) n i=1 e two sequences of positive rel numers (i) If the sequences re similrly sorted, then 1 1 + + + n n n (ii) If the sequences re oppositely sorted, then 1 + + + n n 1 + + + n (1) n 1 1 + + + n n n 1 + + + n n 1 + + + n () n Proof We will only prove (1) Since the sequences re similrly sorted, Rerrngement inequlity implies Adding the ove inequlities we get which ws wht we wnted 1 1 + + + n n = 1 1 + + + n n, 1 1 + + + n n 1 + + + n 1, 1 1 + + + n n 1 + 4 + + n, 1 1 + + + n n 1 n + 1 + + n n 1 n( 1 1 + + + n n ) ( 1 + + + n )( 1 + + + n ), Exmple 1 For,, c > 0 prove tht ( + + c ) ( + + c) Solution Applying Cheyshev s inequlity for (,, c) nd (,, c) we conclude tht ( + + c c) ( + + c)( + + c) Exmple Let,, c > 0 Prove tht 8 + 8 + c 8 c 1 + 1 + 1 c Solution From Cheyshev s inequlity we conclude tht ( 8 + 8 + c 8 ) ( 6 + 6 + c 6 )( + + c ) c ( + + c ) c ( + c + c),
4 CHAPTER REARRANGEMENT AND CHEBYSHEV S INEQUALITIES hence 8 + 8 + c 8 c + c + c c = 1 + 1 + 1 c Exmple Let c 0 nd 0 x y z Prove tht x + y + c z + + c xyz ( + + c x + y + z ) () Solution Applying Cheyshev s inequlity for c nd 1 x 1 y 1 z we deduce tht ( x + y + c ) ( 1 ( + + c) z x + 1 y + 1 ) ( ) ( + + c) + + c 9, z xyz x + y + z which ws wht we wnted Here the lst two inequlities follow from AM-GM Exmple 4 Let,, c > 0 Prove tht c + c + c + + c Solution WLOG ssume tht c Then c nd c c Hence using () nd (1) we conclude tht c ( + + c ) + c + c ( + + c)( + + c ) + c + c Exercise 1 Prove the second Cheyshev s inequlity () Exercise Let 1,,, n 0 nd k 1 Prove tht k k 1 + k + + k n 1 + + + n n n + + c Exercise Deduce proof of Nesitt s inequlity from Cheyshev s inequlity (Hint: you my use Exmple ) Exercise 4 Let 1 n nd 1 n e positive Prove tht n i=1 i i 1 + + + n n 1 n n( 1 + + + n ) 1 + + + n Exercise 5 Let c 0 nd d 0 Prove tht ( + + c + d) 8(d + c) Exercise 6 (Rdu Titiu) Let,, c > 0 such tht + + c Show tht + c + c + + c +
MORE CHELLENGING PROBLEMS 5 Exercise 7 Let,, c > 0 such tht c = 1 Prove tht 9 Exercise 8 Let,, c > 0 Prove tht 1 c ( + ) c c (c) ++c c ( + ) More Chellenging Prolems Exercise 1 (Smin Rist) Let,, c > 0 Prove tht mx{,, c} min{,, c} + min{,, c} mx{,, c} + + c c 1 Exercise Let 1 n nd 1 n e positive (i) If (c i ) n i=1 is permuttion of ( i) n i=1 prove tht (ii) Let = 1 + + + n Prove tht n 1 c 1 1 c n cn n 1 1 1 n n 1 Exercise Let x, y, z R + Prove tht x + y + z xyz + xyz x + y + z Exercise 4 (Smin Rist) Let,, c e positive rel numers Prove tht + + c + c + c ( + c + c + + c ) + *Exercise 5 (Smin Rist) Let,, c e positive rel numers nd n e positive integer Prove tht ( ) n ( ) ( ) n 1 n + c n + n 1 c c +
6 CHAPTER REARRANGEMENT AND CHEBYSHEV S INEQUALITIES
Chpter 4 Other Useful Strtegies 41 Schur s Inequlity Let,, c e positive rel numers, nd n e positive Then the following inequlity holds: n ( )( c) + n ( c)( ) + c n (c )(c ) 0, with equlity if nd only if = = c or =, c = 0 nd permuttions The ove inequlity is known s Schur s inequlity, fter Issi Schur Proof Since the inequlity is symmetric in,, c WLOG we my ssume tht c Then the inequlity is equivlent to which is oviously true ( )( n ( c) n ( c)) + c n ( c)( c) 0, 4 Jensen s Inequlity Suppose f is convex function in [, ] Then the inequlity ( ) 1 + + + n f f( 1) + f( ) + + f( n ) n n is true for ll i [, ] Similrly, if f is concve in the intervl the sign of inequlity turns over This is clled Jensen s inequlity The convexity is usully determined y checking if f (x) 0 holds for ll x [, ] concvity one cn check if f (x) 0 for ll x [, ] Here is n exmple: Similrly for Exmple 41 Let,, c > 0 Prove tht ( ) + + c ++c c c Solution Consider the function f(x) = x ln x Verify tht f (x) = 1/x > 0 for ll x R + Thus f is 7
8 CHAPTER 4 OTHER USEFUL STRATEGIES convex in R + nd y Jensen s inequlity we conclude tht ( ) ( + + c + + c f() + f() + f(c) f ln + ln + ln c c ln which is equivlent to which ws wht we wnted 4 Minkowski s Inequlity ( ) + + c ++c ln( c c ) ln, ) ++c Minkowski s inequlity sttes tht for positive numers x i, y i nd p the following inequlity holds:, ( n ) 1 p (x i + y i ) p i=1 ( n i=1 x p i ) 1 ( p n + i=1 y p i ) 1 p 44 Rvi Trnsformtion Suppose tht,, c re the side lengths of tringle Then positive rel numers x, y, z exist such tht = x + y, = y + z nd c = z + x To verify this, let s e the semi-perimeter Then denote z = s, x = s, y = s c nd the conclusion is ovious since s = + c > 0 nd similrly for the others Geometriclly, let D, E, F denote the points of tngency of BC, CA, AB, respectively, with the incircle of tringle ABC Then BD = BF = x, CD = CE = y nd AE = AF = z implies the conclusion Here re some exmples of how the Rvi trnsformtion cn trnsform geometric inequlity into n lgeric one: Exmple 441 (IMO 1964) Let,, c e the side lengths of tringle Prove tht ( + c ) + (c + ) + c ( + c) c First Solution Verify tht the inequlity cn e written s ( + c)( + c )(c + ) c Let = x + y, = y + z nd c = z + x Then the ove inequlity ecomes 8xyz (x + y)(y + z)(z + x), which is Exmple 11 Second Solution The inequlity is equivlent to + + c + c + + c + c + c + c, or, ( )( c) + ( c)( ) + c(c )(c ) 0,
45 NORMALIZATION 9 which is Schur s inequlity Exmple 44 Let,, c e the lengths of the sides of tringle Prove tht ( ) + c + c + c + + c + c + Solution Let x, y, z > 0 such tht = x + y, = y + z, c = z + x Then our inequlity is equivlent to ( ) (x + y)(y + z) x From Cuchy-Schwrz inequlity, 45 Normliztion (x + y)(y + z) (y + zx) y + 4 zx ( ) = x Homogeneous inequlities cn e normlized, eg pplied restrictions with homogeneous expressions in the vriles For exmple, in order to show tht + + c c 0, one my ssume, WLOG, tht c = 1 or + + c = 1 etc The reson is explined elow Suppose tht c = k Let = k, = k, c = kc This implies c = 1, nd our inequlity ecomes + + c c 0, which is the sme s efore Therefore the restriction c = 1 doesn t chnge nything of the inequlity Similrly one might lso ssume + + c = 1 The reder is requested to find out how it works 46 Homogeniztion This is the opposite of Normliztion It is often useful to sustitute = x/y, = y/z, c = z/x, when the condition c = 1 is given Similrly when + + c = 1 we cn sustitute = x/x + y + z, = y/x + y + z, c = z/x + y + z to homogenize the inequlity For n exmple of homogeniztion note tht we cn write the inequlity in exercise 11 in the following form: + c + c + + c c On the other hnd, if we sustitute = x/y, = y/z, c = z/x the inequlity ecomes, zx y + xy z + yz x x y + y z + z x, which clerly looks esier to del with (Hint: Rerrngement) Mny such sustitutions exist, nd the reder is urged to study them nd find them using his/her own ides
0 CHAPTER 4 OTHER USEFUL STRATEGIES
Chpter 5 Supplementry Prolems Exercise 511 Let,, c e nonnegtive rels Prove tht + c + c ( + ) ( + c) (c + ) 8 Exercise 51 For,, c > 0 prove tht ( + c) 4 + (c + ) 4 + c ( + ) 4 ( + )( + c)(c + ) Exercise 51 Let,, c e rel numers Prove tht + (c) + + + c ( + c + c) Exercise 514 (Michel Rozenerg) Let,, c e non-negtive numers such tht + + c = Prove tht + c + c + + c + Exercise 515 For ny cute-ngled tringle ABC show tht tn A + tn B + tn C s r, where s nd r denote the semi-perimeter nd the inrduis, respectively Exercise 516 (Irn 008) Find the smllest rel K such tht for ech x, y, z R + : x y + y z + z x K (x + y)(y + z)(z + x) Exercise 517 (USA 1997) Prove the following inequlity for,, c > 0 1 + + c + 1 + c + c + 1 + c + c 1 c Exercise 518 Let x y z > 0 e rel numers Prove tht x y z + y z x + z x y x + y + z 1
CHAPTER 5 SUPPLEMENTARY PROBLEMS Exercise 519 (Greece 007) Let,, c e sides of tringle Show tht (c + ) 4 ( + c)4 ( + c )4 + + + c + c ( + c) ( + c ) c(c + ) Exercise 5110 (Smin Rist) Let,, c e sides of tringle Show tht (c + ) 4 ( + c)4 ( + c )4 + + ( + c) ( + c ) c(c + ) + + c Exercise 5111 (Crux Mthemticorum) If, nd c re the sidelengths of tringle, then prove the inequlity ( + c) + c + (c + ) + c + ( + ) c + 6 Exercise 511 Let,, c e the side-lengths of tringle Prove tht ( + )( + c) + c 4( + + c) ( + + c)( + c)( + c) Exercise 511 Prove tht if,, c > 0 then c( + + c) + ( + + c) 4 c( + + c) Exercise 5114 Let,, c e non-negtive rel numers such tht + + c = 1 Prove tht + c + c 1 (1 )(1 c) + + c 8 Exercise 5115 Let, nd c e nonnegtive rel numers such tht 1 + 1 + 1 + 1 + 1 c + 1 = Prove tht + c + c Exercise 5116 (Kore 1998) Let I e the incenter of tringle ABC Prove tht (IA + IB + IC ) AB + BC + CA Exercise 5117 (Smin Rist) Let,, c e positive rel numers such tht 6 + 6 + c 6 = Prove tht 7 + 7 c + c 7 Exercise 5118 (Smin Rist) Let x, y, z e positive rel numers Prove tht x y + y z + z x + y y + z z + x x + z x + y Exercise 5119 (Smin Rist) Let x, y, z e positive rel numers Prove tht xy (x + y)(y + z) + yz (y + z)(z + x) + zx (z + x)(x + y)
Chpter 6 Hints nd Solutions to Selected Prolems 11 Expnd nd use Exmple 111 11 + + 11 Use + + c to prove c + c + c + + c c 14 See hint for 11 15 Prove nd use the following: 1 +c = +c 15 Use Exmple 15 6 Solution: Note tht +c c+ Therefore our inequlity is equivlent to By AM-GM, + c + c ( + + c)( + + c ) + c + c (c + )( + ) ( + c)(c + ) = (c+)(+) (+c)(c+) + + c c ( + ) c(c + )( + ) 1 ( + + c)( + )( + c)(c + ) ( + ) c(c + )( + ) ( + )((c + ) + c( + )) = ( + )( + c + c) Now ( + )( + c + c) = ( + c + c) + c + ( + c + c) + c = ( + + c + + c + c)( + c + c) + c( + + c) = ( + + c) ( + c + c) ( + c + c) + c( + + c) = ( + + c) ( + c + c) ( + c + c ) ( + + c) ( + c + c) c( + + c) = ( + + c)( + )( + c)(c + ),
4 CHAPTER 6 HINTS AND SOLUTIONS TO SELECTED PROBLEMS which ws wht we wnted 7 Solution: For the right prt, from Hölder s Inequlity we hve ( ) ( x ) x ( ) x(x + y) (x + y + z) x + y x + y So it remins to show tht ( x (x + y + z) x + y x + y + z + xy + yz + zx (x + y + z) x + y + z + xy + yz + zx ) 7 (yz + zx + xy) 4 Let x + y + z = 1 nd xy + yz + zx = t Thus we need to prove tht ( ) 1 7t 1 t 4 t(1 t) 4 7 (4 t)(1 t) 0, which is ovious using t 1/ Now for the left prt,we need to prove tht x x + y 4 7 4 (x + y + z ) We hve x = ( 4 x + y ( x x + y x x + y x ) ( 4 x x + y ) ( x x + y ) x x + y ) Thus we need to show tht Or, ( ( x x x x + y x x + y ) 7 4 (x + y + z ) ) (x + y + z )
5 But we hve ( ) x x = x + y ( ) x x y + z z + x (x + y)(y + z)(z + x) ( ) (x y + z)( x z + x) = (x + y)(y + z)(z + x) ( x (y + z) ) ( x(z + x)) (x + y)(y + z)(z + x) Let p = x + y + z, q = xy + yz + zx, r = xyz Then it remins to show tht (pq r)(p q) pq r (p q) (p q pq p r + qr) (p q p r pq + qr) p q + p r 4pq p q + pr 4q (x + y + z) (xy + yz + zx) + xyz(x + y + z) 4(xy + yz + zx) (x + y + z )(xy + yz + zx) + (xy + yz + zx) + xyz(x + y + z) 4(xy + yz + zx) (x + y + z )(xy + yz + zx) + xyz(x + y + z) (xy + yz + zx) (x + y + z )(xy + yz + zx) (x y + y z + z x ) + xyz(x + y + z) x y x y sym sym sym xy(x y) 0, which is oviously true 15 c implies / + c /c + c/ + 6 Use Exmple 4 Solution: WLOG ssume tht c This implies + c + + c Therefore + c c + c + On the other hnd, we hve c c Therefore ( ( + c) ) (c + ) c( + ) Applying Cheyshev s inequlity for the similrly sorted sequences +c, c+, c + nd (( + c), (c + ), c( + )) we get which ws wht we wnted ( ) ( ) ( + c) ( + c) + c + c ( + + c ) ( + c + c) + c
6 CHAPTER 6 HINTS AND SOLUTIONS TO SELECTED PROBLEMS 5 Solution: Let x = n +c, y = n c+, z = cn + Then + c = n n ( + c) n = x n ( + c) n 1 = n n 1 x (c + ) n 1 But the sequences ( n n 1 x, n n 1 y, n ( c n 1 z) nd n 1, (c+) n 1 n 1, (+c) n 1 n 1 re oppositely sorted, since the sequences (,, c) nd (x, y, z) re similrly sorted Hence y Rerrngement inequlity we get n n 1 x (c + ) n 1 n n 1 x ( + c) n 1 = n x n 1 n 1 (c + ) n 1 Finlly using Hölder s inequlity + c n x n 1 n 1 (c + ) n 1 n which ws wht we wnted 51 The following stronger inequlity holds: ( x ) ( ) n 1 n 1, c + ( + c) 4 + (c + ) 4 + c ( + ) 4 ( + + c) ( + c + c)( + )( + c)(c + ) You my use Cheyshev s inequlity to prove it (c+c) n 1 ) 51 First Solution: Consider the numers 1, 1, c 1 Two of them must e of the sme sign ie either positive or negtive WLOG suppose tht 1 nd 1 re of the sme sign Then ( 1)( 1) 0 + 1 + (+) Now the inequlity cn e written s c ( + 1) c( + ) + + ( ) 0 Using the ove rgument, we ll e done if we cn show tht Or, which is oviously true c ( + ) c( + ) + + ( ) 0 1 (c + c ) + ( ) 0, Second Solution: WLOG we my ssume tht,, c re positive First we hve the following inequlity + + c + (c) ( + c + c)
7 This follows from Schur nd AM-GM ( ) + ( ) + (c ) + (c) ( ( ) ( ) ) + So we ll e done if we cn show tht Let (c) = t We need to show tht or, which is oviously true + (c) (c) + t t, (t 1) (t + ) 0, 517 Solution: The inequlity is equivlent to Verify tht which is oviously true Hence we conclude tht 511 Solution: The inequlity is equivlent to + + + c + + + c + + + c c( + ) c c( ) 0 + + + c + + + c = ( + )( + c) ( + c )(c + ) 4( + + c) From AM-GM we get ( + )( + c) ( + c )(c + ) ( + )( + c) + c + c + ( + )( + c) = c Therefore it remins to show tht ( + )( + c) c 4( + + c) Since the sequences { 1, 1, 1 c } nd {(c + )( + ), ( + )( + c), ( + c)(c + )} re oppositely sorted, from Rerrngement we get ( + )( + c) c ( + )( + c) = + + c + c Therefore it remins to show tht c + + c,
8 CHAPTER 6 HINTS AND SOLUTIONS TO SELECTED PROBLEMS which follows from Rerrngement c c c = + + c 5115 Solution: Let x = 1 +, y = 1 +, z = 1 from the given condition x + y + z = 1 nd we need to prove tht (1 x)(1 y) x y, or But we hve Hence we re done (y + z x)(z + x y) xy c + Thus = (y + z x)(z + x y) xy ( y + z x Remrk: The inequlity holds even if,, c re rel numers y 1 x x, = 1 y y, c = + z + x y ) = 6 x 1 z z Then
References 1 Phm Kim Hung, Secrets in Inequlities (Volume 1), GIL Pulishing House, 007 Rdmil Buljich Mnfrino, José Antonio Gómez Orteg, Rogelio Vlcez Delgdo, Inequlities, Institudo de Mtemtics, Universidd Ncionl Autónom de México, 005 http://wwwmthlinksro 9