INEQUALITIES. Ozgur Kircak

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1 INEQUALITIES Ozgur Kirk Septemer 27, 2009

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3 Contents MEANS INEQUALITIES 5. EXERCISES CAUCHY-SCHWARZ INEQUALITY 2. EXERCISES REARRANGEMENT INEQUALITY 7. EXERCISES CHEBYSHEV S INEQUALITY 2 4. EXERCISES MIXED PROBLEMS 25 6 PROBLEMS FROM OLYMPIADS Yers Yers Supplementry Prolems

4 4 CONTENTS

5 Chpter MEANS INEQUALITIES Definition Arithmeti men of, 2,..., n is AM = 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 n n n Definition 5 The rth power men of, 2,..., n is P r = r r + r 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 Solution: By AM-GM inequlity we hve Similrly, nd Adding these three inequlities we get 2( ) 2( + + ) = Let s link this result sine we will use it in mny prolems (.) 5

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

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

9 .. EXERCISES Prove tht if,, > 0 then Prove tht if + + =, then Prove tht if + + =, then Let,, e positive rel numers suh tht =. Prove tht 28. Prove tht the inequlity 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 Prove tht if,, > 0, then ++d +. Prove tht if +, then Let,, > 0. Prove tht d + ++d Prove tht for positive rel numers x, y, z we hve x x + 2y + 2z + 6. Prove tht if,, > 0, then y y + 2z + 2x z z + 2x + 2y d Prove the inequlity 2 x + 2 x < x < 2 x 2 x for x. 8. Prove tht if,, > 0 then ( + ) + ( + ) + ( + ) 27 2( + + ) 2.

10 0 CHAPTER. MEANS INEQUALITIES 9. Prove tht for x, y > 0, ( + x) + 2 ( + y) 2 2 x + y Prove tht if,, > 0 then 4. Prove tht if,, > 0, then 42. Solve the system in R d = 2 d = 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.

11 Chpter 2 CAUCHY-SCHWARZ INEQUALITY Theorem (Cuhy-Shwrz)If x i, y i re rel numers, then (x 2 + x x 2 n)(y 2 + y y 2 n) (x y + x 2 y 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 for ny rel numers,,. Solution: By Cuhy-Shwrz we hve tht ( )( ) ( + + ) 2 = ( + + ) 2. Therefore, (++)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.

12 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 x2 n y n (x + x x n ) 2 y + y y n (2.) Exmple 8 Let,, > 0. Prove tht Solution: We n write + = 4 +. So y (2.) we hve + = 4 + nd similrly + = 4 + nd LHS = y (.) s desired ( ) 2 2( + + )

13 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 ( + + )( + + ) Prove tht if x, y > 0 nd x + y = then x 2 + y ,,, x, y, z re rel numers nd = 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 x + x x k = 9 x + x x k = 52. Prove tht if,, re positive rel numers with =, then 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) Prove tht if,, > 0 then ( + ) 2 + ( + ) 2 + ( + ) 2 9 4( + + ). 55. Prove tht if,,, d re positive rel numers, then d d. 56. Let,,, d e positive rel numers. Prove tht ( + )( + d) + d. 57. Let > > 0 nd > > 0. Prove tht ( ) + ( ).

14 4 CHAPTER 2. CAUCHY-SCHWARZ INEQUALITY 58. Let,, > 0 with + + =. Prove tht Let,, > 0 with =. Prove tht Let,, e positive rel numers suh tht =. Prove tht Let,,, x, y, z e positive rel numers suh tht x + y + z =. Prove tht x + y + z + 2 (xy + yz + zx)( + + ) Let,, e positive rel numers. Prove tht 6. Prove tht if,,, x, y, z > 0, then x 4 + y4 + z4 (x + y + z)4 ( + + ). 64. Prove tht if,, re positive rel numers, then Prove tht if,, re positive rel numers, then Prove tht if,, > 0, then Let,,, d e positive rel numers suh tht ( ) = 2 + d 2. Prove tht + d.

15 2.. EXERCISES Let, e positive rel numers. Prove tht Let,, > 0 with =. Prove tht Prove tht if,,, x, y, z > 0 nd ( ) = x 2 + y 2 + z 2, then x + y + z. 7. Prove tht if,, re positive rel numers, then ( + ) Let,,, d 0 with + + d + d =. Prove tht + + d d d + d Let, 2,..., n ;, 2,..., n e positive rel numers suh tht n = n. Show tht Let,, > 0. Prove tht ( + ) 2 + ( + ) Let,, > 0 with + + =. Prove tht 2 n n n + n 2 ( + ) 2 9 4( ). 2 ( + ) ( + ) ( + )

16 6 CHAPTER 2. CAUCHY-SCHWARZ INEQUALITY

17 Chpter REARRANGEMENT INEQUALITY Definition 6 Let 2... n nd 2... n. The numer A = 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 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 holds. Solution: Sine the inequlity is symmetri, WLOG we n ssume tht. So y Rerrngement inequlity, eing the ordered sum = Exmple 0 Prove tht if,, re positive rel numers, then Solution: WLOG, we n ssume tht. Then nd. By Rerrngement inequlity we hve, LHS = =

18 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 = = And gin y Rerrngement inequlity, eing the reversed sum + + = Comining these two inequlities we get tht. EXERCISES Let,, e positive rel numers. Prove tht () + ( + ) () ( + ) () (d) (e) (f) (g) ( + + ) (h) Prove tht if,, re positive rel numers, then 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 x2 n y n x + x 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 x n.

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

20 20 CHAPTER. REARRANGEMENT INEQUALITY

21 Chpter 4 CHEBYSHEV S INEQUALITY Theorem 5 Let 2... n nd 2... n. Then n n n ( n )( n ) n + 2 n n. Exmple 2 Prove tht ( + + )2. Solution: By Cheyshev s inequlity, = + + ( + + )( + + ) = ( + + )2. Exmple Let,, > 0. Prove tht 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, (4.2)

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

23 4.. EXERCISES Prove tht if,, > 0 with =, then Let,,, d 0 with + + d + d =. Prove tht + + d d d + d 90. Prove tht if,, > 0, then Prove tht if,, > 0, then Prove tht if,, > 0 nd = then Let x, y, z e positive rel numers with xyz = nd. Prove tht x y + z + y z + x + z x + y Prove tht if,, > 0 with + + =, then Prove tht if,, > 0 then Let,, > 0. Prove tht ( + + )

24 24 CHAPTER 4. CHEBYSHEV S INEQUALITY

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

26 26 CHAPTER 5. MIXED PROBLEMS 04. Let,, > 0 with + + =. Prove tht (IMO2000/2) Let,, e positive rel numers with produt. Prove tht ( + )( + )( + ). 06. Prove tht if,, > 0 nd =, then Let,, e positive rel numers suh tht. Prove tht Let,, e positive rel numers suh tht =. Prove tht If,, (0, ), then prove tht + ( )( )( ) <. 0. Prove tht 2 + ( ) ( ) ( ) for ritrry rel numers,,.. Prove tht ( ) 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 for positive rel numers,,.

27 27 5. Prove tht if,, > 0, then ( + + ) (IMO200/2)Let,, e positive rel numers. Prove tht Prove tht if,, re positive, then 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 Let,, > 0. Prove tht ( + + ). 2. Let,, > 0 with =. Prove tht ( + )( + )( + ) If x, y > 0 nd x 2 + y x + y 4. Prove tht x + y Let,, e positive rel numers. Prove tht ( ). 24. Prove tht if,, re positive rel numers with + + =, then Let,, > 0 with =. Prove tht

28 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 ) (MMO/2009) Let,, > 0 suh tht + + =. Prove tht Prove tht x, y, z R + we hve: x Let x, y, z > 0. Prove tht 0. Prove tht if,, > 0, then x yz + yz 4 2. x y(x 2 + 2y 2 ) + y z(y 2 + 2z 2 ) + z x(z 2 + 2x 2 ) Prove tht if,, > 0 with =, then 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 Let x, y, z e positive rels. Prove tht yz 2x 2 + yz 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 +.

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

30 0 CHAPTER 6. PROBLEMS FROM OLYMPIADS 6 (Romni 2005, Roert Szsz) ( + + =,,, > 0) ( 2)( 2)( 2) 7 (Romni 2005) (,,, > 0) (Romni 2005, Unused) ( =,,, > 0) 2 ( + ) + 2 ( + ) + 2 ( + ) 2 9 (Romni 2005, Unused) ( ,,, > 0) ( + ) + ( + ) + ( + ) 2 0 (Romni 2005, Unused) ( + + =,,, > 0) (Romni 2005, Unused) ( =,,, > 0) (Chzeh nd Solvk 2005) ( =,,, > 0) ( + )( + ) + ( + )( + ) + ( + )( + ) 4 (Jpn 2005) ( + + =,,, > 0) ( + ) + ( + ) + ( + )

31 4 (Germny 2005) ( + + =,,, > 0) ( ) (Vietnm 2005) (,, > 0) ( ) ( ) ( ) (Chin 2005) ( + + =,,, > 0) 0( + + ) 9( ) 7 (Chin 2005) (d =,,,, d > 0) ( + ) 2 + ( + ) 2 + ( + ) 2 + ( + d) 2 8 (Chin 2005) ( + + =,,, 0) (Polnd 2005) (0,, ) (Polnd 2005) ( + + =,,, > 0) (Blti Wy 2005) ( =,,, > 0)

32 2 CHAPTER 6. PROBLEMS FROM OLYMPIADS 22 (Seri nd Montenegro 2005) (,, > 0) ( + + ) 2 2 (Seri nd Montenegro 2005) ( + + =,,, > 0) (Bosni nd Heregovin 2005) ( + + =,,, > 0) (Irn 2005) (,, > 0) ( + + ) 2 ( ( + + ) + + ) 26 (Austri 2005) (,,, d > 0) d d d 27 (Moldov 2005) ( =,,, > 0) (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

33 0 (Polnd 2004) ( + + = 0,,, R) (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) (APMO 2004) (,, > 0) ( 2 + 2)( 2 + 2)( 2 + 2) 9( + + ) 5 (USA 2004) (,, > 0) ( )( )( ) ( + + ) 6 (Junior BMO 200) (x, y, z > ) + x 2 + y + z y2 + z + x z2 + x + y (USA 200) (,, > 0) (2 + + ) 2 (2 + + )2 (2 + + ) ( + ) ( + ) ( + ) 2 8

34 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) d =,,,, d > 0 4 d 40 (Alni 2002) (,, > 0) + ( ( ) + + ) (Belrus 2002) (,,, d > 0) ( + )2 + ( + d) 2 2 d + ( + )2 + ( + d) d 2 ( + ) 2 + ( + d) (Cnd 2002) (,, > 0) (Vietnm 2002, Dung Trn Nm) ( = 9,,, R) 2( + + ) 0 44 (Bosni nd Heregovin 2002) ( =,,, R) (Junior BMO 2002) (,, > 0) ( + ) + ( + ) + ( + ) 27 2( + + ) 2

35 5 46 (Greee 2002) ( =,,, > 0) ( + + ) (Greee 2002) ( 0, 2 0,,, R) 0( ) (Tiwn 2002) (,,, d ( 0, 2 ]) d ( )( )( )( d) 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) (USA 200) ( = 4,,, 0) (Columi 200) (x, y R) (x + y + ) 2 + xy

36 6 CHAPTER 6. PROBLEMS FROM OLYMPIADS 54 (KMO Winter Progrm Test 200) (,, > 0) ( ) ( ) + ( + ) ( + ) ( + ) 55 (KMO Summer Progrm Test 200) (,, > 0) (IMO 200) (,, > 0) Yers (IMO 2000, Titu Andreesu) ( =,,, > 0) ( + ) ( + ) ( + ) 58 (Czeh nd Slovki 2000) (, > 0) ( 2( + ) + ) + 59 (Hong Kong 2000) ( =,,, > 0) (Czeh Repuli 2000) (m, n N, x [0, ]) ( x n ) m + ( ( x) m ) n

37 6.. YEARS (Medoni 2000) (x, y, z > 0) x 2 + y 2 + z 2 2 (xy + yz) 62 (Russi 999) (,, > 0) > 6 (Belrus 999) ( =,,, > 0) (Czeh-Slovk Mth 999) (,, > 0) (Moldov 999) (,, > 0) ( + ) + ( + ) + ( + ) (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 (Proposed for 999 USAMO, [AB, pp.25]) (x, y, z > ) x x2 +2yz y y2 +2zx z z2 +2xy (xyz) xy+yz+zx

38 8 CHAPTER 6. PROBLEMS FROM OLYMPIADS 69 (Turkey, 999) ( 0) ( + )( + 4)( + 2) (Medoni 999) ( =,,, > 0) (Polnd 999) ( + + =,,, > 0) (Cnd 999) (x + y + z =, x, y, z 0) x 2 y + y 2 z + z 2 x (Irn 998) ( x + y + z = 2, x, y, z > ) x + y + z x + y + z 74 (Belrus 998, I. Gorodnin) (,, > 0) (APMO 998) (,, > 0) ( + ) ( + ) ( + ) 2 ( ) 76 (Polnd 998) ( d + e + f =, e + df 08,,, d, e, f > 0) + d + de + def + ef + f 6

39 6.. YEARS (Kore 998) (x + y + z = xyz, x, y, z > 0) x 2 + y 2 + z (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) (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 40 CHAPTER 6. PROBLEMS FROM OLYMPIADS 85 (Bulgri 997) ( =,,, > 0) (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 (Romni 997) (,, > 0) (USA 997) (,, > 0) (Jpn 997) (,, > 0) ( + ) 2 ( + )2 ( + )2 ( + ) ( + ) ( + ) (Estoni 997) (x, y R) x 2 + y 2 + > x y y x (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) ( )( )

41 6.. YEARS (IMO Short List 996) ( =,,, > 0) (Polnd 996) ( + + =,,, ) (Hungry 996) ( + =,, > 0) (Vietnm 996) (,, R) ( + ) 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 + ( + ) (Vietnm 996) (2( + + d + + d + d) + + d + d + d = 6,,,, d 0) d 2 ( + + d + + d + d)

42 42 CHAPTER 6. PROBLEMS FROM OLYMPIADS 6.2 Yers Any good ide n e stted in fifty words or less. S. M. Ulm 00 (Blti Wy 995) (,,, d > 0) d d + d + d (Cnd 995) (,, > 0) (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) (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

43 6.2. YEARS (IMO Short List 99) (,,, d > 0) d + + 2d (APMC 99) (, 0) ( ) d d ( ) 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 d (Polnd 992) (,, R) ( + ) 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,,, R) (IMO Short List 990) ( + + d + d =,,,, d > 0) + + d + + d + + d d + +

44 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 (IMO 984) (x + y + z =, x, y, z 0) 0 xy + yz + zx 2xyz (USA 980) (,, [0, ]) ( )( )( ) (USA 979) (x + y + z =, x, y, z > 0) x + y + z + 6xyz 4. 2 (IMO 974) (,,, d > 0) < + + d d + d + + d < 2

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

46 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 ( 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

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

48 48 CHAPTER 6. PROBLEMS FROM OLYMPIADS 46 (C252) ( =,,, > 0) ( + + ) 47 (C02, Vsile Cirtoje) ( =,,, > 0) (C2645) (,, > 0) 2( + + ) + 9( + + )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)

49 6.. SUPPLEMENTARY PROBLEMS (x + y + z =, x, y, z > 0) x x + y y + z z 2 55 (,, R) 2 + ( ) ( ) ( ) (,, > 0) (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 50 CHAPTER 6. PROBLEMS FROM OLYMPIADS 62 (Drij Grinerg) (,, > 0) 2 ( + ) ( ) (2 + + ) + 2 ( + ) ( ) (2 + + ) + 2 ( + ) ( ) (2 + + ) > 2. 6 (Drij Grinerg) (,, > 0) ( + ) ( + ) ( + ) 2 < (Vsile Cirtoje) (,, R) ( ) 2 ( + + )

INEQUALITIES THROUGH PROBLEMS. 1. Heuristics of problem solving

INEQUALITIES THROUGH PROBLEMS. 1. Heuristics of problem solving INEQUALITIES THROUGH PROBLEMS Hojoo Lee. Heuristis of prolem solving Strtegy or ttis in prolem solving is lled heuristis. Here is summry tken from Prolem-Solving Through Prolems y Loren C. Lrson.. Serh