TOPICS IN INEQUALITIES. Hojoo Lee

Transcription

1 TOPICS IN INEQUALITIES Hojoo Lee Version 0.5 [005/08/5] Introdution Inequlities re useful in ll fields of Mthemtis. The purpose in this ook is to present stndrd tehniques in the theory of inequlities. The reders will meet lssil theorems inluding Shur s inequlity, Muirhed s theorem, the Cuhy-Shwrtz inequlity, AM-GM inequlity, nd Ho lder s theorem, et. There re mny prolems from Mthemtil olympids nd ompetitions. The ook is ville t idehitme/eng.html I wish to express my ppreition to Stnley Rinowitz who kindly sent me his pper On The Computer Solution of Symmetri Homogeneous Tringle Inequlities. This is n unfinished mnusript. I would gretly ppreite hering out ny errors in the ook, even minor ones. You n send ll omments to the uthor t hojoolee@kore.om. To Students The given tehniques in this ook re just the tip of the inequlities ieerg. Wht young students red this ook should e wre of is tht they should find their own retive methods to ttk prolems. It s impossile to present ll tehniques in smll ook. I don t even lim tht the methods in this ook re mthemtilly eutiful. For instne, lthough Muirhed s theorem nd Shur s theorem whih n e found t hpter re extremely powerful to ttk homogeneous metri polynomil inequlities, it s not good ide for eginners to lern how to pply them to prolems. (Why?) However, fter mstering homogeniztion method using Muirhed s theorem nd Shur s theorem, you n hve more rod mind in the theory of inequlities. Tht s why I inlude the methods in this ook. Hve fun! Reommended Reding List. K. S. Kedly, A < B, I. Niven, Mxim nd Minim Without Clulus, MAA. T. Andreesu, Z. Feng, 0 Trigonometry Prolems From the Trining of the USA IMO Tem, Birkhuser 4. O. Bottem, R. Z. Djordjević, R. R. Jnić, D. S. Mitrinović, P. M. Vsić, Geometri Inequlities, Wolters-Noordhoff Pulishing, Groningen 969

2 Contents 00 Prolems Sustitutions. Euler s Theorem nd the Rvi Sustitution Trigonometri Sustitutions Algeri Sustitutions Supplementry Prolems for Chpter Homogeniztions 6. Homogeneous Polynomil Inequlities Shur s Theorem Muirhed s Theorem Polynomil Inequlities with Degree Supplementry Prolems for Chpter Normliztions 7 4. Normliztions Clssil Theorems : Cuhy-Shwrtz, (Weighted) AM-GM, nd Hölder Homogeniztions nd Normliztions Supplementry Prolems for Chpter Multivrile Inequlities 45 6 Referenes 5

3 Chpter 00 Prolems Eh prolem tht I solved eme rule, whih served fterwrds to solve other prolems. Rene Desrtes I. (Hungry 996) ( + =,, > 0) I. (Columi 00) (x, y R) I. (0 < x, y < ) (x + y + ) + xy x y + y x > I 4. (APMC 99) (, 0) ( ) I 5. (Czeh nd Slovki 000) (, > 0) 4 ( + ) + + ( + ) + I 6. (Die W URZEL, Heinz-Jürgen Seiffert) (xy > 0, x, y R) xy x + y + x + y xy + x + y I 7. (Crux Mthemtiorum, Prolem 645, Hojoo Lee) (,, > 0) ( + + ) + 9( + + ) ( + + ) ( + ) I 8. (x, y, z > 0) I 9. (,,, x, y, z > 0) xyz + x y + y z + z x x + y + z ( + x)( + y)( + z) + xyz

4 I 0. (x, y, z > 0) x x + (x + y)(x + z) + I. (x + y + z =, x, y, z > 0) I. (Irn 998) x x + ( x + y + z =, x, y, z > ) y y + (y + z)(y + x) + y y + z z x + y + z x + y + z z z + (z + x)(z + y) I. (KMO Winter Progrm Test 00) (,, > 0) ( + + ) ( + + ) + ( + ) ( + ) ( + ) I 4. (KMO Summer Progrm Test 00) (,, > 0) I 5. (Gzet Mtemtiã, Hojoo Lee) (,, > 0) I 6. (,, R) I 7. (,, > 0) + ( ) + + ( ) + + ( ) I 8. (Belrus 00) (,,, d > 0) ( + ) + ( + d) + d ( + ) + ( + d) d ( + ) + ( + d) I 9. (Hong Kong 998) (,, ) + + ( + ) I 0. (Crlson s inequlity) (,, > 0) ( + )( + )( + ) I. (Kore 998) (x + y + z = xyz, x, y, z > 0) x + y + z I. (IMO 00) (,, > 0) I. (IMO Short List 004) ( + + =,,, > 0)

5 I 4. (,, > 0) ( + ) + ( + ) + ( + ) 4 + ( + )( + )( + ) I 5. (Medoni 995) (,, > 0) I 6. (Nesitt s inequlity) (,, > 0) I 7. (IMO 000) ( =,,, > 0) ( + ) ( + ) ( + ) I 8. ([ONI], Vsile Cirtoje) (,, > 0) ( + ) ( + ) + ( + ) ( + ) + ( + ) ( + ) I 9. (IMO Short List 998) (xyz =, x, y, z > 0) x ( + y)( + z) + y ( + z)( + x) + z ( + x)( + y) 4 I 0. (IMO Short List 996) ( =,,, > 0) I. (IMO 995) ( =,,, > 0) I. (IMO Short List 99) (,,, d > 0) + + d ( + ) + ( + ) + ( + ) + d + + d I. (IMO Short List 990) ( + + d + d =,,,, d > 0) + + d + + d + + d d d I 4. (IMO 968) (x, x > 0, y, y, z, z R, x y > z, x y > z ) I 5. (Romni 997) (,, > 0) I 6. (Cnd 00) (,, > 0) x y z + x y z 8 (x + x )(y + y ) (z + z )

6 I 7. (USA 997) (,, > 0) I 8. (Jpn 997) (,, > 0) I 9. (USA 00) (,, > 0) ( + ) ( + ) ( + ) ( + ) + + ( + ) + + ( + ) + 5 ( + + ) ( + + ) ( + + ) + + ( + ) + + ( + ) + ( + ) 8 I 40. (Crux Mthemtiorum, Prolem 580, Hojoo Lee) (,, > 0) I 4. (Crux Mthemtiorum, Prolem 58, Hojoo Lee) (,, > 0) I 4. (Crux Mthemtiorum, Prolem 5, Hojoo Lee) ( + + =,,, > 0) I 4. (Belrus 999) ( + + =,,, > 0) ( + + ) I 44. (Crux Mthemtiorum, Prolem 0, Vsile Cirtoje) ( + + =,,, > 0) I 45. (Greee 00) ( + + =,,, > 0) ( + + ) 4 I 46. (Irn 996) (,, > 0) ( ) ( + + ) ( + ) + ( + ) + ( + ) 9 4 I 47. (Alni 00) (,, > 0) + ( ( + + ) + + ) I 48. (Belrus 997) (,, > 0) I 49. (Belrus 998, I. Gorodnin) (,, > 0)

7 I 50. (Polnd 996) ( + + =,,, I 5. (Bulgri 997) ( =,,, > 0) I 5. (Romni 997) (xyz =, x, y, z > 0) I 5. (Vietnm 99) (x y z > 0) I 54. (Ukrine 99) ( 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 x y z + y z x + + z x y x + y + z I 55. (Irn 997) (x x x x 4 =, x, x, x, x 4 > 0) ) x + x + x + x 4 mx (x + x + x + x 4, x + x + x + x4 I 56. (Hong Kong 000) ( =,,, > 0) + I 57. (Hong Kong 997) (x, y, z > 0) I 58. (Czeh-Slovk Mth 999) (,, > 0) I 59. (Moldov 999) (,, > 0) ( + ) + I 60. (Blti Wy 995) (,,, d > 0) xyz(x + y + z + x + y + z ) (x + y + z )(xy + yz + zx) + + ( + ) + I 6. ([ONI], Vsile Cirtoje) (,,, d > 0) I 6. (Polnd 99) (x, y, u, v > 0) + + ( + ) d d + d + d d + d d + + d + 0 xy + xv + uy + uv x + y + u + v + + xy x + y + uv u + v + 7

8 I 6. (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 I 64. (Lithuni 987) (x, y, z > 0) x x + xy + y + y y + yz + z + z z + zx + x x + y + z I 65. (Klmkin s inequlity) ( < x, y, z < ) I 66. (xy + yz + zx =, x, y, z > 0) ( x)( y)( z) + ( + x)( + y)( + z) x + x + y + y + z + z x( x ) ( + x ) + y( y ) ( + y ) + z( z ) ( + z ) I 67. (Russi 00) (x + y + z =, x, y, z > 0) x + y + z xy + yz + zx I 68. (APMO 998) (,, > 0) ( + ) ( + ) ( + ) ( ) I 69. (Elemente der Mthemtik, Prolem 07, Sefket Arslngić) (x, y, z > 0) x y + y z + z x x + y + z xyz I 70. (Die W URZEL, Wlther Jnous) (x + y + z =, x, y, z > 0) ( + x)( + y)( + z) ( x ) + ( y ) + ( z ) I 7. (United Kingdom 999) (p + q + r =, p, q, r > 0) I 7. (USA 979) (x + y + z =, x, y, z > 0) I 7. (IMO 984) (x + y + z =, x, y, z 0) 7(pq + qr + rp) + 9pqr x + y + z + 6xyz 4. 0 xy + yz + zx xyz 7 7 I 74. (IMO Short List 99) ( d =,,,, d > 0) I 75. (Polnd 99) (,, R) + d + d + d d ( + ) ( + ) ( + ) ( + )( + )( + ) 8

9 I 76. (Cnd 999) (x + y + z =, x, y, z 0) x y + y z + z x 4 7 I 77. (Hong Kong 994) (xy + yz + zx =, x, y, z > 0) x( y )( z ) + y( z )( x ) + z( x )( y ) 4 9 I 78. (Vietnm 996) (( + + d + + d + d) + + d + d + d = 6,,,, d 0) d ( + + d + + d + d) I 79. (Polnd 998) ( d + e + f =, e + df 08,,, d, e, f > 0) I 80. (Itly 99) (0,, ) + d + de + def + ef + f I 8. (Czeh Repuli 000) (m, n N, x [0, ]) I 8. (Irelnd 997) ( + +,,, 0) I 8. (BMO 00) ( + +,,, 0) ( x n ) m + ( ( x) m ) n I 84. (Berus 996) (x + y + z = xyz, x, y, z > 0) xy + yz + zx 9(x + y + z) I 85. (Polnd 99) (x + y + z =, x, y, z R) x + y + z + xyz I 86. (Mongoli 99) ( + + =,,, R) + + I 87. (Vietnm 00, Dung Trn Nm) ( + + = 9,,, R) I 88. (Vietnm 996) (,, > 0) ( + + ) 0 ( + ) 4 + ( + ) 4 + ( + ) 4 4 ( ) 7 I 89. (x, y, z 0) xyz (y + z x)(z + x y)(x + y z) ( ) I 90. (Ltvi 00) d =,,,, d > 0 4 d 9

10 I 9. (Proposed for 999 USAMO, [AB, pp.5]) (x, y, z > ) I 9. (APMO 004) (,, > 0) I 9. (USA 004) (,, > 0) x x +yz y y +zx z z +xy (xyz) xy+yz+zx ( + )( + )( + ) 9( + + ) ( 5 + )( 5 + )( 5 + ) ( + + ) I 94. (USA 00) ( = 4,,, 0) I 95. (Turkey, 999) ( 0) ( + )( + 4)( + ) 60 I 96. (Medoni 999) ( + + =,,, > 0) I 97. (Polnd 999) ( + + =,,, > 0) I 98. (Medoni 000) (x, y, z > 0) I 99. (APMC 995) (m, n N, x, y > 0) x + y + z (xy + yz) (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 ) I 00. ([ONI], Griel Dospinesu, Mire Lsu, Mrin Tetiv) (,, > 0) ( + )( + )( + ) 0

11 Chpter Sustitutions. Euler s Theorem nd the Rvi Sustitution Mny inequlities re simplified y some suitle sustitutions. tringle geometry. We egin with lssil inequlity in In 765, Euler showed tht Wht is the first nontrivil geometri inequlity? Theorem. Let R nd r denote the rdii of the irumirle nd inirle of the tringle ABC. Then, we hve R r nd the equlity holds if nd only if ABC is equilterl. Proof. Let BC =, CA =, AB =, s = ++ nd S = [ABC]. Rell the well-known identities : S = 4R, S = rs, S = s(s )(s )(s ). Hene, R r is equivlent to 4S S S s or 8 s or 8(s )(s )(s ). We need to prove the following. Theorem. ([AP], A. Pdo) Let,, e the lengths of tringle. Then, we hve 8(s )(s )(s ) or ( + )( + )( + ) nd the equlity holds if nd only if = =. First Proof. We use the Rvi Sustitution : Sine,, re the lengths of tringle, there re positive rels x, y, z suh tht = y + z, = z + x, = x + y. (Why?) Then, the inequlity is (y + z)(z + x)(x + y) 8xyz for x, y, z > 0. However, we get (y + z)(z + x)(x + y) 8xyz = x(y z) + y(z x) + z(x y) 0. Seond Proof. ([RI]) We my ssume tht. It s equivlent to ( + ) + ( + ) + ( + ). Sine ( + ) ( + ) ( + ), pplying the Rerrngement inequlity, we otin ( + ) + ( + ) + ( + ) ( + ) + ( + ) + ( + ), ( + ) + ( + ) + ( + ) ( + ) + ( + ) + ( + ). Adding these two inequlities, we get the result. Exerise. Let ABC e right tringle. Show tht R ( + )r. When does the equlity hold? It s nturl to sk tht the inequlity in the theorem holds for ritrry positive rels,,? Yes! It s possile to prove the inequlity without the dditionl ondition tht,, re the lengths of tringle : The first geometri inequlity is the Tringle Inequlity : AB + BC AC In this ook, [P ] stnds for the re of the polygon P. For exmple, we hve ( + ) ( + ) = ( )( + ) 0.

12 Theorem. Let x, y, z > 0. Then, we hve xyz (y + z x)(z + x y)(x + y z). The equlity holds if nd only if x = y = z. Proof. Sine the inequlity is metri in the vriles, without loss of generlity, we my ssume tht x y z. Then, we hve x + y > z nd z + x > y. If y + z > x, then x, y, z re the lengths of the sides of tringle. And y the theorem, we get the result. Now, we my ssume tht y + z x. Then, xyz > 0 (y + z x)(z + x y)(x + y z). The inequlity in the theorem holds when some of x, y, z re zeros : Theorem 4. Let x, y, z 0. Then, we hve xyz (y + z x)(z + x y)(x + y z). Proof. Sine x, y, z 0, we n find positive sequenes {x n }, {y n }, {z n } for whih (For exmple, tke x n = x + n lim x n = x, n lim y n = y, lim z n = z. n n (n =,, ), et.) Applying the theorem yields x n y n z n (y n + z n x n )(z n + x n y n )(x n + y n z n ) Now, tking the limits to oth sides, we get the result. Clerly, the equlity holds when x = y = z. However, xyz = (y+z x)(z +x y)(x+y z) nd x, y, z 0 does not gurntee tht x = y = z. In ft, for x, y, z 0, the equlity xyz = (y +z x)(z +x y)(x+y z) is equivlent to x = y = z or x = y, z = 0 or y = z, x = 0 or z = x, y = 0. It s strightforwrd to verify the equlity xyz (y + z x)(z + x y)(x + y z) = x(x y)(x z) + y(y z)(y x) + z(z x)(z y). Hene, the theorem 4 is prtiulr se of Shur s inequlity. 4 Prolem. (IMO 000/) Let,, e positive numers suh tht =. Prove tht ( + ) ( + ) ( + ). First Solution. Sine =, we mke the sustitution = x y, = y z, = z x for x, y, z > 0.5 We rewrite the given inequlity in the terms of x, y, z : ( x y + z ) (y y z + x ) ( z z x + y ) xyz (y + z x)(z + x y)(x + y z). x The Rvi Sustitution is useful for inequlities for the lengths,, of tringle. After the Rvi Sustitution, we n remove the ondition tht they re the lengths of the sides of tringle. Prolem. (IMO 98/6) Let,, e the lengths of the sides of tringle. Prove tht ( ) + ( ) + ( ) 0. Solution. After setting = y + z, = z + x, = x + y for x, y, z > 0, it eomes x z + y x + z y x yz + xy z + xyz or x y + y z + z x x + y + z, whih follows from the Cuhy-Shwrtz inequlity ( ) x (y + z + x) y + y z + z (x + y + z). x 4 See the theorem 0 in the hpter. Tke r =. 5 For exmple, tke x =, y =, z =.

13 Prolem. (IMO 96/, Weitzenök s inequlity) Let,, e the lengths of tringle with re S. Show tht S. Solution. Write = y + z, = z + x, = x + y for x, y, z > 0. It s equivlent to whih n e otined s following : ((y + z) + (z + x) + (x + y) ) 48(x + y + z)xyz, ((y + z) + (z + x) + (x + y) ) 6(yz + zx + xy) 6 (xy yz + yz zx + xy yz). 6 Exerise. (Hdwiger-Finsler inequlity) Show tht, for ny tringle with sides,, nd re S, + + ( + + ) 4 S. Exerise. (Pedoe s inequlity) Let,, denote the sides of the tringle A B C with re F. Let,, denote the sides of the tringle A B C with re F. Show tht ( + ) + ( + ) + ( + ) 6F F. 6 Here, we used the well-known inequlities p + q pq nd (p + q + r) (pq + qr + rp).

14 . Trigonometri Sustitutions If you re fed with n integrl tht ontins squre root expressions suh s x dx, + y dy, z dz then trigonometri sustitutions suh s x = sin t, y = tn t, z = se t re very useful. When deling with squre root expressions, mking suitle trigonometri sustitution simplifies the given inequlity. Prolem 4. (Ltvi 00) Let,,, d e the positive rel numers suh tht Prove tht d d 4 =. Solution. We n write = tn A, = tn B, = tn C, d = tn D, where A, B, C, D ( 0, π ). Then, the lgeri identity eomes the following trigonometri identity : Applying the AM-GM inequlity, we otin Similrly, we otin os A + os B + os C + os D =. sin A = os A = os B + os C + os D (os B os C os D). sin B (os C os D os A), sin C (os D os A os B), nd sin D (os A os B os C). Multiplying these inequlities, we get the result! Exerise 4. ([ONI], Titu Andreesu, Griel Dosinesu) Let,,, d e the rel numers suh tht ( + )( + )( + )( + d ) = 6. Prove tht + + d + + d + d d 5. Prolem 5. (Kore 998) Let x, y, z e the positive rels with x + y + z = xyz. Show tht + x + + y + + z. Sine the funtion f is not onve down on R +, we nnot pply Jensen s inequlity to the funtion f(t) = +t. However, the funtion f(tn θ) is onve down on ( 0, π )! Solution. We n write x = tn A, y = tn B, z = tn C, where A, B, C ( 0, π ). Using the ft tht + tn θ = ( os θ ), where os θ 0, we rewrite it in the terms of A, B, C : os A + os B + os C. It follows from tn(π C) = z = x+y xy = tn(a + B) nd from π C, A + B (0, π) tht π C = A + B or A + B + C = π. Hene, it suffies to show the following. Theorem 5. In ny ute tringle ABC, we hve os A + os B + os C. Proof. Sine os x is onve down on ( 0, π ), it s diret onsequene of Jensen s inequlity. We note tht the funtion os x is not onve down on (0, π). In ft, it s onve up on ( π, π). One my think tht the inequlity os A + os B + os C doesn t hold for ny tringles. However, it s known tht it lso holds for ny tringles. 4

15 Theorem 6. In ny tringle ABC, we hve os A + os B + os C. First Proof. It follows from π C = A + B tht os C = os(a + B) = os A os B + sin A sin B or (os A + os B + os C) = (sin A sin B) + (os A + os B ) 0. Seond Proof. Let BC =, CA =, AB =. Use the Cosine Lw to rewrite the given inequlity in the terms of,, : + Clering denomintors, this eomes ( + ) + ( + ) + ( + ), whih is equivlent to ( + )( + )( + ) in the theorem. In se even when there is no ondition suh s x + y + z = xyz or xy + yz + zx =, the trigonometri sustitutions re useful. Prolem 6. (APMO 004/5) Prove tht, for ll positive rel numers,,, ( + )( + )( + ) 9( + + ). Proof. Choose A, B, C ( ) 0, π with = tn A, = tn B, nd = tn C. Using the well-known trigonometri identity + tn θ = os θ, one my rewrite it s 4 os A os B os C (os A sin B sin C + sin A os B sin C + sin A sin B os C). 9 One my esily hek the following trigonometri identity os(a + B + C) = os A os B os C os A sin B sin C sin A os B sin C sin A sin B os C. Then, the ove trigonometri inequlity tkes the form 4 os A os B os C (os A os B os C os(a + B + C)). 9 Let θ = A+B+C. Applying the AM-GM inequlity nd Jesen s inequlity, we hve ( ) os A + os B + os C os A os B os C os θ. We now need to show tht Using the trigonometri identity 4 9 os θ(os θ os θ). os θ = 4 os θ os θ or os θ os θ = os θ os θ, it eomes 4 7 os4 θ ( os θ ), whih follows from the AM-GM inequlity ( os θ os θ ( os θ )) ( os θ + os θ + ( os θ )) =. One find tht the equlity holds if nd only if tn A = tn B = tn C = if nd only if = = =. 5

16 Exerise 5. ([TZ], pp.7) Let x, y, z e rel numers suh tht 0 < x, y, z < nd xy + yz + zx =. Prove tht x x + y y + z z. Exerise 6. ([TZ], pp.7) Let x, y, z e positive rel numers suh tht x + y + z = xyz. Prove tht x + x + y + y + z + z. Exerise 7. ([ONI], Florin Crln, Mrin Tetiv) Prove tht if x, y, z > 0 stisfy the ondition x + y + z = xyz then xy + yz + zx + + x + + y + + z. Exerise 8. ([ONI], Griel Dospinesu, Mrin Tetiv) Let x, y, z e positive rel numers suh tht x + y + z = xyz. Prove tht (x )(y )(z ) 6 0. Exerise 9. ([TZ], pp.) Let,, e rel numers. Prove tht ( + )( + )( + ) ( + + ). Exerise 0. ([TZ], pp.49) Let nd e positive rel numers. Prove tht if either () 0 <, or (). In the theorem nd, we see tht the geometri inequlity R r is equivlent to the lgeri inequlity ( + )( + )( + ). We now find tht, in the proof of the theorem 6, ( + )( + )( + ) is equivlent to the trigonometri inequlity os A + os B + os C. One my sk tht In ny tringles ABC, is there nturl reltion etween os A + os B + os C nd R r, where R nd r re the rdii of the irumirle nd inirle of ABC? Theorem 7. Let R nd r denote the rdii of the irumirle nd inirle of the tringle ABC. Then, we hve os A + os B + os C = + r R. Proof. Use the identity ( + )+( + )+( + ) = +(+ )(+ )(+ ). We leve the detils for the reders. Exerise. Let R nd r e the rdii of the irumirle nd inirle of the tringle ABC with BC =, CA =, AB =. Let s denote the semiperimeter of ABC. Verify the follwing identities 7 : () + + = s + 4Rr + r, () = 4Rrs, () os A os B + os B os C + os C os A = s 4R +r 4R, (4) os A os B os C = s (R+r) 4R Exerise. () Let p, q, r e the positive rel numers suh tht p + q + r + pqr =. Show tht there exists n ute tringle ABC suh tht p = os A, q = os B, r = os C. () Let p, q, r 0 with p + q + r + pqr =. Show tht there re A, B, C [ 0, π ] with p = os A, q = os B, r = os C, nd A + B + C = π. 7 For more identities, see the exerise 0. 6

17 Exerise. ([ONI], Mrin Tetiv) Let x, y, z e positive rel numers stisfying the ondition Prove tht () xyz 8, () xy + yz + zx 4, () x + y + z 4, nd (4) xy + yz + zx xyz +. x + y + z + xyz =. Exerise 4. ([ONI], Mrin Tetiv) Let x, y, z e positive rel numers stisfying the ondition Prove tht () xyz 7, () xy + yz + zx 7, () x + y + z 9, nd (4) xy + yz + zx (x + y + z) + 9. x + y + z = xyz. Prolem 7. (USA 00) Let,, nd e nonnegtive rel numers suh tht = 4. Prove tht Solution. Notie tht,, > implies tht > 4. If, then we hve ++ ( ) 0. We now prove tht + +. Letting = p, = q, = r, we get p + q + r + pqr =. By the exerise, we n write [ = os A, = os B, = os C for some A, B, C 0, π ] with A + B + C = π. We re required to prove os A os B + os B os C + os C os A os A os B os C. One my ssume tht A π or os A 0. Note tht os A os B + os B os C + os C os A os A os B os C = os A(os B + os C) + os B os C( os A). We pply Jensen s inequlity to dedue os B + os C os A. Note tht os B os C = os(b C) + os(b + C) os A. These imply tht ( ) ( ) os A os A(os B + os C) + os B os C( os A) os A os A + ( os A). However, it s esy to verify tht os A ( os A) + ( ) os A ( os A) =. In the ove solution, we showed tht os A os B + os B os C + os C os A os A os B os C holds for ll ute tringles. Using the results () nd (d) in the exerise (4), we n rewrite it in the terms of R, r, s : R + 8Rr + r s. In 965, W. J. Blundon found the est possile inequlities of the form A(R, r) s B(R, r), where A(x, y) nd B(x, y) re rel qudrti forms αx + βxy + γy : 8 Exerise 5. Let R nd r denote the rdii of the irumirle nd inirle of the tringle ABC. Let s e the semiperimeter of ABC. Show tht 6Rr 5r s 4R + 4Rr + r. 8 For proof, see [WJB]. 7

18 . Algeri Sustitutions We know tht some inequlities in tringle geometry n e treted y the Rvi sustitution nd trigonometri sustitutions. We n lso trnsform the given inequlities into esier ones through some lever lgeri sustitutions. Prolem 8. (IMO 00/) Let,, e positive rel numers. Prove tht First Solution. To remove the squre roots, we mke the following sustitution : x = + 8, y = + 8, z = + 8. Clerly, x, y, z (0, ). Our im is to show tht x + y + z. We notie tht 8 = x x, 8 = Hene, we need to show tht y y, 8 = z z = ( ) ( ) ( ) x y z 5 = x y z. x + y + z, where 0 < x, y, z < nd ( x )( y )( z ) = 5(xyz). However, > x + y + z implies tht, y the AM-GM inequlity, ( x )( y )( z ) > ((x + y + z) x )((x + y + z) y )((x + y + z) z ) = (x + x + y + z)(y + z) (x + y + y + z)(z + x)(x + y + z + z)(x + y) 4(x yz) 4 (yz) 4(y zx) 4 (zx) 4(z xy) 4 (xy) = 5(xyz). This is ontrdition! Prolem 9. (IMO 995/) Let,, e positive numers suh tht =. Prove tht ( + ) + ( + ) + ( + ). First Solution. After the sustitution = x, = y, = z, we get xyz =. The inequlity tkes the form x y + z + y z + x + It follows from the Cuhy-shwrtz inequlity tht ( x [(y + z) + (z + x) + (x + y)] y + z + so tht, y the AM-GM inequlity, x y + z + y z + x + z x + y. y z + x + z x + y x + y + z (xyz) ) z (x + y + z) x + y =. We offer n lterntive solution of the prolem 5 : (Kore 998) Let x, y, z e the positive rels with x + y + z = xyz. Show tht x + y + z. 8

19 Seond Solution. The strting point is letting = x, = y, = z. We find tht + + = is equivlent to = xy + yz + zx. The inequlity eomes x x + + y y + + z z + or or x x + xy + yz + zx + x (x + y)(x + z) + y y + xy + yz + zx + y (y + z)(y + x) + z z + xy + yz + zx z (z + x)(z + y). By the AM-GM inequlity, we hve x = x (x + y)(x + z) x[(x + y) + (x + z)] = ( x (x + y)(x + z) (x + y)(x + z) (x + y)(x + z) x + z + x ). x + z In like mnner, we otin y (y + z)(y + x) ( y y + z + Adding these three yields the required result. y ) y + x nd z ( z (z + x)(z + y) z + x + z ). z + y We now prove lssil theorem in vrious wys. Theorem 8. (Nesitt, 90) For ll positive rel numers,,, we hve Proof. After the sustitution x = +, y = +, z = +, it eomes yli y + z x x whih follows from the AM-GM inequlity s following: or yli y + z x 6, yli y + z x = y x + z x + z y + x y + x z + y ( y z 6 x z x z y x y x z y ) 6 = 6. z Proof. We mke the sustitution It follows tht x = f(x) = yli +, y = yli +, z = +. =, where f(t) = t + + Sine f is onve down on (0, ), Jensen s inequlity shows tht ( ) f = = ( ) x + y + z f(x) f or f yli ( ) f + t. ( x + y + z ). Sine f is monotone deresing, we hve x + y + z or yli + = x + y + z. 9

20 Proof. As in the previous proof, it suffies to show tht T where T = x + y + z nd yli x + x =. One n esily hek tht the ondition yli x + x = eomes = xyz + xy + yz + zx. By the AM-GM inequlity, we hve = xyz + xy + yz + zx T + T T + T 0 (T )(T + ) 0 T. Proof 4. Sine the inequlity is metri in the three vriles, we my ssume tht. After the sustitution x =, y =, we hve x y. It eomes We pply the AM-GM inequlity to otin It suffies to show tht y + + x + x + y + or x y + + y x + x + y. x + y + + y + x + or x y + + y x + y + + x +. y + x + x + y However, the lst inequlity lerly holds for x y. Proof 5. As in the previous proof, we need to prove Let A = x + y nd B = xy. It eomes x y + + y x + + x + y where x y. y ( + y) y (x + )(x + y). x + y + x + y (x + )(y + ) + x + y or A B + A A + B + + A or A A A + B(7A ). Sine 7A > (x + y ) > 0 nd A = (x + y) 4xy = 4B, it s enough to show tht 4(A A A + ) A (7A ) A A 4A However, it s esy to hek tht A A 4A + 8 = (A ) (A + ) 0. We now present lterntive solutions of prolem. (IMO 000/) Let,, e positive numers suh tht =. Prove tht ( + ) ( + ) ( + ). Seond Solution. ([IV], Iln Vrdi) Sine =, we my ssume tht. 9 It follows tht ( + ) ( + ) ( + ) = ( + ) ( + ) ( )( ) Why? Note tht the inequlity is not metri in the three vriles. Chek it! 0 For verifition of the identity, see [IV]. 0

21 Third Solution. As in the first solution, fter the sustitution = x y, = y z, = z x for x, y, z > 0, we n rewrite it s xyz (y + z x)(z + x y)(x + y z). Without loss of generlity, we n ssume tht z y x. Set y x = p nd z x = q with p, q 0. It s strightforwrd to verify tht xyz (y + z x)(z + x y)(x + y z) = (p pq + q )x + (p + q p q pq ). Sine p pq + q (p q) 0 nd p + q p q pq = (p q) (p + q) 0, we get the result. Fourth Solution. (sed on work y n IMO 000 ontestnt from Jpn) Putting =, it eomes ( + ) ( ( + ) + ) or Setting x =, it eomes f(x) 0, where f (t) = t + t t + t t t +. Fix positive numer. We need to show tht F (t) := f (t) 0 for ll t 0. It s esy to hek tht the ui polynomil F / (t) = t ( + )t ( + ) hs two rel roots nd λ = Sine F hs lol minimum t t = λ, we find tht F (t) Min {F (0), F (λ)} for ll t 0. We hve to prove tht F (0) 0 nd F (λ) 0. Sine F (0) = + = ( ) ( + ) 0, it remins to show tht F (λ) 0. Notie tht λ is root of F / (t). After long division, we get ( F (t) = F / (t) t + ) + ( ( )t ). 9 9 Putting t = λ, we hve F (λ) = ( ( )λ ). 9 Thus, our jo is now to estlish tht, for ll 0, ( ( ) ) , whih is equivlent to ( ) Sine oth nd re positive, it s equivlent to or ( ) ( ) ( ) or Let G(x) = 864x 4 75x + 50x 75x We prove tht G(x) 0 for ll x R. We find tht G / (x) = 456x 05x x 75 = (x )(456x 6669x + 75). Sine 456x 6669x + 75 > 0 for ll x R, we find tht G(x) nd x hve the sme sign. It follows tht G(x) is monotone deresing on (, ] nd monotone inresing on [, ). We onlude tht G(x) hs the glol minimum t x =. Hene, G(x) G() = 0 for ll x R. It s esy to hek tht = 6( + ) + + > 6( )( ) 0 nd = 8( ) + > 0.

22 Fifth Solution. (From the IMO 000 Short List) Using the ondition =, it s strightforwrd to verify the equlities = ( + ) ( + + ), = = ( + ) ( + + ), ( + ) ( + + ). In prtiulr, they show tht t most one of the numers u = +, v = +, w = + is negtive. If there is suh numer, we hve ( + ) ( + ) ( + ) = uvw < 0 <. And if u, v, w 0, the AM-GM inequlity yields = u + v uv, = v + w vw, = w + w wu. Thus, uv, vw, wu, so (uvw) =. Sine u, v, w 0, this ompletes the proof. It turns out tht the sustitution p = x + y + z, q = xy + yz + zx, r = xyz is powerful for the three vriles inequlities. We need the following lemm. Lemm. Let x, y, z e non-negtive rel numers numers. Set p = x + y + z, q = xy + yz + zx, nd r = xyz. Then, we hve () p 4pq + 9r 0, () p 4 5p q + 4q + 6pr 0, () pq 9r 0. Proof. They re equivlent to ( ) x(x y)(x z) + y(y z)(y x) + z(z x)(z y) 0, ( ) x (x y)(x z) + y (y z)(y x) + z (z x)(z y) 0, ( ) x(y z) + y(z x) + z(x y) 0. We leve the detils for the reders. Prolem 0. (Irn 996) Let x, y, z e positive rel numers. Prove tht ( ) (xy + yz + zx) (x + y) + (y + z) + (z + x) 9 4. First Solution. We mke the sustitution p = x + y + z, q = xy + yz + zx, r = xyz. Notie tht (x + y)(y + z)(z + x) = (x + y + z)(xy + yz + zx) xyz = pq r. One my esily rewrite the given inequlity in the terms of p, q, r : ( (p + q) ) 4p(pq r) q (pq r) 9 4 or 4p 4 q 7p q + 4q + 4pqr 9r 0 or pq(p 4pq + 9r) + q(p 4 5p q + 4q + 6pr) + r(pq 9r) 0. We find tht every term on the left hnd side is nonnegtive y the lemm. When does equlity hold in eh inequlity? For more p-q-r inequlities, visit the site [ESF]. See the theorem 0.

23 Prolem. Let x, y, z e nonnegtive rel numers with xy + yz + zx =. Prove tht x + y + y + z + z + x 5. First Solution. Rewrite the inequlity in the terms of p = x + y + z, q = xy + yz + zx, r = xyz: It n e rewritten s 4p 4 q + 4q 7p q 5r + 50pqr 0. pq(p 4pq + 9r) + q(p 4 5p q + 4q + 6pr) + 7r(pq 9r) + 8r 0. However, the every term on the left hnd side is nonnegtive y the lemm. Exerise 6. (Crlson s inequlity) Prove tht, for ll positive rel numers,,, ( + )( + )( + ) Exerise 7. (Bulgri 997) Let,, e positive rel numers suh tht =. Prove tht We lose this setion y presenting prolem whih n e solved y two lgeri sustitutions nd trigonometri sustitution. Prolem. (Irn 998) Prove tht, for ll x, y, z > suh tht x + y + z =, x + y + z x + y + z. First Solution. We egin with the lgeri sustitution = x, = y, = z. Then, the ondition eomes nd the inequlity is equivlent to = = Let p =, q =, r =. Our jo is to prove tht p + q + r where p + q + r + pqr =. By the exerise, we n mke the trigonometri sustitution [ p = os A, q = os B, r = os C for some A, B, C 0, π ) with A + B + C = π. Wht we need to show is now tht os A+os B+os C. However, it follows from Jensen s inequlity!

24 .4 Supplementry Prolems for Chpter Exerise 8. Let x, y, nd z e positive numers. Let p = x + y + z, q = xy + yz + zx, nd r = xyz. Prove the following inequlities : () p q () p 7r () q pr (d) p + 9r 7pq (e) p q + pr 4q (f) p q pr + q (g) p 4 + q 4p q (h) pq p r + qr (i) q + 9r 7pqr (j) q + 9r 4pqr (k) p r + q 6pqr Exerise 9. ([ONI], Mire Lsu, Mrin Tetiv) Let x, y, z e positive rel numers stisfying the ondition xy + yz + zx + xyz =. Prove tht () xyz 8, () x + y + z, () x + y + z 4(x + y + z), nd (4) x + y + (z ) z 4(x + y + z) z(z+), where z x, y. Exerise 0. Let f(x, y) e rel polynomil suh tht, for ll θ R, f(os θ, sin θ) = 0. Show tht the polynomil f(x, y) is divisile y x + y. Exerise. Let f(x, y, z) e rel polynomil. Suppose tht f(os α, os β, os γ) = 0, for ll α, β, γ R with α + β + γ = π. Show tht f(x, y, z) is divisile y x + y + z + xyz. 4 Exerise. (IMO Unused 986) Let,, e positive rel numers. Show tht ( + ) ( + ) ( + + ) ( + )( + )( + + ). 5 Exerise. With the usul nottion for tringle, verify the following identities : () sin A + sin B + sin C = s R () sin A sin B + sin B sin C + sin C sin A = s +4Rr+r 4R () sin A sin B sin C = sr R (4) sin A + sin B + sin C = s(s 6Rr r ) 4R (5) os A + os B + os C = (R+r) rs 4R 4R rs (6) tn A + tn B + tn C = tn A tn B tn C = s (R+r) (7) tn A tn B + tn B tn C + tn C tn A = s 4Rr r s (R+r) (8) ot A + ot B + ot C = s 4Rr r sr (9) sin A sin B sin C = r 4R (0) os A os B os C = s 4R 4 For proof, see [JmhMh]. 5 If we ssume tht there is tringle ABC with BC =, CA =, AB =, then it s equivlent to the inequlity s 4R + 4Rr + r in the exerise 6. 4

25 Exerise 4. Let,, e the lengths of the sides of tringle. Let s e the semi-perimeter of the tringle. Then, the following inequlities holds. () ( + + ) ( + + ) ( < 4( ) + + ) () [JfdWm] s + s () [AP] 8(s )(s )(s ) (d) [EC] 8 ( + )( + )( + ) (e) [AP] ( + )( + )( + ) 8( + + ) (f) [MC] ( + + )( + + ) ( ) (g) < (s ) + (s ) + (s ) (h) ( + ) + ( + ) + ( + ) 48(s )(s )(s ) (i) s + s + s 9 s (j) [AMN], [MP] < (k) 5 4 s+ + + s+ + + s+ + < 9 (l) [SR] ( + + ) 5[( + ) + ( + ) + ( + )] Exerise 5. ([RS], R. Sondt) Let R, r, s e positive rel numers. Show tht neessry nd suffiient ondition for the existene of tringle with irumrdius R, inrdius r, nd semiperimeter s is s 4 (R + 0Rr r )s + r(4r + r) 0. Exerise 6. With the usul nottion for tringle, show tht 4R + r s. 6 Exerise 7. ([WJB],[RAS], W. J. Blundon) Let R nd r denote the rdii of the irumirle nd inirle of the tringle ABC. Let s e the semiperimeter of ABC. Show tht s R + ( 4)r. Exerise 8. Let G nd I e the entroid nd inenter of the tringle ABC with inrdius r, semiperimeter s, irumrdius R. Show tht GI = ( s + 5r 6Rr ). 7 9 Exerise 9. Show tht, for ny tringle with sides,,, > It s equivlent to the Hdwiger-Finsler inequlity. 7 See the exerise 6. For solution, see [KWL]. 5

26 Chpter Homogeniztions. Homogeneous Polynomil Inequlities Mny inequlity prolems ome with onstrints suh s =, xyz =, x+y +z =. A non-homogeneous metri inequlity n e trnsformed into homogeneous one. Then we pply two powerful theorems : Shur s inequlity nd Muirhed s theorem. We egin with simple exmple. Prolem. (Hungry 996) Let nd e positive rel numers with + =. Prove tht Solution. Using the ondition + =, we n redue the given inequlity to homogeneous one, i. e., ( + )( + ( + )) + ( + )( + ( + )) or + +, whih follows from ( + ) ( + ) = ( ) (+) 0. The equlity holds if nd only if = =. The ove inequlity + + n e generlized s following : Theorem 9. Let,,, e positive rel numers suh tht + = + nd mx(, ) mx(, ). Let x nd y e nonnegtive rel numers. Then, we hve x y + x y x y + x y. Proof. Without loss of generlity, we n ssume tht,,. If x or y is zero, then it lerly holds. So, we lso ssume tht oth x nd y re nonzero. It s esy to hek x y + x y x y x y = x y ( x + y x y x y ) = x y ( x y ) ( x y ) ( = x y x y ) ( x y ) 0. Remrk. When does the equlity hold in the theorem 8? We now introdue two summtion nottions yli nd. Let P (x, y, z) e three vriles funtion of x, y, z. Let us define : P (x, y, z) = P (x, y, z) + P (y, z, x) + P (z, x, y), yli P (x, y, z) = P (x, y, z) + P (x, z, y) + P (y, x, z) + P (y, z, x) + P (z, x, y) + P (z, y, x) 6

27 For exmple, we know tht yli x y = x y + y z + z x, x = (x + y + z ) x y = x y + x z + y z + y x + z x + z y, xyz = 6xyz. Prolem 4. (IMO 984/) Let x, y, z e nonnegtive rel numers suh tht x + y + z =. Prove tht 0 xy + yz + zx xyz 7 7. Solution. Using the ondition x + y + z =, we redue the given inequlity to homogeneous one, i. e., 0 (xy + yz + zx)(x + y + z) xyz 7 7 (x + y + z). The left hnd side inequlity is trivil euse it s equivlent to 0 xyz + x y. The right hnd side inequlity simplifies to 7 yli x + 5xyz 6 x y 0. In the view of 7 yli x + 5xyz 6 x y = yli x x y + 5 xyz + yli x it s enough to show tht yli x x y nd xyz + yli x x y. Note tht yli x x y, x y = (x + y ) (x y + xy ) = (x + y x y xy ) 0. yli The seond inequlity n e rewritten s yli yli x(x y)(x z) 0, whih is prtiulr se of the theorem 0 in the next setion. yli 7

28 . Shur s Theorem Theorem 0. (Shur) Let x, y, z e nonnegtive rel numers. For ny r > 0, we hve x r (x y)(x z) 0. yli Proof. Sine the inequlity is metri in the three vriles, we my ssume without loss of generlity tht x y z. Then the given inequlity my e rewritten s (x y)[x r (x z) y r (y z)] + z r (x z)(y z) 0, nd every term on the left-hnd side is lerly nonnegtive. Remrk. When does the equlity hold in Theorem 0? The following speil se of Shur s inequlity is useful : x(x y)(x z) 0 xyz + yli yli x x y xyz + x Exerise 0. ([TZ], pp.4) Prove tht for ny ute tringle ABC, ot A + ot B + ot C + 6 ot A ot B ot C ot A + ot B + ot C. Exerise. (Kore 998) Let I e the inenter of tringle ABC. Prove tht IA + IB + IC BC + CA + AB. Exerise. ([IN], pp.0) Let,, e the lengths of tringle. Prove tht > We present nother solution of the prolem : (IMO 000/) Let,, e positive numers suh tht =. Prove tht ( + ) ( + ) ( + ). x y. Seond Solution. It is equivlent to the following homogeneous inequlity : ) ) ) ( () / + ( ()/ () / + ( ()/ () / + ()/. After the sustitution = x, = y, = z with x, y, z > 0, it eomes ) ) ) (x xyz + (y (xyz) xyz + (z (xyz) xyz + (xyz) x y z, whih simplifies to y z ( x y y z + z x ) ( y z z x + x y ) ( z x x y + y z ) x y z x or x y z + x 6 y x 4 y 4 z + x 5 y z yli yli yli or (x y)(y z)(z x) + yli(x y) (x y) (y z) whih is speil se of Shur s inequlity. For n lterntive homogeniztion, see the prolem in the hpter. 8

29 Here is nother inequlity prolem with the onstrint =. Prolem 5. (Tournment of Towns 997) Let,, e positive numers suh tht =. Prove tht Solution. We n rewrite the given inequlity s following : + + () + / + + () + / + + () / (). / We mke the sustitution = x, = y, = z with x, y, z > 0. Then, it eomes x + y + xyz + y + z + xyz + z + x + xyz xyz whih is equivlent to xyz (x + y + xyz)(y + z + xyz) (x + y + xyz)(y + z + xyz)(z + x + xyz) yli nd hene to x6 y x5 y z, whih is speil se of theorem in the next setion. 9

30 . Muirhed s Theorem Theorem. (Muirhed) Let,,,,, e rel numers suh tht 0, 0,, + +, + + = + +. Let x, y, z e positive rel numers. Then, we hve x y z x y z. Proof. Cse. : It follows from + nd from tht mx( +, ) so tht mx(, ) = mx( +, ). From + + = nd +, we hve mx( +, ) mx(, ). Apply the theorem 8 twie to otin x y z = z (x y + x y ) yli z (x + y + x y + ) yli = x (y + z + y z + ) yli x (y z + y z ) yli = x y z. Cse. : It follows from + + = tht + nd tht +. Therefore, we hve mx(, ) mx(, + ) nd mx(, + ) mx(, ). Apply the theorem 8 twie to otin x y z = x (y z + y z ) yli x (y z + + y + z ) yli = y (x z + + x + z ) yli y (x z + x z ) yli = x y z. Remrk. The equlity holds if nd only if x = y = z. However, if we llow x = 0 or y = 0 or z = 0, then one my esily hek tht the equlity holds if nd only if x = y = z or x = y, z = 0 or y = z, x = 0 or z = x, y = 0. We n use Muirhed s theorem to prove Nesitt s inequlity. (Nesitt) For ll positive rel numers,,, we hve Note the equlity in the finl eqution. However, in this se, we ssume tht 0 0 = in the sense tht lim x 0 + x 0 =. In generl, 0 0 is not defined. Note lso tht lim x x = 0. 0

31 Proof 6. Clering the denomintors of the inequlity, it eomes yli ( + )( + ) ( + )( + )( + ) or Prolem 6. ((IMO 995) Let,, e positive numers suh tht =. Prove tht Solution. It s equivlent to ( + ) + ( + ) + ( + ). ( + ) + ( + ) + ( + ) (). 4/ Set = x, = y, = z with x, y, z > 0. Then, it eomes yli denomintors, this eomes or ( x y x y + x y 9 z + x 9 y 9 z 6 x y 8 z 5 ) + ( x y 9 z x y 8 z 5 ) +. x 9 (y +z ) x 4 y 4 z. 4 x y 8 z 5 + 6x 8 y 8 z 8 ( nd every term on the left hnd side is nonnegtive y Muirhed s theorem. x 9 y 9 z 6 x 8 y 8 z 8 ) 0, We n lso ttk prolem 0 nd prolem with Shur s inequlity nd Muirhed s theorem. (Irn 996) Let x, y, z e positive rel numers. Prove tht ( ) (xy + yz + zx) (x + y) + (y + z) + (z + x) 9 4. Seond Solution. It s equivlent to 4 x 5 y + x 4 yz + 6x y z x 4 y 6 x y x y z 0. yli yli We rewrite this s following ( x 5 y x 4 y ) + ( x 5 y ) x y + xyz x(x y)(x z) 0. By Muirhed s theorem nd Shur s inequlity, it s sum of three terms whih re nonnegtive. yli Let x, y, z e nonnegtive rel numers with xy + yz + zx =. Prove tht x + y + y + z + z + x 5. Seond Solution. Using xy + yz + zx =, we homogenize the given inequlity s following : Clering or ( (xy + yz + zx) x + y + y + z + ) z + x 4 x 5 y + x 4 yz + 4 x y z + 8x y z ( ) 5 x 4 y + x y

32 or ( x 5 y x 4 y ) + ( x 5 y x y ) + xyz ( x + 4 x y + 8xyz By Muirhed s theorem, we get the result. In the ove inequlity, without the ondition xy + yz + zx =, the equlity holds if nd only if x = y, z = 0 or y = z, x = 0 or z = x, y = 0. Sine xy + yz + zx =, the equlity ours when (x, y, z) = (,, 0), (, 0, ), (0,, ). Now, we pply Muirhed s theorem to otin geometri inequlity [ZsJ] : Prolem 7. If m,m,m re medins nd r,r,r the exrdii of tringle, prove tht r r m m + r r m m + r r m m. An Impossile Verifition. Let s = + +. Using the well-known identities s(s )(s ) r =, m = + s, et. we hve yli r r = m m yli Applying the AM-GM inequlity, we otin yli r r m m yli We now give moonshine proof of the inequlity 4s(s ) ( + )( + ). 8s(s ) ( + ) + ( + ) = yli ( + + )( + ) yli ) ( + + )( + ) After expnding the ove inequlity, it eomes yli yli yli We see tht this nnot e diretly proven y pplying Muirhed s theorem. Sine,, re the sides of tringle, we n mke the Rvi Sustitution = y + z, = z + x, = x + y, where x, y, z > 0. After some rute-fore lger, we n rewrite the ove inequlity s yli 5 x x 5 y + 5 x 4 y + 0 x y + 80 x 4 yz yli 0. Now, y Muirhed s theorem, we get the result! 6 x y z + 4 x y z.

33 .4 Polynomil Inequlities with Degree The solution of prolem shows us diffiulties in pplying Muirhed s theorem. Furthermore, there exist homogeneous metri polynomil inequlities whih nnot e verified y just pplying Muirhed s theorem. See the following inequlity : 5 x 5 y + 8 x 4 yz + 6 x y yli x x 4 y + x y z + x y z 8 This holds for ll positive rel numers x, y, nd z. However, it is not diret onsequene of Muirhed s theorem euse the oeffiients of x5 y nd yli x y re too ig. In ft, it is equivlent to Another exmple is 6 yli (y z) 4 (x + 5y + 5z + 8xy + 4yz + 8zx) 0. 4 yli x 4 + x y x y. yli yli We relized tht the ove inequlity is stronger thn x (x y)(x z) 0 or x 4 + yli yli yli x y It n e proved y the identities 4 x 4 + x y x y = (x y) 4 + (y z) 4 + (z x) 4 yli yli or x y. x 4 + x y x y = (x + y + z xy yz zx). yli yli As I know, there is no generl riterion to ttk the metri polynomil inequlities. However, there is result for the homogeneous metri polynomil inequlities with degree. It s diret onsequene of Muirhed s theorem nd Shur s inequlity. Theorem. Let P (u, v, w) R[u, v, w] e homogeneous metri polynomil with degree. Then the following two sttements re equivlent. () P (,, ), P (,, 0), P (, 0, 0) 0. () P (x, y, z) 0 for ll x, y, z 0. Proof. (See [SR].) We only prove tht () implies (). Let P (u, v, w) = A yli u + B u v + Cuvw. Let p = P (,, ) = A+6B +C, q = P (,, 0) = A+B, nd r = P (, 0, 0) = A. We hve A = r, B = q r, C = p 6q + r, nd p, q, r 0. Let x, y, z 0. It follows tht P (x, y, z) = r x y + (p 6q + r)xyz. However, we see tht P (x, y, z) = r yli x + (q r) yli x + xyz x y + q yli ( ) x y 6xyz + pxyz 0. 4 See [JC].

34 Here is n lterntive wy to prove of the ft tht P (x, y, z) 0. Cse. q r : We find tht P (x, y, z) = r ( x xyz ) + (q r) nd tht the every term on the right hnd side is nonnegtive. Cse. q r : We find tht P (x, y, z) = q ( x ) xyz + (r q) nd tht the every term on the right hnd side is nonnegtive. ( x y xyz yli x + xyz ) + pxyz. x y + pxyz. For exmple, we n pply the theorem to hek the inequlity in the prolem 4. (IMO 984/) Let x, y, z e nonnegtive rel numers suh tht x + y + z =. Prove tht 0 xy + yz + zx xyz 7 7. Solution. Using x + y + z =, we homogenize the given inequlity s following : Let us define L(u, v, w), R(u, v, w) R[u, v, w] y 0 (xy + yz + zx)(x + y + z) xyz 7 (x + y + z) 7 L(u, v, w) = (uv + vw + wu)(u + v + w) uvw, R(u, v, w) = 7 7 (u + v + w) (uv + vw + wu)(u + v + w) + uvw. However, one my esily hek tht L(,, ) = 7, L(,, 0) =, L(, 0, 0) = 0, R(,, ) = 0, R(,, 0) = 7, nd R(, 0, 0) = 7 7. Exerise. (M. S. Klmkin [MEK]) Determine the mximum nd minimum vlues of x + y + z + λxyz where x + y + z =, x, y, z 0, nd λ is given onstnt. Exerise 4. (Wlter Jnous [MC]) let x, y, z 0 with x + y + z =. For fixed rel numers 0 nd, determine the mximum = (, ) suh tht + xyz (xy + yz + zx). Here is the riterion for homogeneous metri polynomil inequlities for the tringles : Theorem. (K. B. Stolrsky) Let P (u, v, w) e rel metri form of degree. 5 If P (,, ), P (,, 0), P (,, ) 0, then we hve P (,, ) 0, where,, re the lengths of the sides of tringle. Proof. Mke the Rvi sustitution = y + z, = z + x, = x + y nd pply the ove theorem. We leve the detils for the reders. For n lterntive proof, see [KBS]. As noted in [KBS], we n pply Stolrsky s theorem to prove ui inequlities in tringle geometry. We rell the exerise. 5 P (x, y, z) = ( px + qx y + rxyz ) (p, q, r R.) 4

35 Let,, e the lengths of the sides of tringle. Let s e the semi-perimeter of the tringle. Then, the following inequlities holds. () ( + + ) ( + + ) ( < 4( ) + + ) () [JfdWm] s + s () [AP] 8(s )(s )(s ) (d) [EC] 8 ( + )( + )( + ) (e) [AP] ( + )( + )( + ) 8( + + ) (f) [MC] ( + + )( + + ) ( ) (g) < (s ) + (s ) + (s ) (h) ( + ) + ( + ) + ( + ) 48(s )(s )(s ) (i) s + s + s 9 s (j) [AMN], [MP] < (k) 5 4 s+ + + s+ + + s+ + < 9 (l) [SR] ( + + ) 5[( + ) + ( + ) + ( + )] Proof. For exmple, we show the right hnd side inequlity in (j). It s equivlent to the ui inequlity T (,, ) 0, where ( T (,, ) = ( + )( + )( + ) ( + )( + )( + ) ). + Sine T (,, ) = 4, T (,, 0) = 0, nd T (,, ) = 6, the result follows from Stolrsky s theorem. lterntive proofs of the ove inequlities, see [GI]. For 5

36 .5 Supplementry Prolems for Chpter Exerise 5. Let x, y, z e positive rel numers. Prove tht (x + y z)(x y) + (y + z x)(y z) + (z + x y)(z x) 0. Exerise 6. Let x, y, z e positive rel numers. Prove tht (x + y z )(x y) + (y + z x )(y z) + (z + x y )(z x) 0. Exerise 7. (APMO 998) Let,, e positive rel numers. Prove tht ( + ) ( + ) ( + ) ( ). Exerise 8. (Irelnd 000) Let x, y 0 with x + y =. Prove tht x y (x + y ). Exerise 9. (IMO Short-listed 998) Let x, y, z e positive rel numers suh tht xyz =. Prove tht x ( + y)( + z) + y ( + z)( + x) + z ( + x)( + y) 4. Exerise 40. (United Kingdom 999) Some three nonnegtive rel numers p, q, r stisfy p + q + r =. Prove tht 7(pq + qr + rp) + 9pqr. 6

37 Chpter 4 Normliztions 4. Normliztions In the previous hpter, we trnsformed non-homogeneous inequlities into homogeneous ones. On the other hnd, homogeneous inequlities lso n e normlized in vrious wys. We offer two lterntive solutions of the prolem 8 y normliztions : (IMO 00/) Let,, e positive rel numers. Prove tht Seond Solution. We mke the sustitution x = ++, y = ++, z = xf(x + 8yz) + yf(y + 8zx) + zf(z + 8xy), ++. The prolem is where f(t) = t. Sine the funtion f is onvex down on R + nd x + y + z =, we pply (the weighted) Jensen s inequlity to hve xf(x + 8yz) + yf(y + 8zx) + zf(z + 8xy) f(x(x + 8yz) + y(y + 8zx) + z(z + 8xy)). Note tht f() =. Sine the funtion f is stritly deresing, it suffies to show tht x(x + 8yz) + y(y + 8zx) + z(z + 8xy). Using x + y + z =, we homogenize it s (x + y + z) x(x + 8yz) + y(y + 8zx) + z(z + 8xy). However, this is esily seen from (x + y + z) x(x + 8yz) y(y + 8zx) z(z + 8xy) = [x(y z) + y(z x) + z(x y) ] 0. In the ove solution, we normlized to x + y + z =. We now prove it y normlizing to xyz =. Third Solution. We mke the sustitution x =, y =, z =. Then, we get xyz = nd the inequlity eomes x + 8y + 8z whih is equivlent to ( + 8x)( + 8y) ( + 8x)( + 8y)( + 8z) yli Dividing y + + gives the equivlent inequlity yli ++. (++) + 8 (++) 7

38 hene, fter squring oth sides, equivlent to 8(x + y + z) + ( + 8x)( + 8y)( + 8z) + 8x 50. Rell tht xyz =. The AM-GM inequlity gives us x + y + z, ( + 8x)( + 8y)( + 8z) 9x 8 9 9y 8 9 9z 8 9 = 79 nd Using these three inequlities, we get the result. We now present nother proofs of Nesitt s inequlity. (Nesitt) For ll positive rel numers,,, we hve yli yli x yli 9x 8 9 9(xyz) 4 7 = 9. Proof 7. We my normlize to + + =. Note tht 0 <,, <. The prolem is now to prove yli + = yli f(), where f(x) = x x. Sine f is onve down on (0, ), Jensen s inequlity shows tht ( ) ( ) + + f() f = f = or f(). yli Proof 8. As in the previous proof, we need to prove yli yli, where + + =. It follows from 4x ( x)(9x ) = (x ) or 4x ( x)(9x ) tht yli yli 9 4 = =. yli 8

39 4. Clssil Theorems : Cuhy-Shwrtz, (Weighted) AM-GM, nd Hölder We now illustrte the normliztion tehnique to estlish lssil theorems. Theorem 4. (The Cuhy-Shwrtz inequlity) Let,, n,,, n e rel numers. Then, we hve ( + + n )( + + n ) ( + + n n ). Proof. Let A = + + n nd B = + + n. In the se when A = 0, we get = = n = 0. Thus, the given inequlity lerly holds. So, we now my ssume tht A, B > 0. Now, we mke the sustitution x i = i A (i =,, n). Then, it s equivlent to (x + + x n )( + + n ) (x + + x n n ). However, we hve x + + x n =. (Why?). Hene, it s equivlent to Next, we mke the sustitution y i = i B Hene, we need to to show tht + + n (x + + x n n ). (i =,, n). Then, it s equivlent to = y + + y n (x y + + x n y n ) or x y + + x n y n. x y + + x n y n, where x + + x n = y + + y n =. However, it s very esy. We pply the AM-GM inequlity to dedue x y + + x n y n x y + + x n y n x + y + + x n + y n = A + B =. Exerise 4. Prove the Lgrnge s identity : n i= i ( n n ) i i i = i= i= i<j n ( i j j i ). Exerise 4. Let,, n,,, n e positive rel numers. Show tht ( + + n )( + + n ) + + n n. Exerise 4. Let,, n,,, n e positive rel numers. Show tht + + n ( + + n ). n + + n Exerise 44. Let,, n,,, n e positive rel numers. Show tht + + ( n + + ) n. n + + n n Exerise 45. Let,, n,,, n e positive rel numers. Show tht + + n n ( + + n ) + + n n. We now pply the Cuhy-Shwrtz inequlity to prove Nesitt s inequlity. 9

40 (Nesitt) For ll positive rel numers,,, we hve Proof 9. Applying the Cuhy-Shwrtz inequlity, we hve ( (( + ) + ( + ) + ( + )) ). + It follows tht Proof 0. The Cuhy-Shwrtz inequlity yields ( + ) + yli yli yli Here is n extremely short proof of the prolem : or 9 or + yli yli (Irn 998) Prove tht, for ll x, y, z > suh tht x + y + z =, x + y + z x + y + z ( + + ) + ( + + ). Seond Solution. We notie tht x + y + z = x + y + z =. x y z We now pply the Cuhy-Shwrtz inequlity to dedue ( x x + y + z = (x + y + z) + y + z ) x + y + z. x y z Prolem 8. (Gzet Mtemtiã, Hojoo Lee) Prove tht, for ll,, > 0, Solution. We otin the hin of equlities nd inequlities = ( ) ( ) yli yli ( ) (Cuhy Shwrtz) yli ( = ) yli ( ) ( ) (AM GM) yli 4 + (Cuhy Shwrtz) yli = 4 +. yli 40

41 Using the sme ide in the proof of the Cuhy-Shwrtz inequlity, we find nturl generliztion : Theorem 5. Let ij (i, j =,, n) e positive rel numers. Then, we hve ( n + + n n ) ( n n + + nn n ) ( n + + n n nn ) n. Proof. Sine the inequlity is homogeneous, s in the proof of the theorem, we n normlize to ( i n + + in n ) n = or i n + + in n = (i =,, n). Then, the inequlity tkes the form n + + n n nn or n i= i in. Hene, it suffies to show tht, for ll i =,, n, i in n, where i + + in =. To finish the proof, it remins to show the following homogeneous inequlity : Theorem 6. (AM-GM inequlity) Let,, n e positive rel numers. Then, we hve + + n n ( n ) n. Proof. Sine it s homogeneous, we my resle,, n so tht n =. Hene, we wnt to show tht n = = + + n n. The proof is y indution on n. If n =, it s trivil. If n =, then we get + = + = ( ) 0. Now, we ssume tht it holds for some positive integer n. And let,, n+ e positive numers suh tht n n+ =. We my ssume tht. (Why?) Sine ( ) n =, y the indution hypothesis, we hve n+ n. Thus, it suffies to show tht + +. However, we hve + = ( )( ) 0. The following simple oservtion is not triky : Let, > 0 nd m, n N. Tke x = = x m = nd x m+ = = x xm+n =. Applying the AM-GM inequlity to x,, x m+n > 0, we otin m + n m + n (m n ) m+n or m m + n + n m + n m n m+n m+n. Hene, for ll positive rtionls ω nd ω with ω + ω =, we get We immeditely hve ω + ω ω ω. Theorem 7. Let ω, ω > 0 with ω + ω =. Then, for ll x, y > 0, we hve ω x + ω y x ω y ω. Proof. We n hoose positive rtionl sequene,,, suh tht And letting i = i, we get From the previous oservtion, we hve lim n = ω. n lim n = ω. n n x + n y x n y n Now, tking the limits to oth sides, we get the result. i Set x i = ( n ) n (i =,, n). Then, we get x x n = nd it eomes x + + x n n. 4