INEQUALITIES. Ozgur Kircak

INEQUALITIES Ozgur Kirk Septemer 27, 2009

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Contents MEANS INEQUALITIES 5. EXERCISES............................. 7 2 CAUCHY-SCHWARZ INEQUALITY 2. EXERCISES............................. REARRANGEMENT INEQUALITY 7. EXERCISES............................. 8 4 CHEBYSHEV S INEQUALITY 2 4. EXERCISES............................. 22 5 MIXED PROBLEMS 25 6 PROBLEMS FROM OLYMPIADS 29 6. Yers 996 2000.......................... 6 6.2 Yers 990 995.......................... 42 6. Supplementry Prolems....................... 44

4 CONTENTS

Chpter MEANS INEQUALITIES Definition Arithmeti men of, 2,..., n is AM = +2+...+n n Definition 2 Geometri men of, 2,..., n is GM = n 2... n Definition Hrmoni men of, 2,..., n is HM = Definition 4 Qudrti men of, 2,..., n is QM = n + +...+ 2 n 2 +2 2 +...+2 n n Definition 5 The rth power men of, 2,..., n is P r = r r + r 2 +...+r n n Theorem Let i R + QM(, 2,..., n ) AM(, 2,..., n ) GM(, 2,..., n ) HM(, 2,..., n ) nd equlity holds if nd only if = 2 =... = n. Theorem 2 Let i R + then P r P r2 whenever r r 2. Exmple Let,, > 0, prove tht 2 + 2 + 2 + +. Solution: By AM-GM inequlity we hve 2 + 2 2. Similrly, 2 + 2 2 nd 2 + 2 2. Adding these three inequlities we get 2( 2 + 2 + 2 ) 2( + + ) = 2 + 2 + 2 + +. Let s link this result sine we will use it in mny prolems. 2 + 2 + 2 + + (.) 5

6 CHAPTER. MEANS INEQUALITIES Exmple 2 Prove tht if,, > 0, then + +. Solution: By AM-GM we hve + + =. So, + +. Exmple Prove tht for ny positive rel numers,, we hve + + + + + 9 2( + + ). Solution: By AM-HM inequlity, we hve + + + + + + + + + + = 2( + + ). Therefore, + + + + + 9 2( + + ). Exmple 4 Prove tht if,, > 0 then + + + + Solution: By the previous exmple we hve, + 2. (.2) ( + + )[ + + + + + ] 9 2 + + + + + + + + 9 + + + 2 + + + + + + + + 9 2 + + + + + 2. This inequlity is lled Nesitt s inequlity. Exmple 5 Prove tht + + < 2. Solution: By AM P we hve, + + 2 ( + ) + ( ) 2 =. Sine the terms re not equl we hve strit inequlity. So, + + < 2.

.. EXERCISES 7. EXERCISES. Prove tht for ny positive rel numers,, we hve ( + 2)( + )( + 6) 48. 2. Prove tht if,, > 0, then ( + )( + )( + ) 8.. Prove tht if,, > 0, then 4. Prove tht if,, > 0, then 5. Prove tht if x, y > 0 then, 2 + 2 + 2. + + + +. x x 4 + y 2 + y x 2 + y 4 xy. 6. Prove tht ( + )( + )( + ) if (),, re sides of tringle (),, re positive rel numers. 7. Prove tht if,, > 0 nd 2 + 2 + 2 =, then + + + + + 2. 8. Prove tht if x, y, z re rel numers with z > 0, then x 2 + y 2 + 2z 2 + 4z x + y +. 9. Prove tht the inequlity ( + + ) 2 2( + ) holds for ny rel numers,,. 0. Prove tht if x, y, z > 0, then () xy + yz + zx x + y + z. () x2 y + y2 2 z + z2 2 x x 2 z + z y + y x. () x2 y + y2 z + z2 x x y z + y z x + z x y. (d) x 4 + y 4 + z 4 xyz( xy + yz + zx).

8 CHAPTER. MEANS INEQUALITIES. Prove tht if x, y > 0, then x + y 4x + 4y. 2. Let,, 0 nd + +. Prove tht + 2 + + 2 + + 2 2 + + + + +.. Prove the inequlity x 4 + y 4 + 8 8xy for positive rel numers x, y. 4. Prove tht if nd re positive rel numers, then ( + )n + ( + )n 2 n+. 5. Prove tht if p, q > 0 nd p + q =, then (p + p )2 + (q + q )2 25 2. 6. Prove tht if x + y + z = then, 8( ( ) 2 xy yz zx) (x + y) 2 + (y + z) 2 + (z + x) 2 9. 7. Prove tht if, 2,..., n re distint positive rel numers nd + 2 +... + n = S, then S S + 8. Find the minimum vlue of S S 2 +... + 2 S S n > n2 n. 2009 + +...+ + 2 + +... + 2009 + +... + 2009 + + 2 +... + 2008 where, 2,..., 2009 > 0 nd + 2 +... + 2009 =. 9. Prove tht for ny x R we hve x 2 +x 4 2. 20. Let x, y. Prove tht x y + y x xy. 2. Prove the inequlity x2 +2 x2 2 for ny x R. + 22. Let x > y > 0 nd xy =. Prove tht x2 +y 2 x y > 2 2. 2. Prove tht if x > y 0, then x + 4 (x y)(y+) 2.

.. EXERCISES 9 24. Prove tht if,, > 0 then + + + +. 25. Prove tht if + + =, then 2 + 2 + 2. 26. Prove tht if + + =, then 2 + 2 + 2 + +. 27. Let,, e positive rel numers suh tht 2 + 2 + 2 =. Prove tht 28. Prove tht the inequlity + + + + + + + +. 2 + 2 + 2 holds for ny rel numers,,. 29. Prove tht if x, y, z > 0, then ( + + )2 + + x 2 y 2 + y2 z 2 + z2 x 2 x y + y z + z x. 0. Prove tht if x + y = 2, then x + y 2.. Let,, e positive rel numers with + + = 24. Prove tht + + 6. 2. Prove tht if,, > 0, then ++d +. Prove tht if +, then 4 + 4 8. 4. Let,, > 0. Prove tht + 2 + + + 2 + ++d + ++d + + + 2 + 5. Prove tht for positive rel numers x, y, z we hve x x + 2y + 2z + 6. Prove tht if,, > 0, then 4 2 2 + y y + 2z + 2x + 4 2 2 +. z z + 2x + 2y 5. 4 2 2 2 + 2 + 2. d ++ 4. 7. Prove the inequlity 2 x + 2 x < x < 2 x 2 x for x. 8. Prove tht if,, > 0 then ( + ) + ( + ) + ( + ) 27 2( + + ) 2.

0 CHAPTER. MEANS INEQUALITIES 9. Prove tht for x, y > 0, ( + x) + 2 ( + y) 2 2 x + y + 2. 40. Prove tht if,, > 0 then 4. Prove tht if,, > 0, then 42. Solve the system in R + + 2 + + + 2 + + + 2 + 2. + + + 6 + +. + + + d = 2 d = 27 + + + d + + d + d. 4. Solve the system in the set of rel numers 44. Solve the system where x, y, z R 4x 2 4x 2 + = y, 4y 2 4y 2 + = z, 4z 2 4z 2 + = x. x + 2 x = 2y, y + 2 y = 2z, z + 2 z = 2x. 45. Find the positive rel numers x, y, z, t stisfying the system 6xyzt = (x 2 + y 2 + z 2 + t 2 )(xyz + xyt + xzt + yzt), 8 = 2xy + 2zt + xz + xt + yz + yt.

Chpter 2 CAUCHY-SCHWARZ INEQUALITY Theorem (Cuhy-Shwrz)If x i, y i re rel numers, then (x 2 + x 2 2 +... + x 2 n)(y 2 + y 2 2 +... + y 2 n) (x y + x 2 y 2 +... + x n y n ) 2 nd equlity holds if nd only if x y = x2 y 2 =... = xn y n. Exmple 6 Prove tht 2 + 2 + 2 (++)2 for ny rel numers,,. Solution: By Cuhy-Shwrz we hve tht ( 2 + 2 + 2 )( 2 + 2 + 2 ) ( + + ) 2 = ( + + ) 2. Therefore, 2 + 2 + 2 (++)2. Exmple 7 Let x, y, z e positive rel numers. Prove tht x 2 x + y + y2 y + z + z2 z + x x + y + z 2 Solution: By Cuhy-Shwrz inequlity we hve [( x + y) 2 + ( y + z) 2 + ( ( ) z + x) 2 x ] ( ) 2 y + ( ) 2 z + ( ) 2 x + y y + z z + x ( x + y x x + y + y + z y y + z + z + x. z z + x ) 2 = (x + y + z) 2 ( ) x So, wht we got is (2x + 2y + 2z) 2 x+y + y2 y+z + z2 z+x x Therefore, 2 x+y + y2 y+z + z2 z+x x+y+z 2. (x + y + z) 2.

2 CHAPTER 2. CAUCHY-SCHWARZ INEQUALITY This is very useful trik tht n e pplied to mny prolems. We n generlize this result s: x 2 y + x2 2 y 2 +... + x2 n y n (x + x 2 +... + x n ) 2 y + y 2 +... + y n (2.) Exmple 8 Let,, > 0. Prove tht + + + + + 2 + 2 + 2 2. Solution: We n write + = 4 +. So y (2.) we hve + = 4 + nd similrly + = 4 + nd LHS = 4 + + y (.) s desired. 4 + + 4 + (2 + 2 + 2 ) 2 2( + + ) 2 + 2 + 2 2

2.. EXERCISES 2. EXERCISES 46. Prove tht the inequlity x 2 +y 2 +z 2 (x+2y+z)2 4 holds for ny x, y, z R. 47. Prove tht if,, > 0, then ( + + )( + + ) 9. 48. Prove tht if x, y > 0 nd x + y = then x 2 + y 2 0. 49.,,, x, y, z re rel numers nd 2 + 2 + 2 = 6, x 2 + y 2 + z 2 = 25 nd x + y + z = 20. Compute ++ x+y+z. 50. Prove tht if, > 0, then + 5. Solve the system where x i R +. 2 + 2. x + x 2 +... + x k = 9 x + x 2 +... + x k = 52. Prove tht if,, re positive rel numers with =, then 2 + + 2 + + 2 + 2. 5. Prove tht if x, y, z > 0 nd x + y + z =, then 8( ( ) 2 xy yz zx) (x + y) 2 + (y + z) 2 + (z + x) 2 9. 54. Prove tht if,, > 0 then ( + ) 2 + ( + ) 2 + ( + ) 2 9 4( + + ). 55. Prove tht if,,, d re positive rel numers, then + + 4 + 6 d 64 + + + d. 56. Let,,, d e positive rel numers. Prove tht ( + )( + d) + d. 57. Let > > 0 nd > > 0. Prove tht ( ) + ( ).

4 CHAPTER 2. CAUCHY-SCHWARZ INEQUALITY 58. Let,, > 0 with + + =. Prove tht 2 + + + 2 + + + 2 + +. 59. Let,, > 0 with =. Prove tht 2 + + + 2 + + + 2 + +. 60. Let,, e positive rel numers suh tht 2 + 2 + 2 =. Prove tht + 2 + + 2 + + 2. 6. Let,,, x, y, z e positive rel numers suh tht x + y + z =. Prove tht x + y + z + 2 (xy + yz + zx)( + + ) + +. 62. Let,, e positive rel numers. Prove tht 6. Prove tht if,,, x, y, z > 0, then 2 + 2 + 2 2 + 2 + 2. x 4 + y4 + z4 (x + y + z)4 ( + + ). 64. Prove tht if,, re positive rel numers, then + 2 + + 2 + + 2. 65. Prove tht if,, re positive rel numers, then + 2 + 66. Prove tht if,, > 0, then 2 + + + 2 + 2 + + + 2. 2 +. 67. Let,,, d e positive rel numers suh tht ( 2 + 2 ) = 2 + d 2. Prove tht + d.

2.. EXERCISES 5 68. Let, e positive rel numers. Prove tht 4 + 4 + 2 + 2 +. 69. Let,, > 0 with =. Prove tht 2 + 2 + + + + 2 + 2 + + + + 2 + 2 + + +. 70. Prove tht if,,, x, y, z > 0 nd ( 2 + 2 + 2 ) = x 2 + y 2 + z 2, then x + y + z. 7. Prove tht if,, re positive rel numers, then 27 2 + ( + )2 2. 72. Let,,, d 0 with + + d + d =. Prove tht + + d + + + d + + + d + d + +. 7. Let, 2,..., n ;, 2,..., n e positive rel numers suh tht + 2 +... + n = + 2 +... + n. Show tht 2 + + 2 2 2 + 2 +... + 74. Let,, > 0. Prove tht ( + ) 2 + ( + ) 2 + 75. Let,, > 0 with + + =. Prove tht 2 n + 2 +... + n n + n 2 ( + ) 2 9 4( 2 + 2 + 2 ). 2 ( + ) + + + 2 ( + ) + + + 2 ( + ) + + 2..

6 CHAPTER 2. CAUCHY-SCHWARZ INEQUALITY

Chpter REARRANGEMENT INEQUALITY Definition 6 Let 2... n nd 2... n. The numer A = + 2 2 +... + n n is lled ordered sum nd the numer B = n + 2 n +... + n is lled reversed sum. And if x, x 2,..., x n is permuttion of, 2,..., n then X = x + 2 x 2 +...+ n x n is lled mixed sum. Theorem 4 Let 2... n nd 2... n e given. For ny mixed sum X we hve B X A. Exmple 9 Prove tht for positive rel numers,, the inequlity 2 + 2 + 2 + + holds. Solution: Sine the inequlity is symmetri, WLOG we n ssume tht. So y Rerrngement inequlity, eing the ordered sum 2 + 2 + 2 = + + + +. Exmple 0 Prove tht if,, re positive rel numers, then + + + +. Solution: WLOG, we n ssume tht. Then nd. By Rerrngement inequlity we hve, LHS = + + + + = + +. 7

8 CHAPTER. REARRANGEMENT INEQUALITY Exmple Prove tht if,, re positive rel numers, then + + + +. Solution: By symmetry ssume tht then nd. By Rerrngement inequlity we hve tht LHS = + + + + = 2 + 2 + 2 And gin y Rerrngement inequlity, eing the reversed sum + + = 2 + 2 + 2 2 + 2 + 2 Comining these two inequlities we get tht. EXERCISES + + + +. 76. Let,, e positive rel numers. Prove tht () + ( + ) () 5 + 5 ( + ) () + + 2 + 2 + 2 (d) + + + + (e) 2 + 2 + 2 + + (f) 2 + 2 2 + 2 2 2 + + (g) ( + + ) 2 + 2 + 2 (h) + + + + + + + + + +. 77. Prove tht if,, re positive rel numers, then + + + + + 2. 78. Let x i R + nd y, y 2,..., y n e permuttion of x, x 2,..., x n. Prove tht x 2 y + x2 2 y 2 +... + x2 n y n x + x 2 +... + x n. 79. Let x, x 2,..., x n e positive rel numers. Prove tht x 2 x 2 + x2 2 x +... + x2 n x x + x 2 +... + x n.

.. EXERCISES 9 80. Let, 2,..., n e distint positive integers. Prove tht 8. Prove tht if for ny,, we hve 2 + + 2 + + then = =. 2 + 2 2 2 +... + n n 2 + 2 +... + n. 2 + 2 + + 2 + + 2 + 82. Let 2 nd 2. Prove tht 2 + + + 2 2 + ( + 2 + )( + 2 + ) 2 + + 2 +

20 CHAPTER. REARRANGEMENT INEQUALITY

Chpter 4 CHEBYSHEV S INEQUALITY Theorem 5 Let 2... n nd 2... n. Then + 2 2 +...+ n n n ( + 2 +...+ n )( + 2 +...+ n ) n + 2 n +...+ n. Exmple 2 Prove tht 2 + 2 + 2 ( + + )2. Solution: By Cheyshev s inequlity, 2 + 2 + 2 = + + ( + + )( + + ) = ( + + )2. Exmple Let,, > 0. Prove tht + + + + + 2. Solution: Due to symmetry, we n ssume tht then + + +. So y Cheyshev s inequlity we hve + + + + And y AM-HM we hve + ( + + )( + + + + + ) (4.) ( + + + + + ) + + + + + = 2( + + ) Comining (4.) nd (4.2) we get, + + + + 2 + 2. (4.2)

22 CHAPTER 4. CHEBYSHEV S INEQUALITY Exmple 4 Prove tht if,, > 0 nd =, then 2 + + 2 + + 2 + 2. Solution: WLOG, suppose tht. Then 2 2 2 nd + + +. So y Cheyshev s inequlity we hve, 2 + + 2 + + 2 + (2 + 2 + 2 )( + + + + + ) By Exmple nd (4.2) we hve tht (2 + 2 + 2 )( + + + + ( + + )2 ) + 4. EXERCISES 8. Prove tht if,, > 0. then () + (+)(2 + 2 ) 2 () n+ + n+ n + n + 2 () 4 + 4 + 4 ( + + ) (d) + + (++) 9 84. Prove tht if,, > 0, then + 2 + 2 + 85. Prove tht if,, > 0, then + + + + 2 + 2 + + + + 86. Prove tht if,, > 0 nd =, then 87. Prove tht if,, > 0, then = + + 2 + 2 + 2 5. + + 7. 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 5. + + + + + + +. 2 2( + + ) = 2

4.. EXERCISES 2 88. Prove tht if,, > 0 with =, then + + + + + + + +. 89. Let,,, d 0 with + + d + d =. Prove tht + + d + + + d + + + d + d 90. Prove tht if,, > 0, then + + + + 9. Prove tht if,, > 0, then + + 2 + 2 + + + + + + + 2 + 2 + 92. Prove tht if,, > 0 nd = then + + + + + + + +. + + + 6 7. + + 2 + 2 6 5. + +. 9. Let x, y, z e positive rel numers with xyz = nd. Prove tht x y + z + y z + x + z x + y 2. 94. Prove tht if,, > 0 with + + =, then 2 + 2 + 95. Prove tht if,, > 0 then 5 + + 96. Let,, > 0. Prove tht 2 + 2 + 5 + + 5 + 2 + 2 2. ( + + )2. 6 + 2 + 2 + + 2 + 2 + + 2 + 2 + +.

24 CHAPTER 4. CHEBYSHEV S INEQUALITY

Chpter 5 MIXED PROBLEMS 97. Prove tht if,, > 0 nd =, then + + + + + + + +. 98. Prove tht if,, > 0 nd =, then 99. Prove tht if x, y, z > 0 then + + + + + + + +. xy x 2 + xy + yz + 00. Prove tht if x, y, z > 0 then xy x 2 + 2y 2 + z 2 + yz y 2 + yz + zx + yz y 2 + 2z 2 + x 2 + 0. Prove tht if,, > 0 nd =, then 5 + 5 + + 5 + 5 + + zx z 2 + zx + xy. zx z 2 + 2x 2 + y 2 2. 5 + 5 +. 02. Let,, e rel numers suh tht 2 + 2 + 2 =. Prove tht 2 + 2 + 2 + 2 + 2 + 2 5. 0. (IMO95/2) Let,, nd e positive rel numers suh tht =. Prove tht ( + ) + ( + ) + ( + ) 2. 25

26 CHAPTER 5. MIXED PROBLEMS 04. Let,, > 0 with + + =. Prove tht 2 + 2 + 2 + 2 + 2 + 2 2. 05. (IMO2000/2) Let,, e positive rel numers with produt. Prove tht ( + )( + )( + ). 06. Prove tht if,, > 0 nd =, then + + + 6 + +. 07. Let,, e positive rel numers suh tht. Prove tht + + + +. 08. Let,, e positive rel numers suh tht =. Prove tht + + + + + + + +. 09. If,, (0, ), then prove tht + ( )( )( ) <. 0. Prove tht 2 + ( ) 2 + 2 + ( ) 2 + 2 + ( ) 2 2 2 for ritrry rel numers,,.. Prove tht ( 2 + 2 ) 2 + + 2 for ny rel numers nd. 2. (Russi2002) Let x, y, z e positive rel numers with sum. Prove tht x + y + z xy + yz + zx.. Let,, e positive rel numers. Prove tht 4. Prove tht + + 2 + + + 2 + + + 2 + +. 4 2 + + 2 + 2 + + 2 + 2 + + 2 + + for positive rel numers,,.

27 5. Prove tht if,, > 0, then 2 + 2 + 2 + 2 + ( + + ) 2. + 2 + + 6. (IMO200/2)Let,, e positive rel numers. Prove tht 2 + 8 + 2 + 8 + 2 + 8. 7. Prove tht if,, re positive, then + 2 + 2 + + 2 + 2 + + 2 + 2 + +. 8. Prove tht for positive rel numers,,, d the inequlity holds + d ( + + )( + + d). 9. Let x, y, z e positive rels with x + y + z = nd let = x 2 + xy + y 2, = y 2 + yz + z 2, = z 2 + zx + x 2. Prove tht + +. 20. Let,, > 0. Prove tht + + + + + 9 + + + ( + + ). 2. Let,, > 0 with =. Prove tht + + + 2 + + + + + 2 + + + + ( + )( + )( + ) + + 2. + + + 22. If x, y > 0 nd x 2 + y x + y 4. Prove tht x + y 2. 2. Let,, e positive rel numers. Prove tht + + + 2(2 + 2 + 2 ). 24. Prove tht if,, re positive rel numers with + + =, then 5 4 + 4 + 5 4 + 4 + 25. Let,, > 0 with =. Prove tht 2 + 2 + 2 + 2 + 5 4 + 4 2. 2 + 2.

28 CHAPTER 5. MIXED PROBLEMS 26. (Mthemtil Exliur) Let x, y, z >. Prove tht x 4 (y ) 2 + y 4 (z ) 2 + z 4 (x ) 2 48. 27. (MMO/2009) Let,, > 0 suh tht + + =. Prove tht 2 + + 2 + + 28. Prove tht x, y, z R + we hve: x 5 29. Let x, y, z > 0. Prove tht 0. Prove tht if,, > 0, then 2 + + +. x yz + yz 4 2. x y(x 2 + 2y 2 ) + y z(y 2 + 2z 2 ) + z x(z 2 + 2x 2 ). 2 + + + 2 + + +. Prove tht if,, > 0 with =, then + + 2 + 2 +. 2 + +. 2. Prove tht if x, y, z > 0, then x 2 z 2 + y 4 + y 2 z 2 yz(xz + y 2 + yz).. Let,, > 0 with ( + )( + )( + ) = 8. Prove tht + + 4. Let x, y, z e positive rels. Prove tht yz 2x 2 + yz + 27 + +. zx 2y 2 + zx + 5. Let x, y, z e positive rels. Prove tht 6. Let,, x, y, z R +. Prove tht xy 2z 2 + xy. x 2 2x 2 + yz + y 2 2y 2 + zx + z 2 2z 2 + xy. x y + z + y z + x + z x + y +.

Chpter 6 PROBLEMS FROM OLYMPIADS (From Inequlities Through Prolems -Hojoo Lee) (BMO 2005, Proposed y Seri nd Montenegro) (,, > 0) 2 + 2 + 2 + + + 4( )2 + + 2 (Romni 2005, Cezr Lupu) (,, > 0) + 2 + + 2 + + 2 + + (Romni 2005, Trin Tmin) (,, > 0) + 2 + d + + 2d + + d + 2 + + d + 2 + 4 (Romni 2005, Cezr Lupu) ( + + + +,,, > 0) + + 5 (Romni 2005, Cezr Lupu) ( = ( + )( + )( + ),,, > 0) + + 4 29

0 CHAPTER 6. PROBLEMS FROM OLYMPIADS 6 (Romni 2005, Roert Szsz) ( + + =,,, > 0) 2 2 2 ( 2)( 2)( 2) 7 (Romni 2005) (,,, > 0) + + + + + + + + 8 (Romni 2005, Unused) ( =,,, > 0) 2 ( + ) + 2 ( + ) + 2 ( + ) 2 9 (Romni 2005, Unused) ( + + + +,,, > 0) ( + ) + ( + ) + ( + ) 2 0 (Romni 2005, Unused) ( + + =,,, > 0) + + + + + 2 (Romni 2005, Unused) ( + + + 2 =,,, > 0) + + 2 2 (Chzeh nd Solvk 2005) ( =,,, > 0) ( + )( + ) + ( + )( + ) + ( + )( + ) 4 (Jpn 2005) ( + + =,,, > 0) ( + ) + ( + ) + ( + )

4 (Germny 2005) ( + + =,,, > 0) ( 2 + + ) + + + + + 5 (Vietnm 2005) (,, > 0) ( ) ( ) ( ) + + + + + 8 6 (Chin 2005) ( + + =,,, > 0) 0( + + ) 9( 5 + 5 + 5 ) 7 (Chin 2005) (d =,,,, d > 0) ( + ) 2 + ( + ) 2 + ( + ) 2 + ( + d) 2 8 (Chin 2005) ( + + =,,, 0) 2 + + 2 + + 2 + 9 (Polnd 2005) (0,, ) + + + + + 2 20 (Polnd 2005) ( + + =,,, > 0) + + + 6 9 2 (Blti Wy 2005) ( =,,, > 0) 2 + 2 + 2 + 2 + 2 + 2

2 CHAPTER 6. PROBLEMS FROM OLYMPIADS 22 (Seri nd Montenegro 2005) (,, > 0) + + + + + ( + + ) 2 2 (Seri nd Montenegro 2005) ( + + =,,, > 0) + + + + 24 (Bosni nd Heregovin 2005) ( + + =,,, > 0) + + 25 (Irn 2005) (,, > 0) ( + + ) 2 ( ( + + ) + + ) 26 (Austri 2005) (,,, d > 0) + + + d + + + d d 27 (Moldov 2005) ( 4 + 4 + 4 =,,, > 0) 4 + 4 + 4 28 (APMO 2005) ( = 8,,, > 0) 2 ( + )( + ) + 2 ( + )( + ) + 2 ( + )( + ) 4 29 (IMO 2005) (xyz, x, y, z > 0) x 5 x 2 x 5 + y 2 + z 2 + y5 y 2 y 5 + z 2 + x 2 + z5 z 2 z 5 + x 2 + y 2 0

0 (Polnd 2004) ( + + = 0,,, R) 2 2 + 2 2 + 2 2 + 6 (Blti Wy 2004) ( =,,, > 0, n N) n + n + + n + n + + n + n + 2 (Junior Blkn 2004) ((x, y) R 2 {(0, 0)}) 2 2 x 2 + y 2 x + y x 2 xy + y 2 (IMO Short List 2004) ( + + =,,, > 0) + 6 + + 6 + + 6 4 (APMO 2004) (,, > 0) ( 2 + 2)( 2 + 2)( 2 + 2) 9( + + ) 5 (USA 2004) (,, > 0) ( 5 2 + )( 5 2 + )( 5 2 + ) ( + + ) 6 (Junior BMO 200) (x, y, z > ) + x 2 + y + z 2 + + y2 + z + x 2 + + z2 + x + y 2 2 7 (USA 200) (,, > 0) (2 + + ) 2 (2 + + )2 (2 + + )2 2 2 + + ( + ) 2 2 2 + + ( + ) 2 2 2 + ( + ) 2 8

4 CHAPTER 6. PROBLEMS FROM OLYMPIADS 8 (Russi 2002) (x + y + z =, x, y, z > 0) x + y + z xy + yz + zx ( ) 9 (Ltvi 2002) + + 4 + + 4 + + 4 +d =,,,, d > 0 4 d 40 (Alni 2002) (,, > 0) + ( (2 + 2 + 2 ) + + ) + + + 2 + 2 + 2 4 (Belrus 2002) (,,, d > 0) ( + )2 + ( + d) 2 2 d + ( + )2 + ( + d) 2 + 2 + 2 + d 2 ( + ) 2 + ( + d) 2 2 42 (Cnd 2002) (,, > 0) + + + + 4 (Vietnm 2002, Dung Trn Nm) ( 2 + 2 + 2 = 9,,, R) 2( + + ) 0 44 (Bosni nd Heregovin 2002) ( 2 + 2 + 2 =,,, R) 2 + 2 + 2 + 2 + 2 + 2 5 45 (Junior BMO 2002) (,, > 0) ( + ) + ( + ) + ( + ) 27 2( + + ) 2

5 46 (Greee 2002) ( 2 + 2 + 2 =,,, > 0) 2 + + 2 + + 2 + ( + + ) 2 4 47 (Greee 2002) ( 0, 2 0,,, R) 0( 2 + 2 + 2 ) 2 + 5 48 (Tiwn 2002) (,,, d ( 0, 2 ]) d ( )( )( )( d) 4 + 4 + 4 + d 4 ( ) 4 + ( ) 4 + ( ) 4 + ( d) 4 49 (APMO 2002) ( x + y + z =, x, y, z > 0) x + yz + y + zx + z + xy xyz + x + y + z 50 (Irelnd 200) (x + y = 2, x, y 0) x 2 y 2 (x 2 + y 2 ) 2. 5 (BMO 200) ( + +,,, 0) 2 + 2 + 2 52 (USA 200) ( 2 + 2 + 2 + = 4,,, 0) 0 + + 2 5 (Columi 200) (x, y R) (x + y + ) 2 + xy

6 CHAPTER 6. PROBLEMS FROM OLYMPIADS 54 (KMO Winter Progrm Test 200) (,, > 0) (2 + 2 + 2 ) ( 2 + 2 + 2 ) + ( + ) ( + ) ( + ) 55 (KMO Summer Progrm Test 200) (,, > 0) 4 + 4 + 4 + 2 2 + 2 2 + 2 2 + + + + + 56 (IMO 200) (,, > 0) 2 + 8 + 2 + 8 + 2 + 8 6. Yers 996 2000 57 (IMO 2000, Titu Andreesu) ( =,,, > 0) ( + ) ( + ) ( + ) 58 (Czeh nd Slovki 2000) (, > 0) ( 2( + ) + ) + 59 (Hong Kong 2000) ( =,,, > 0) + 2 + + 2 + + 2 8 + + 60 (Czeh Repuli 2000) (m, n N, x [0, ]) ( x n ) m + ( ( x) m ) n

6.. YEARS 996 2000 7 6 (Medoni 2000) (x, y, z > 0) x 2 + y 2 + z 2 2 (xy + yz) 62 (Russi 999) (,, > 0) 2 + 2 2 + 2 + 2 + 2 2 + 2 + 2 + 2 2 + 2 > 6 (Belrus 999) ( 2 + 2 + 2 =,,, > 0) + + + + + 2 64 (Czeh-Slovk Mth 999) (,, > 0) + 2 + + 2 + + 2 65 (Moldov 999) (,, > 0) ( + ) + ( + ) + ( + ) + + + + + 66 (United Kingdom 999) (p + q + r =, p, q, r > 0) 7(pq + qr + rp) 2 + 9pqr 67 (Cnd 999) (x + y + z =, x, y, z 0) x 2 y + y 2 z + z 2 x 4 27 68 (Proposed for 999 USAMO, [AB, pp.25]) (x, y, z > ) x x2 +2yz y y2 +2zx z z2 +2xy (xyz) xy+yz+zx

8 CHAPTER 6. PROBLEMS FROM OLYMPIADS 69 (Turkey, 999) ( 0) ( + )( + 4)( + 2) 60 70 (Medoni 999) ( 2 + 2 + 2 =,,, > 0) + + + 4 7 (Polnd 999) ( + + =,,, > 0) 2 + 2 + 2 + 2 72 (Cnd 999) (x + y + z =, x, y, z 0) x 2 y + y 2 z + z 2 x 4 27 7 (Irn 998) ( x + y + z = 2, x, y, z > ) x + y + z x + y + z 74 (Belrus 998, I. Gorodnin) (,, > 0) + + + + + + + + 75 (APMO 998) (,, > 0) ( + ) ( + ) ( + ) 2 ( + + + ) 76 (Polnd 998) ( + + + d + e + f =, e + df 08,,, d, e, f > 0) + d + de + def + ef + f 6

6.. YEARS 996 2000 9 77 (Kore 998) (x + y + z = xyz, x, y, z > 0) + + + x 2 + y 2 + z 2 2 78 (Hong Kong 998) (,, ) + + ( + ) 79 (IMO Short List 998) (xyz =, x, y, z > 0) x ( + y)( + z) + y ( + z)( + x) + z ( + x)( + y) 4 80 (Belrus 997) (, x, y, z > 0) + y + x x + + z + x y + + x + y z x + y + z + z + z x + + x + y y + + y + z z 8 (Irelnd 997) ( + +,,, 0) 2 + 2 + 2 82 (Irn 997) (x x 2 x x 4 =, x, x 2, x, x 4 > 0) ) x + x 2 + x + x 4 mx (x + x 2 + x + x 4, x + x2 + x + x4 8 (Hong Kong 997) (x, y, z > 0) + 9 xyz(x + y + z + x 2 + y 2 + z 2 ) (x 2 + y 2 + z 2 )(xy + yz + zx) 84 (Belrus 997) (,, > 0) + + + + + + + + + +

40 CHAPTER 6. PROBLEMS FROM OLYMPIADS 85 (Bulgri 997) ( =,,, > 0) + + + + + + + + 2 + + 2 + + 2 + 86 (Romni 997) (xyz =, x, y, z > 0) x 9 + y 9 x 6 + x y + y 6 + y 9 + z 9 y 6 + y z + z 6 + z 9 + x 9 z 6 + z x + x 6 2 87 (Romni 997) (,, > 0) 2 2 + 2 + 2 2 + 2 + 2 2 + 2 2 + 2 + 2 + 2 + 2 + 2 88 (USA 997) (,, > 0) + + + + + + + +. 89 (Jpn 997) (,, > 0) ( + ) 2 ( + )2 ( + )2 ( + ) 2 + + 2 ( + ) 2 + + 2 ( + ) 2 + 2 5 90 (Estoni 997) (x, y R) x 2 + y 2 + > x y 2 + + y x 2 + 9 (APMC 996) (x + y + z + t = 0, x 2 + y 2 + z 2 + t 2 =, x, y, z, t R) xy + yz + zt + tx 0 92 (Spin 996) (,, > 0) 2 + 2 + 2 ( )( )

6.. YEARS 996 2000 4 9 (IMO Short List 996) ( =,,, > 0) 5 + 5 + + 5 + 5 + + 5 + 5 + 94 (Polnd 996) ( + + =,,, 4 2 + + 2 + + ) 2 + 9 0 95 (Hungry 996) ( + =,, > 0) 2 + + 2 + 96 (Vietnm 996) (,, R) ( + ) 4 + ( + ) 4 + ( + ) 4 4 7 ( 4 + 4 + 4) 97 (Berus 996) (x + y + z = xyz, x, y, z > 0) xy + yz + zx 9(x + y + z) 98 (Irn 996) (,, > 0) ( ) ( + + ) ( + ) 2 + ( + ) 2 + ( + ) 2 9 4 99 (Vietnm 996) (2( + + d + + d + d) + + d + d + d = 6,,,, d 0) + + + d 2 ( + + d + + d + d)

42 CHAPTER 6. PROBLEMS FROM OLYMPIADS 6.2 Yers 990 995 Any good ide n e stted in fifty words or less. S. M. Ulm 00 (Blti Wy 995) (,,, d > 0) + + + + d + + + + d + d + d + 4 0 (Cnd 995) (,, > 0) ++ 02 (IMO 995, Nzr Agkhnov) ( =,,, > 0) ( + ) + ( + ) + ( + ) 2 0 (Russi 995) (x, y > 0) xy x x 4 + y 2 + y y 4 + x 2 04 (Medoni 995) (,, > 0) + + + + + 2 05 (APMC 995) (m, n N, x, y > 0) (n )(m )(x n+m +y n+m )+(n+m )(x n y m +x m y n ) nm(x n+m y+xy n+m ) 06 (Hong Kong 994) (xy + yz + zx =, x, y, z > 0) x( y 2 )( z 2 ) + y( z 2 )( x 2 ) + z( x 2 )( y 2 ) 4 9

6.2. YEARS 990 995 4 07 (IMO Short List 99) (,,, d > 0) + 2 + d + + 2d + + 08 (APMC 99) (, 0) ( ) 2 + + 2 + 2 + 2 4 d + 2 + + + + d + 2 + 2 ( 2 + 2 2 ) 09 (Polnd 99) (x, y, u, v > 0) xy + xv + uy + uv x + y + u + v xy x + y + uv u + v 0 (IMO Short List 99) ( + + + d =,,,, d > 0) (Itly 99) (0,, ) + d + d + d 27 + 76 27 d 2 + 2 + 2 2 + 2 + 2 + 2 (Polnd 992) (,, R) ( + ) 2 ( + ) 2 ( + ) 2 ( 2 + 2 2 )( 2 + 2 2 )( 2 + 2 2 ) (Vietnm 99) (x y z > 0) x 2 y z + y2 z x + z2 x y x2 + y 2 + z 2 4 (Polnd 99) (x 2 + y 2 + z 2 = 2, x, y, z R) x + y + z 2 + xyz 5 (Mongoli 99) ( 2 + 2 + 2 = 2,,, R) + + 2 2 6 (IMO Short List 990) ( + + d + d =,,,, d > 0) + + d + + d + + d + + + d + +

44 CHAPTER 6. PROBLEMS FROM OLYMPIADS 6. Supplementry Prolems 7 (Lithuni 987) (x, y, z > 0) x x 2 + xy + y 2 + y y 2 + yz + z 2 + z z 2 + zx + x 2 x + y + z 8 (Yugoslvi 987) (, > 0) 2 ( + )2 + 4 ( + ) + 9 (Yugoslvi 984) (,,, d > 0) + + + d + d + + d + 2 20 (IMO 984) (x + y + z =, x, y, z 0) 0 xy + yz + zx 2xyz 7 27 2 (USA 980) (,, [0, ]) + + + + + + + ( )( )( ). + + 22 (USA 979) (x + y + z =, x, y, z > 0) x + y + z + 6xyz 4. 2 (IMO 974) (,,, d > 0) < + + d + + + + + + d + d + + d < 2

6.. SUPPLEMENTARY PROBLEMS 45 24 (IMO 968) (x, x 2 > 0, y, y 2, z, z 2 R, x y > z 2, x 2 y 2 > z 2 2 ) x y z 2 + x 2 y 2 z 2 8 2 (x + x 2 )(y + y 2 ) (z + z 2 ) 2 25 (Nesitt s inequlity) (,, > 0) + + + + + 2 26 (Poly s inequlity) (,, > 0) ( 2 + + ) ln ln 2 27 (Klmkin s inequlity) ( < x, y, z < ) ( x)( y)( z) + ( + x)( + y)( + z) 2 28 (Crlson s inequlity) (,, > 0) ( + )( + )( + ) + + 8 29 ([ONI], Vsile Cirtoje) (,, > 0) ( + ) ( + ) + ( + ) ( + ) + ( + ) ( + ) 0 ([ONI], Vsile Cirtoje) (,,, d > 0) + + + d + d d + + d + 0

46 CHAPTER 6. PROBLEMS FROM OLYMPIADS (Elemente der Mthemtik, Prolem 207, Sefket Arslngić) (x, y, z > 0) x y + y z + z x x + y + z xyz 2 ( W URZEL, Wlther Jnous) (x + y + z =, x, y, z > 0) ( + x)( + y)( + z) ( x 2 ) 2 + ( y 2 ) 2 + ( z 2 ) 2 ( W URZEL, Heinz-Jürgen Seiffert) (xy > 0, x, y R) 2xy x + y + x2 + y 2 xy + x + y 2 2 4 ( W URZEL, Šefket Arslngić) (,, > 0) x + y + ( + + ) z (x + y + z) 5 ( W URZEL, Šefket Arslngić) ( =,,, > 0) 2 ( + ) + 2 ( + ) + 2 ( + ) 2. 6 ( W URZEL, Peter Strek, Donuwörth) ( =,,, > 0) + + ( + ) ( + ) ( + ). 2 7 ( W URZEL, Peter Strek, Donuwörth) (x+y+z =, x 2 +y 2 +z 2 = 7, x, y, z > 0) + 6 xyz ( x z + y x + z ) y

6.. SUPPLEMENTARY PROBLEMS 47 8 ( W URZEL, Šefket Arslngić) (,, > 0) + + + + ( + + ) + + + +. 9 (, 0) ( + ) + ( + ) 2 ( + 2 ) + 2 ( + 2 ) 40 (Ltvi 997) (n N,,, > 0) + + + 2 + + + n < n ( + n) 4 ([ONI], Griel Dospinesu, Mire Lsu, Mrin Tetiv) (,, > 0) 2 + 2 + 2 + 2 + ( + )( + )( + ) 42 (Gzet Mtemtiã) (,, > 0) 4 + 2 2 + 4 + 4 + 2 2 + 4 + 4 + 2 2 + 4 2 2 + + 2 2 + + 2 2 + 4 (C 262, Mohmmed Assil) (,, > 0) + + + + + + 44 (C2580) (,, > 0) + + + 2 + + + 2 + + + 2 + 45 (C258) (,, > 0) 2 + + CRUX with MAYHEM + 2 + + + 2 + + + +

48 CHAPTER 6. PROBLEMS FROM OLYMPIADS 46 (C252) ( 2 + 2 + 2 =,,, > 0) 2 + 2 + 2 + 2( + + ) 47 (C02, Vsile Cirtoje) ( 2 + 2 + 2 =,,, > 0) + + 9 2 48 (C2645) (,, > 0) 2( + + ) + 9( + + )2 ( 2 + 2 + 2 ) 49 (x, y R) 2 (x + y)( xy) ( + x 2 )( + y 2 ) 2 50 (0 < x, y < ) x y + y x > 5 (x, y, z > 0) xyz + x y + y z + z x x + y + z 52 (,,, x, y, z > 0) ( + x)( + y)( + z) + xyz 5 (x, y, z > 0) x x + (x + y)(x + z) + y y + (y + z)(y + x) + z z + (z + x)(z + y)

6.. SUPPLEMENTARY PROBLEMS 49 54 (x + y + z =, x, y, z > 0) x x + y y + z z 2 55 (,, R) 2 + ( ) 2 + 2 + ( ) 2 + 2 + ( ) 2 2 2 56 (,, > 0) 2 + 2 + 2 + 2 2 + + 2 57 (xy + yz + zx =, x, y, z > 0) x + x 2 + y + y 2 + z + z 2 2x( x2 ) ( + x 2 ) 2 + 2y( y2 ) ( + y 2 ) 2 + 2z( z2 ) ( + z 2 ) 2 58 (x, y, z 0) xyz (y + z x)(z + x y)(x + y z) 59 (,, > 0) ( + ) + ( + ) + ( + ) 4 + ( + )( + )( + ) 60 (Drij Grinerg) (x, y, z 0) ( x (y + z) + y (z + x) + z (x + y) ) x + y + z 2 (y + z) (z + x) (x + y). 6 (Drij Grinerg) (x, y, z > 0) y + z z + x x + y + + x y z 4 (x + y + z) (y + z) (z + x) (x + y).

50 CHAPTER 6. PROBLEMS FROM OLYMPIADS 62 (Drij Grinerg) (,, > 0) 2 ( + ) ( 2 + 2 ) (2 + + ) + 2 ( + ) ( 2 + 2 ) (2 + + ) + 2 ( + ) ( 2 + 2 ) (2 + + ) > 2. 6 (Drij Grinerg) (,, > 0) 2 2 2 + ( + ) 2 + 2 2 2 + ( + ) 2 + 2 2 2 + ( + ) 2 < 2. 64 (Vsile Cirtoje) (,, R) ( 2 + 2 + 2 ) 2 ( + + )