RMM - Cyclic Inequalities Marathon 1-100

Transcription

1 RMM - Cyclic Inequlities Mrthon - 00 ROMANIAN MATHEMATICAL MAGAZINE Avilble online Founding Editor DANIEL SITARU ISSN-L

2 RMM CYCLIC INEQUALITIES MARATHON 00

3 Proposed by Dniel Sitru Romni, George Apostolopoulos - Messolonghi - Greece Nguyen Viet Hung Hnoi Vietnm, Bbis Stergioiu Greece Erdene Ntsgdorj Ulnbtr-Mongoli, Abhy Chndr-Indi Dorin Mrghidnu Romni, Vggelis Stmtidis-Greece Le Viet Hung Hi Lng-Vietnm, Sldjn Stnkovic-Mcedoni Richdd Phuc-Hnoi-Vietnm, Hong Le Nht Tung Hnoi Vietnm Anish Ry-Sntrgchi-Indi, Mrin Chirciu Romni Nguyen Phuc Tng - Dong Thp Vietnm, Mihlce Andrei Ștefn Romni Adil Abdullyev Bku Azerbidin

4 Solutions by Dniel Sitru Romni, Henry Ricrdo - New York USA Kunihiko Chiky Tokyo Jpn, Imd Zk Sid Lebnon Ngoc Minh Ngoc Bo-Gi Ln Province-Vietnm, Fotini Kldi Greece Redwne El Mellss-Csblnc-Morroco, Sk Rejun-West Bengl-Indi Mygmrsuren Ydmsuren Mongoli, Sfl Ds Bisws Chinsurh Indi Kevin Soto Plcios Hurmey-Peru, Nguyen Viet Hung Hnoi Vietnm Soumitr Mndl-Chndr Ngore-Indi, Abhy Chndr Indi Mrin Dincă Romni, Rvi Prksh - New Delhi Indi Leonrd Giugiuc-Romni, Seyrn Ibrhimov Msilli Azerbidin Abdllh El Frisi-Bechr-Algerie, Anh Ti Trn Hnoi Vietnm Le Viet Hung- Qung Tri Vietnm, Nirpd Pl-Indi, Soumv Chkrborty Kolkt Indi, Ans Adlny-El Jdid-Morroco, Phn Loc So'n-Quy - Nhon City Vietnm, Sptk Bhttchry-Kolkt-Indi, Nguyen Minh Triet-Qung Ngi-Vietnm Mihlce Andrei Stefn-Romni, Soumitr Mndl-Chndr Ngore-Indi Hong Le Nht Tung Hnoi Vietnm, Khung Long Xnh - D Nng Vietnm Nguyen Phuc Tng - Dong Thp Vietnm, Sldjn Stnkovic Mcedoni Adity Nryn Shrm Knchrpr Indi, Richdd Phuc Hnoi Vietnm Phm Quy Qung Ngi- Vietnm, Hmz Mhmood-Lhore-Pkistn

5 . Given, b, c, d be positive rel number such tht 4 Prove tht: b bc cd d b c d. Proposed by Sldjn Stnkovic-Mcedoni Solution by Ngoc Minh Ngoc Bo-Gi Ln Province-VietNm Without loss of generlity, ssume tht: b c d. Considering function: f, b, c, d b bc cd d b c d We hve: f.c f, b, c, d f c, b,.c, d 4.c b 4.c d.c c, b,, d c c c We need to prove tht: c.c b d 0 b d b d Use AM-GM inequlity: 4 b c d 4. 4 bcd bcd c And b d c c.c c > 0 b d c c We prove tht: f t, t, t, d t td t d 0. With t d 4. Indeed t dt t d t7 6748t.t.8 t 47t t7 t 47t. Ming t 0 (With g t 6 48t t 8, nd Ming t g > 0) 4

6 . If x, y, z, t,, b, c 0,, xyzt 4 then: Solution by Dniel Sitru-Romni x b y c y z t 4 b c Solution by Mygmrsuren Ydmsuren Mongoli Proposed by Dniel Sitru Romni Solution by Dniel Sitru-Romni x y z t 4 4 xyzt (AM-GM) x b y z t x b 4 b7 x x b7 4 b7 4 bf 4 bf 4bF bf 4 bf 4 bf () y c y z t x c 4 c7 x x 47 4 c7 4 cf 4 cf 4cF cf 4 cf 4 cf By dding (); (): () x y c y z t 4 b7 c7 Solution by Mygmrsuren Ydmsuren Mongoli x, y, z, t,, b 0, xyzt 4 Prove tht: LHS x b.y c y.z.t 4 b c 5

7 x b y c Chebyshev y z t 4 xb x c x y z Cuchy7Schwrz 4 xb x c 4 x y z t 4 x b x c x y z t Chebyshev 4 4 x y z t x b7 x c7 x y z t Cuchy xb7 x c7 4 4 xyzt b7 4 c7 xyzt LHS 4 b7 c7 x b y c y z t 4 b7 c7 4 b c. Prove tht for ny rel numbers, b nd c the following inequlity holds: 8 bc b c c b b c Solution by Leonrd Giugiuc-Romni Proposed by Richdd Phuc-Vietnm bc b c c b p b c b b c c, where p bc. But b c b c b c b c bc p nd b b c c b bc c p b c b c bc p Cse : b c 0 b c p nd b b c c b bc c p 6

8 8 bc b c c b 8 b bc c. But b c b c bc b c 89 b bc c So ctully we ve got the identity 8 bc b c c b b c, hence proved. Cse : b c 0. Assume WLOG b c b bc c t, t 0. We get: b c t, b c p 9t p nd b b c c 7 t 7p t p. Hence we need to prove 8 t 8p t. It s well known tht p t t. It s enough to show 8 t 8 t t t 6t 4 6t 6t 0 t 8t 8t 0, which is clerly true since t 8t 8t 0 nd > 0. Remrk: From bove, if b c 0 we hve equlity nd if b c 0 inequlity is strict. 4. If, b, c 0, then: b 5 b 5 b c b 5 c 5 c c 5 5 b b c c Proposed by Dniel Sitru Romni 7

9 Solution by Kevin Soto Plcios Hurmey-Peru b 5 b 5 b Si:, b, c < 0, >. Probr que: b c b 5 c 5 b 5 b 5 c c 5 5 b L desiguldd es equivlente: b c b c b 5 c 5 c Tener presente lo siguiente: b c c c 5 0 c5 5 b 5 b 4 b b b b 4 (Cocientes Notbles) Se: A b 5.b 5 7.b b.b 5.b 5 5.b 5.b A b 4 b 4 b b 5 b 5 b 4 b 4 b b 5 b 5 Desde que:, b, c < 0, >. Por: MA MG b 4 4 b b 4 b 4 b 4 b 4 4b Sumndo: 4 b 4 b b 4 b 4 b b 0 Por l tnto: A 4.b 4 7b.b b 5.b 5 b 5 0 b If, b, c 0, then: 8 b 8 b 8 c 8 c b 5 c b 4 c 4 Proposed by Dniel Sitru Romni 8

10 Solution by Kevin Soto Plcios-Hurmey-Peru Si:, b, c < 0, >. Probr que: 8 b 8 b 8 c 8 c b 5 c 5 Desde que:, b, c < 0, > b 4 c 4 8 b 8 b 8 c 8 c b 4 c 4 4 b 4 c 4 Por lo cul nos flt probr que: 4 b 4 c 4 4 b 4 c 4 5 b 5 c b 4 c 4 4 b 4 c 4 bc b 4 c 4 4 b 4 c 4 4 bc b 4 c 4 4 b 4 c 4 7 bc 4 Válido por: (MAMG). Por l tnto: 8 b 8 b 8 c 8 c b 5 c 5 L iguldd se lcnz cundo: b c 6. Let, b, c be positive rel numbers such tht: Prove tht b 4 c. b c Solution by Kevin Soto Plcios Hurmey Peru Sen:, b, c números reles positivos tl que : 7 4 b 4 c 4 Proposed by Nguyen Viet Hung Hnoi Vietnm Probr que: b c bc (B) Se: x.y z > 0, b z.x y (A)..b.c y.z > 0, c > 0 x 9

11 x y z x y z z y x b c x y z yz Reemplzndo en (A) z x y z Por l tnto (B) es equivlente: x y z y x y z x x y z z x y z y x x y z z x y y z x xyz x y z z x y y z x x y z y z x x y z yz y z x y z x x y x z x y y z y x z y z z x y z yz xyz x y z z x y y z x x y z xy x y yz y z zx z x xyz xyz x y z xyz xy x y yz y z zx z x 7. If, b, c, then prove tht: b (Válido por desiguldd de Schur) c 4 b Solution by Hung Nguyen Viet Hnoi Vietnm 4 bc 4 c Proposed by Erdene Ntsgdorj - Ulnbtr Mongoli We hve known tht if x 0, y 0 nd xy then x y xy 0

12 Applying this result step by step we obtin b b b b b b Similrly b c 4 bc 4, c 4 4 c b b Adding up these three reltions we get the desired inequlity. 8. Prove the inequlity holds for ll positive rel numbers, b, c b c c b b c bc Solution by Kevin Soto Plcios Hurmey Peru b4.c 4 b c4. 4 c 4.b 4 Proposed by Nguyen Viet Hung Hnoi Vietnm Siendo:, b, c números R.. Probr l siguiente desiguldd: b c b c c b bc b4.c 4 Prtimos de l siguiente desiguldd: b b 4 6 b 4b 4 b 0 4 b 4 b b b b b b 4 b 4 b b Análogmente pr los siguientes términos: b c4. 4 c 4.b 4 4 b 4 4.b 4 (A)

13 b bc c b4.c 4 (B) c c c4. 4 (C) En consecuenci, l desiguldd inicil es equivlente: bc b4 c 4 b c4 4 c 4 b 4 bc b bc c b c c c b b bc b4 c 4 b c4 4 c 4 b 4 b c b c c b (LQQD) 9. If, b, c, x, y, z 0, b c then: x y b z c x b y c z x c y z b Solution by Soumitr Mndl-Chndr Ngore-Indi 9 x y z Proposed by Dniel Sitru Romni x y b z c x y b z c (Bergstrom s inequlity) x by cz b c.b.c 9 x by cz 9 x y z b c

14 9 x y z (Proved) 0. Let, b, c be positive rel numbers such tht b bc c. Prove tht b bc c b c c c b b Proposed by Nguyen Viet Hung Hnoi Vietnm Solution by Kevin Soto Plcios Hurmey Peru Sen, b, c números R. tl que: b bc c. Probr que: b bc c b c b c b c c L desiguldd es equivlente: b c c c b b c b b Pr todos los R. :, b, c, x, y, z, se cumple l siguiente desiguldd: b c x c y b z b bc c xy yz zx xy yz zx (Demostrdo nteriormente) Sen: x b c b.c, y b b c.c Asimismo: b c, z c.b c b Sen: m, n b, p c c b b c xy yz zx mn p n p m np m p m n pm n m n p 4

15 b bc c (Demostrdo nteriormente) b c c Por l tnto: c xy yz zx b b 9 4. Let, b, c be positive rel numbers such tht b c. Prove tht: b bc c bc b c c b b c Proposed by Nguyen Viet Hung Hnoi - Vietnm Solution by Kevin Soto Plcios Hurmey Peru Siendo, b, c números R., de tl mner que: b c. Probr que: b bc c bc b c c b b c bc b c bc b c c b b c c c c b c b c b b b b b c c L desiguldd se puede expresr de l siguiente mner: bc b c c b.bc.bc b.c c.b b c bc b.c b.c.bc c.b c b b c c b c.b.bc c.b b.c b c b c 4

16 b c b c c b c b c b.c b c b c c b c b c b.c b c c b c b c b c c b c c c c b b c c b b c c b b b c b.b.b c b 0 b c b c b 0 b c b c c b b c b b c b c b b c c 0 c bc c b b c b bc b b c b b c c c 0 c c b c c b b c b b b b c b c b c c b c 0 c b c c c c b b c b b b b b c b b c b c b c c b c c 0 0 5

17 c b b b c.c.b b.c b c b c c 0 (Lo cul probremos) Desde que, b, c > 0: Por: MA MG b 4.b 4 b c b.c 4 c.c (A) (B) (C) Sumndo: (A) (B) (C) b c.c.b. Let, b, c be non-negtive rel numbers such tht b c. Prove tht: b.c b c 6 b c Solution by Kevin Soto Plcios Hurmey Peru (LQQD) Proposed by Nguyen Viet Hung Hnoi Vietnm Siendo:, b, c números reles no negtivos, de tl mner que: b c. Probr que: b c 6 b c L desiguldd se puede expresr como: b b b c c c 6 b c Por l desiguldd de Cuchy Schwrz: b b b c c c b c b c b c 6

18 b b b c c c 6 b c b c 6 b c... (LQQD). Prove tht for ll positive rel numbers, b, c : b b b c bc Solution by Kevin Soto Plcios Hurmey Peru c c 8 b c Proposed by Nguyen Viet Hung Hnoi Vietnm Probr pr todos los números R., b, c l siguiente desiguldd: b b c c 8 b c b bc c Relizmos los siguientes cmbios de vribles: p b c q b bc c Desde que:, b, c > 0 Por l desiguldd de Bergstrom s: b b b c bc c c b q 8p Por lo cul nos qued demostrr que: b 8p q p q 4p 8pq 9 4p q p 4pq 6q 4p 8pq 7

19 4p q 9 4pq p 6q 0 Pero... 4p q 9 4pq p 6q p q 0... (LQQD) 4. Prove tht for ll rel numbers, b, c: b c b c Solution by Abhy Chndr Indi b c Solution by Mrin Dincă Romni b c c b Proposed by Nguyen Viet Hung Hnoi Vietnm Solution by Abhy Chndr Indi For rel numbers we hve b c b c which implies b c b c b c b c Hence we re required to prove the following for ll non-negtive rel numbers x, y b, z c x y z x y z x y z But from Cuchy Schwrz on LHS we hve LHS x y z x y z xy yz zx we re left to prove x y z xy yz zx x y z 0 which is obviously true. Equlity t x y z 0 or b c 0. Solution by Mrin Dincă Romni 8

20 b c b c b c b c c b c b b c b c c b b c 9.9. b. c.. b. c 9 ciclic.. b. b. c.. b. c use hrmonic inequlity. b. c.. b. c b c b c becuse the function f x 5. If, b, c, d 0,, prove tht: x.x b c b c b c b c done!, is incresing for x 0 nd b c.d b c d. c d.b d b.c > Proposed by Dorin Mrghidnu Romni Solution by Mrin Dincă Romni, b, c, d 0, b c.d > use Lem by Dorin Mrghidnu x y x, for x, y 0, x.y 9

21 b c d b 0,, 0, c.d b b nd similrly, we shll obtin: c d b b c d b c.d b b c d 6. If, b, c 0, then: 6 bc Solution by Soumitr Mndl - Chndr Ngore Indi Applying Holder s Inequlity bc Proposed by Dniel Sitru Romni b b c c b c b c bc bc (proved) 7., b, c > 0 b bc c b c.b. b.c. c.. 0

22 Solution by Nguyen Viet Hung Hnoi Vietnm By Cuchy Schwrz inequlity we hve Proposed by Vggelis Stmtidis-Greece b b c b b c b c b c b c b c b c It suffices to show tht b c b c b c or b c (since b c b bc c) But this is true by b c b bc c b c The proof is completed. 8. If, b, c 0,, b c then: 5b 7c b 5c 7 Solution by Mygmrsuren Ydmsuren Mongoli I Chebyshev I 5b 7c 7 b c b 5c 7 c 5 7b 6 Proposed by Dniel Sitru Romni c 5 7b 6 5b 7c 5c 7 5 7b

23 7 9. If, b, c 0, then: 5 7b 5b 7c 5c 7 7 b c 6 Cuchy7Schwrz b b c c b c c b b c Proposed by Dniel Sitru - Romni Solution by Rvi Prksh - New Delhi Indi Solution by Seyrn Ibrhimov Msilli Azerbidin Solution by Rvi Prksh - New Delhi Indi b b b c c b c c b c b b b b b b b b b b b 0 Thus, b b b () Similrly b c c b b c () nd

24 c c c () Adding (), () nd () we get the desired inequlity. Solution by Seyrn Ibrhimov Msilli Azerbidin b b c c b c c b b c bc c c b b b c AM GM c b b c bc c b 4 c c b b c 4 bc b c 4 b LHS bc c b b c b c 0. If, b, c 0, nd m, n N, prove tht: m.n bm.n b m cn7 c m n7 cm.n m b c bn7 Solution by Soumitr Mndl - Chndr Ngore Indi Proposed by Dorin Mărghidnu Romni Applying A.M. G.M.

25 mun b m cnf.mb. n7 c m.n b mun c m nf.mc. n7 m.n b nd now,, similrly c mun m bnf.m. n7 b m.n c m.n b m c n7 l m l n l m n l so,. If, b, c, then: b c b c m.n b m c n7 l b c (proved) c b bc Solution by Kevin Soto Plcios Hurmey Peru Solution by Rvi Prksh - New Delhi Indi Solution by Kevin Soto Plcios Hurmey Peru b c b c c b Si:, b, c <, > c b bc Por desiguldd de Cuchy: b c c b Proposed by Dniel Sitru Romni b c c b c b 4 ( b c 4

26 4 7bc 7bc... (I) b 7c b b 7 7c... (II) c 7 c c 7b 7b... (III) Sumndo... (I)(II)(III):. b 7c b. c 7 c. 7b b c... (LQQD) 7bc 7c 7b Solution by Rvi Prksh - New Delhi Indi Rewrite the inequlity s: b c b c c b bc c c b For < b, c < b b c bc b bc c bc b b c b bc c bc c bc b c bc b c c b b c b c b 0 c Thus, b c bc Similrly for the second nd third expressions on both sides.. If, b, c > 0 then: 4 c e b b 4 e b c c 4 b e c e b c 5

27 Solution Abdllh El Frisi-Bechr-Algerie Solution by Mygmrsuren Ydmsuren Mongoli Solution by Sfl Ds Bisws Chindhr Indi Solution 4 by Soumitr Mndl - Chndr Ngore Indi Solution Abdllh El Frisi-Bechr-Algerie Proposed by Dniel Sitru Romni The function x e x is convex for x 0 then we hve for ll, b, c > 0. b e b b c b c b c e b c by MA-MG c e b. b c. c Solution by Mygmrsuren Ydmsuren Mongoli 4 c e b b 4 c e b c c 4 b c c e c b e AM7GM 6 b 6 c 6 e b.b c.c b c e b.b c.c AM7GM b c e b b c c Solution by Sfl Ds Bisws Chindhr Indi e b c 4 c e b b 4 e b c c 4 b e c k. Then by A.M G.M we hve k 6 b 6 c 6 e b.b c.c b c e b ub c uc gin A.M G.M b.b c.c k e b c then e b ub c uc e, thus 6

28 Solution 4 by Soumitr Moukherjee - Chndr Ngore Indi Applying A.M G. M, 4 c e b 4 c e b b c e b e b c (proved). Let, b, c be rel numbers such tht min b, b c, c > 0. Prove tht b b b bc c b bc c c c c c. b b Solution by Leonrd Giugiuc Romni Proposed by Nguyen Viet Hung Hnoi Vietnm We hve: b 0 b b Similrly, b b > 0. c c c c > 0 nd b bc c b bc c > 0. So tht: b b b bc c b b b bc c b b b bc c 4. If, b, c 0 then: b 7c b c 7 c 7b b c Proposed by Dniel Sitru Romni Solution by Abdllh El Frisi-Bechr Algerie Solution by Rvi Prksh - New Dehi Indi 7

29 Solution by Soumitr Mndl - Chndr Ngore Indi Solution by Abdllh El Frisi-Bechr Algerie The functions x nd 7x re convex for x 0 then for ll, b, c > 0 b c b b c 7c we hve: b b c c b b c 7 b ubuc. bc ubuc. c ubuc 7 b ubuc. bc ubuc. c ubuc c b c c b c 7b by MA-MG Solution by Rvi Prksh - New Dehi Indi b b c c b c bubcuc ubuc () 7c b 7 c 7b b c 7cububc ubuc () Adding (), () we get b 7c b c 7 c 7b b c α 7α b c where α b.bc.c.b.c Solution by Soumitr Mndl - Chndr Ngore Indi We know, e x x nd e 7x x for ll x 0 8

30 b 7c e b log 7c log e b log c log log b c b c log b c b c c b b c (proved) 5. Prove tht for ll positive rel numbers, b, c b c b c c b. Solution by Kevin Soto Plcios Hurmey Peru Proposed by Nguyen Viet Hung Hnoi Vietnm Probr pr todos los numerous R., b, c: b c b c c b Por l desiguldd de Cuchy: b b (A) De form nálog: b c b c (B) c c (C) Multiplicndo (A) (B) (C): b c b c c b b c b c c b De l desiguldd propuest Por: MA MG 9

31 . b.c. b. c.. c.. b. c..b. b.c. c...b. (LQQD) 6. If, b 0, ; m, n N then: m nb n mb 9 m n b c Solution by Rvi Prksh - New Delhi Indi Solution by Mygmrsuren Ydmsuren Mongoli Solution by Rvi Prksh - New Delhi Indi Proposed by Dniel Sitru Romni m nb n mb m nb n mb m n b m nb n mb m nb n mb m n b m nb n mb m n b m n b m n b c [AM HM] Solution by Mygmrsuren Ydmsuren Mongoli m nb n mb Cuchy m n m n b m n b 0

32 Cuchy7Schwrz m n 9 b c 9 m n b c 7. If, b, c 0, then: b 5 b b c 5b c Solution by Imd Zk Sid Lebnon Solution by Kevin Soto Plcios Hurmey Peru Solution by Soumitr Mndl Chndr Ngore Indi Solution by Imd Zk Sid Lebnon f x c 5c 7 x 5 x 9 49 ln x 7 Proposed by Dniel Sitru Romni f x x 6x 56x 45 49x x 5 0 f x 0 f x 0 f x 0 x > 0 Now consider the inequlity b 5 b?? 7 Let x b y c b z c xyz We get g x?? 7

33 g x f x 9 49 ln x 7 f x 9 49 ln xyz 7 f x t x y z or b c Solution by Kevin Soto Plcios Hurmey Peru Siendo:, b, c < 0, >. Probr que: b 5 b b c 5b c Por l desiguldd de Cuchy: c 5c 7 b c b 5.b c 5b.c 5c. b 5 b bc 5b c c 5c b c b c.b.c b 5.b c 5b.c 5c. 5 b.b.c b. c 7 7 b c 5 b b c b c 6 4 b 4 c 4 4 b 4 c 4 4 b 5 b 6b 6c b 6 c 4 b 4 c 4 b c b c Por: MA MG: c 4 4 c (A) b 4 b 4 b 4 4 4b c (B) c 4 c 4 c 4 b 4 4c b (C) Sumndo (A) (B) (C): 4 b 4 c 4 b c b c (LQQD) Solution by Soumitr Mndl Chndr Ngore Indi

34 b 5 b Rdon s Inequlity b c b c b c b b 5b 5b b [Applying Cuchy Schwrz] b c b c b c b bc c.b.c.b.c.b.c..b.c 7 (proved) 8. Prove tht for ny positive rel numbers, b, c: b c b c c b bc bc Solution by Kevin Soto Plcios Hurmey Peru b c b c b c b c c b c b c b Proposed by Nguyen Viet Hung Hnoi Vietnm Probr pr todos los numerous R. : b c c b bc bc c b c Por l desiguldd de Nesbitt: R. : c b b c b c b c c b 4 b c Por trnsitividd: b c b c b 4 b c c b

35 b c b c c b bc bc b c c b c Es suficiente probr: b c b c b c b 4 b c b c b c b c c b 4 c b bc bc bc bc c b c c b c b c b b c b b.c b.c c.b.bc b b.c c c.b b c b b bc c c cb bc Por desiguldd de Cuchy: b b c c c b 9 b.c b b.bc c c.cb.bc b b.c c c.b.bc b b.c c c.b 9 bc.. c.b (A) 9b c.b. b.bc (B) 9c.c. bc.c (C) b c b b bc Sumndo (A) (B) (C): c c cb bc b b c c c b 9 bc c b 9 bc c bc b 9 bc b c 9 (LQQD) 4

36 9. Let, b, x, y, z, u, v, w be positive rel numbers such tht x y z. u uw y bz x v wu Prove tht: z bx y w uv Solution by Soumitr Mndl Chndr Ngore Indi u uw x by z b Proposed by Nguyen Viet Hung Hnoi Vietnm x y z, xy yz zx y bz x y bz x Applying Weighted A.M G.M 9 x y bz x y bz x y z xuyuz y bz x y bz x y bz x x y bz u uw y bz x 0. If, b, c > 0, bc then: b b 9 x y bz b bc c Solution by Imd Zk Sid Lebnon (proved) c c Solution by Soumitr Mndl - Chndr Ngore Indi 9 b xy yz zx b b bc c Proposed by Bbis Stergioiu Greece 5

37 Solution by Le Viet Hung- Qung Tri Vietnm Solution 4 by Fotini Kldi Greece Solution by Imd Zk Sid Lebnon, b, c > 0, bc prove tht?? b b when bc it is known tht p b, q pr pc c LHS c bc c b c b bc c b b c bc b b c c c b c b c We need to prove tht p b b c b b b p 9 b p C7B7S b c b 9 b.p?? b or b?? b p or b?? p or b pr?? p q p true Q.E.D t b c Solution by Soumitr Mndl - Chndr Ngore Indi Let x b, y bc nd c ; xyz b b b zx y xy z x x y () z 6

38 zx y xy z x x y xy Now, yz x y z yz xyz x y xy 9 x y z xy yz zx 9 x y z x y z x y z Hence sttement () is estblished b b (proved) Solution by Le Viet Hung- Qung Tri Vietnm First, we hve: b b b b b b 0 b b b b b bc c b bc c b c 0 b bc c b bc c c c c c c 0 c c c c b b b b 9 b b 7

39 .b.c 7 b.bc.c Cuchy Schwrz b.bc.c b c.b.c 7 b.bc.c b.bc.c ub uc Solution 4 by Fotini Kldi Greece 9.b.c 9 b b c c b b c c AM GM B b bc c b bc c B LHS A b bc c b bc c A x y z x y z xy yz zx, A b bc b c c bc A b c A A b c c c c bc A b c A b c b c b bc c b c b b 9 A b c 9 A A b bc c c c b bc b b b bc c A b c b bc c. Let, b, c be positive rel numbers such tht, b 5 nd b c. Prove tht b c 8

40 Solution by Kevin Soto Plcios Hurmey Peru b b c. Let x; y; z > 0. Prove tht: Proposed by Nguyen Viet Hung Hnoi Vietnm Siendo:, b, c números R. de tl mner que, b 5, b c. Probr que: b c Por l desiguldd de Cuchy: 6 9.4b.c (LQQD) 6.b.c..b.5 L iguldd se lcnz cundo, b, c x y y z z x x y y z z x x xy y z Proposed by Le Viet Hung Hi Lng-Vietnm Solution by Rvi Prksh - New Delhi Indi Solution by Soumv Chkrborty Kolkt Indi Solution by Imd Zk Sid Lebnon Solution 4 by Dniel Sitru Romni Solution by Rvi Prksh - New Delhi Indi x y y z z x x y y z z x x xy y z LHS x y y z z x x y y z z x 9

41 x xy z x y x.xy.y z x x y y () Let E x y x y x x y x y 0 y x y x y y x y x y y Thus, form () 0 x y y z z x x y y z z x x xy y z Solution by Soumv Chkrborty Kolkt Indi LHS x y xy z z y x z y yz z x zx y y x z x RHS x z xy z y z y x yz x z x z y zx y x y LHS RHS x y z y z x y z x z x y () Now, x x y AM7GM x, y y y z Adding the : AM7GM y z, z z x AM7GM z x x y z y z x x y z x y z y z x () 40

42 Agin x y Rdon y z x.y.z z x x.y.z () x y z ()() x y y z z x x y z y z x x y z x y z x Solution by Imd Zk Sid Lebnon y y z z x () is true (Proved) By Cuchy Schwrz B x y z x y z B x y z () A x y z B A B x So it is sufficient to prove B B B x which is true by () x Solution 4 by Dniel Sitru Romni holds for x y z LHS x z xyz x z xyz RHS x z x z xyz xy x xy y xyz x z x z xyz xy x xy y x 5 z xy z 4. If, b, c 0, then: 0 7bc 5, 0,,, 4 (Muirhed) 0b b 7c 0c c b c 7b 4

43 Solution by Ans Adlny-El Jdid-Morroco Solution by Imd Zk Sid Lebnon Solution by Kevin Soto Plcios Hurmey Peru Solution 4 by Soumitr Mndl Chndr Ngore Indi Solution by Ans Adlny-Morroco Proposed by Dniel Sitru Romni WLOG, let. 0 7bc 0 7 b c 0 6 7b 0 7c 7 [Using Chebyshev s inequlity two times] Solution by Imd Zk Sid Lebnon s desired. Homogeneous let bc the inequlity is re-written s f 0 where f x 0x4 x 7x x 7 x.7 x.7 f x g x 0 f which is convex x the tngent t x, we get g

44 q.e.d equlity t b c Solution by Kevin Soto Plcios Hurmey Peru By: Inequlity Holder s: 0 7bc 0b b 7c 0c c 7bc b 7c c 8b 7b 0 b c It is enough prove tht: 7bc b 7c c 0 b c 7b b c b bc c 0 7bc 0b b 7c 0c c 7b 0 b c 7bc b 7c c 7b 0 b c b c 0 b c Solution 4 by Soumitr Mndl -Kolkt Indi 0 b c 7bc bc b c b c 7bc b c 7bc b c 7bc b b c 7bc b b c b b 7c c b b c 7bc b 7c

45 7 b 7c b bc b 7c b 7bc b 7c 0 7 b 7c.b.7c.b.4bc.7bc b.7c 0, which is true Hence, 0 b c 7bc (proved) 4. Let, b, c be positive rel numbers such tht b c. Prove tht b c n b c n Solution by Kevin Soto Plcios Hurmey Peru c b n Solution by Phn Loc So'n-Quy - Nhon City Vietnm Proposed by Nguyen Viet Hung Hnoi Vietnm n Solution by Kevin Soto Plcios Hurmey Peru Siendo:, b, c números R. de tl mner que: b c. Probr que: b c n b c n c b n L desiguldd puede ser equivlente: n nu b.c n bnu b c. n cnu c.b De l desiguldd de Rdon:... (A) n Si: x k, k > 0, k,,,, n, n N m > 0, se cumple lo siguiente: 44

46 x m. m x m. m x m. n m n m. x x x n m n L iguldd se lcnz cundo: x x x x n n Por l desiguldd de Rdon en (A)... : n. b c n b n. b c n c n. c b n b c n. b c b c c b n b c Solution by Phn Loc So'n-Quy Nhon City Vietnm Asume: b c. b c b c c b & b c b c b c By Cebyshev s inequlity: b c n b c b c c b n c c n c b b n b c n b. c n c n n. n 5. If, b, c > 0 then: b 5 b b c c Proposed by Dniel Sitru Romni Solution by Kevin Soto Plcios Hurmey Peru Solution by Soumv Chkrborty Kolkt Indi Solution by Seyrn Ibrhimov Msilli Azerbidjin 45

47 Solution by Kevin Soto Plcios Hurmey Peru Si:, b, c > 0. Probr que: b 5 b b c c b b c c 5 b c b b c c 9 b c 6b b 6bc b c 6c c b b c c 9 b c 6b 6c b 6 c 5 b b c c Desde que:, b, c > 0. Por: MA MG 6 6b b (A) 6b 6c b b c (B), 6c 6 c c (C) Sumndo: (A) (B) (C): 6 b c 6b 6c b 6 c b b c c (D) Por otro ldo. Por: MA MG b b b c c c b c b b c c (E) Finlmente sumndo: (D) (E) 9 b c 6b 6c b 6 c 5 b b c c (LQQD) Solution by Soumv Chkrborty Kolkt Indi Given inequlity b b 5 b b b 7 b (A) 46

48 b b b b c bc c c c A7G b A7G b c A7G c b b b b b 6 b Agin, (A-G) b b b b c b c c c c b b () () () b b 7 b (A) is true (Proved) Solution by Seyrn Ibrhimov Msilli Azerbidjin b b 4 b b b bc bc 4b c c c c c 4c x b c b bc c cc b c c b 5 b b c c l l 9 9b 9c 6b 6bc 6c 5 b 5b c 5c b c b bc c 5 b 5b c 5c b c x 5 b 5b c 5c b c b b c c (proved) 6. If, b, c > 0 then: 47

49 c bc c b b b c c Proposed by Dniel Sitru Romni Solution by Rvi Prksh - New Delhi Indi c bc c b b b c c c b b c c b 0 () Consider c b b c c b c b b 4 b b b c b 6b 4 b b c b 4 b b 4b c b b c b 4 0 Similrly for other two expressions. Thus () is true. 7. If x, y, z 0,, xyz then: x 4 y z x y z Proposed by Dniel Sitru Romni Solution by Imd Zk Sid Lebnon Schur x 4 xyz x y z xy x y but xyz xy z 48

50 x 4 x x.y z Our inequlity: x 4 x x... () x.y z!! but () x.y z x 4 x so we need to prove tht x 4 x x?? x 4 x x?? True by AM-GM x y z 8. If, b, c > 0 then: b c bc b c c b Solution by Kevin Soto Plcios Hurmey Peru Solution by Soumv Chkrborty Kolkt-Indi Proposed by Dniel Sitru Romni Solution by Kevin Soto Plcios Hurmey Peru Si:, b, c > 0. Probr que: b c bc b c c b b c c b b b bc b bc b b bc b bc 49

51 b c bc b b c bc bc LQQD Solution by Soumv Chkrborty Kolkt-Indi b c bc b c b c b c b c bc b b c b c c CBS bc bc b c b c c b c b b b c c b c CBS c b b c b c c b bc b( c) b AM7GM b b etc b bc c b 50

52 bc b bc c bc bc b bc c C7B7S () bc bc b c bc () ().b c c.b.b.c bc (Proved) 9. Let x, y, z 0, such tht xy yz zx. Prove tht: x y z y z x z 9 x y 4 Solution by Kevin Soto Plcios Hurmey Peru Solution by Abhy Chndr Indi Proposed by Nguyen Viet Hung Hnoi Vietnm Solution by Kevin Soto Plcios Hurmey Peru Siendo: x, y, z < 0, > de tl mner que: xy yz zx. Probr que: x y z y z x z 9 x y 4 Por l desiguldd de Cuchy: x x y z y y z x z 9 z x y 4 5

53 Por l desiguldd de Cuchy: x x.y.z x 7y 7z x.y.z.xyz xy.yz.zx 7 x.y.z x x y z y y z x z xy.yz.zx.xyz z x y xy yz zx x y z 4xyz (LQQD) 9 4 Lo cul es cierto y que, por MA MG: xyz xyz x y z x y z xy yz zx L iguldd se lcnz cundo, x y z Solution by Abhy Chndr Indi From AM GM, we hve x 7x y z 8 y 7x 8 z 7x 4 which implies x 7x y z 8 Summing it up, we get y z x 7x 7y 7z 8 y z 7 8 x y z We re left to prove tht xy yz zx xyz 5

54 but x y z x y z xyz 4 nd xyz. Hence we re done. Equlity t x y z. 40. Let, b, c be positive rel numbers such tht: b c b bc c bc b c. Prove tht: 9 9 b 9 9 b9 c 9 9 c9 9. Solution by Kevin Soto Plcios Hurmey Peru Proposed by Nguyen Viet Hung Hnoi Vietnm Siendo:, b, c números R. de tl mner que: b c b bc c bc b c Probr que: 9 9 b 9 9 b9 c 9 9 c9 9 De l condición, se tiene lo siguiente: bc b c c b b b bc b c c c bc b c c b bc b c b c c b b b Demostrremos lo siguiente: b c bc c c 5

55 9 9 b 9 b b b b 6 b 9 b b 8 8 b b 9 70 b 56 9 b b 8 8 b b b 9 56b b 4 8 b 8 8 b b 9 6 b 00 9 b b 8 8 b b 4 0 Dividendo b, de tl mner que el sentido no se ltere, y que:, b > 0: b b 8 9 b9 b b6 b b b 6 0 Se: b b x (Válido por: MA MG). Por lo tnto: 6 b6 b6 6 x, 9 b9 b9 9 x x, b b x x 8 x x 8 x 00x 6 0 x 4 4x 4x x 4 8x 4x 4x 7 0 Lo cul se puede expresr de l siguiente form: x 48x 48x 68x 7x 7 0 x x x x 68 x x x x 68 0 Luego: 54

56 9 9 b 9 9 b9 c 9 9 c9 9 b b b c bc c c (LQQD) 4. Let, b, c be the side lengths of tringle. Prove tht bc b c c b b c b c c b b c. Solution by Kevin Soto Plcios Hurmey Peru Proposed by Nguyen Viet Hung Hnoi Vietnm Siendo, b, c los ldos de un triángulo. Probr que: bc b c c b b c b c c b b c. sin A sin B sin C Recordr lo siguiente: b c c b b c 8bc Dividiendo 8 bc l desiguldd inicil se tiene: b c c b b c 8bc b c bc c b c b c b sin A sin B sin C cos A cos B cos C ) Supongmos que se un triángulo cutángulo: cos A cos B cos A B cos A B cos C sin C, 55

57 cos B cos C sin A cos C cos A sin B Luego, multiplicndo: cos A cos B cos C sin A sin B sin C sin A sin B sin C cos A cos B cos C (LQQD) ) Supongmos que se un triángulo obtusángulo: C B A, C 90, A, B 90 cos A cos B cos C 0 sin A sin B sin C > 0 sin A sin B sin C cos A cos B cos C 0 (LQQD) 4. If, b, c > 0, b c then: b c 6 b bc c Proposed by Le Viet Hung - Qung Tri Vietnm Solution by Hung Nguyen Viet Hnoi Vietnm Solution by Imd Zk Sid Lebnon Solution by Hung Nguyen Viet Hnoi Vietnm Squring both sides, the desired inequlity becomes b c b b c c 6 b bc c 56

58 By AM-GM inequlity we obtin LHS b b c c b b c b b c c b c b b c c b bc c bc b c It s enough to show tht b b c c bc b c b bc c. Indeed, we hve b bc c b c b bc c b b bc b c c c bc b c b b c c Solution by Imd Zk Sid Lebnon nd we re done. bc b c, b, c > 0, Prove tht b 6 b... (E) First let s prove b b b b bc... (F) LHS RHS b b b c c c AM7GM b c bc b c (F) is proved. b b 0 Now by weighted Jensen s of weights, b, c on the concve function f x x, we hve 57

59 b b c f c b b c c b b c b b ccording to (F). b b 6 b Q.E.D. t, b, c,, 4. If, b, c, b bc c 9 then: b b c c bc Solution by Soumv Chkrborty Kolkt Indi b c Solution by Soumitr Mndl Chndr Ngore Indi Solution by Mygmrsuren Ydmsuren Mongoli Solution by Soumv Chkrborty Kolkt Indi b b c c Weighted GM-HM inequlity Bergstrom b b c c b b c c Proposed by Dniel Sitru Romni A7G bc A7G b c bc bc b c 58

60 ( bc, >, b c it suffices to prove: >, s, b, c ) bc b c bc b c x x x x 6 x x x x x bc x x x x x ln x x ln x () Cse () x < x bc x, ln x 0 LHS of 0 x <, ln x < 0 RHS of < 0 Cse x x LHS > RHS () is true Now, b bc c A7G b c x x x x x x nd lso, ln x ln x x ln x x ln x ( x, ln x, x > 0 nd ln x 0) () is true (Proved) 59

61 (Equlity when b c, i.e., t x ) Solution by Soumitr Mndl Kolkt Indi Let f x x log x for ll x, f x x log x x > 0 for ll x, log now b c f x log x > 0 Applying Jensen s Inequlity b c log b c b bc c 9bc b c bc log b b c c log bc.b.c (proved) Solution by Mygmrsuren Ydmsuren Mongoli b b c c bc bc ) ln b ln b c ln c b c LHS f x x ln x f ln bc x 0 JENSEN: ln b ln b c ln c.b.c ln.b.c ) RHS: b c b bc c 9bc (ASSURE) Cuchy Cuchy.b.cª bc b.bc.cª bc.b.c b.bc.c ª9bc b c b bc c 9bc 9 b c 9bc b c bc (*) (*) b c b c 60

62 b c b bc c Cuchy b c b c b bc c b c b c b c (**) (*), (**) RHS: b c ln bc,.b.c ln.b.c LHS 44. Prove the inequlity holds for ll positive rel numbers, b, c b b b bc c b bc c c c c c b b 4 b 4 c 4 Solution by Kevin Soto Plcios Hurmey Peru Proposed by Nguyen Viet Hung Hnoi Vietnm Probr pr todos los números R. :, b, c b b b bc c b bc c c c c c b b 4 b 4 c 4 Ahor bien : x, y, z R, se cumple l siguiente desiguldd: Siendo: xy yz zx x y z x b b, y b bc c, z c c c Por lo cul solo hce flt demostrr lo siguiente: 6

63 b b b bc c c c 4 b 4 b 4 c 4 c b 4 c 4 Lo cul es cierto y que: b b 4 6 b 4 b 4b 0 4 b 4 b b b b 4 b 4 b b 4 b 4 Además por desiguldd de Cuchy: x, y, z > 0, se cumple lo siguiente: x y z x y y z z x 45. Prove the inequlity holds for ll positive rel numbers, b, c b b bc c c c b c c c b b c b b b bc c Proposed by Hung Nguyen Viet-Hnoi-Vietnm Solution by Le Viet Hung-Vietnm Using AM-GM inequlity: b b b b b b b b 4 b 4 Similrly, we hve: 6

64 b bc c b4 c 4 c c c4 4 b b bc c c c b b 4 c 4 c c 4 4 b 4 b 4 c 4 nd 4 c 4 4 b 4 b 4 c 4 4 c 4 4 b4 b 4 c 4 Inequlity holds: b c 46. If x, y, z > 0 then: 9 x y z y z x > 8 x y y z z x x y z y z x Solution by Sptk Bhttchry-Kolkt-Indi Solution by Nguyen Minh Triet-Qung Ngi-Vietnm Proposed by Dniel Sitru Romni Solution by Sptk Bhtchry-Kolkt-Indi Put x y, y z b, z x c So, to show, 6

65 9 b c > 8 b c Using b c bc b c And rerrnging b c bc b c b bc c b c 8 b bc c b c 6 b bc c > 0 b c 4b 4bc 4c > 0 which is true. (Proved) Solution by Nguyen Minh Triet-Qung Ngi-Vietnm Let x ; b y, c z then bc y x The inequlity becomes 9 b c 8 b c 9 b c b c bc 8 b c b c b bc c 9 b c 8 b c b bc c 9t 8 t 9 b c b bc c t 9 (where t b c b bc c 9 b bc c 9t 8t 89t 69 t 49 0 True q.e.d. t 49 The equlity holds t bc 4 b c 4b bc c 5 0 b 5 c bc b 5 c ) 64

66 x y 4 5 ± 0 y 5 ± z x y 5 ± x 5 ± z And permuttions. 47. If, b, c > 0, b c then: b c Solution by Leonrd Giugiuc Romni Solution by Soumv Chkrborty Kolkt Indi Solution by Sptk Bhttchry Kolkt-Indi 4 Solution 4 by Soumitr Mndl Chndr Ngore Indi Solution by Leonrd Giugiuc Romni b 4 c 4 Proposed by Dniel Sitru Romni By AM-QM, b c. The functions x. 4 nd x. 4 4 re convex on 0,. By Jensen, b c nd b c

67 Solution by Soumv Chkrborty Kolkt Indi, b, c > 0, b c (Bergstrom s inequlity) it suffices to prove: () Now, let f x x.. x. 4 x > 0 f x 0x 5x x 5 x > 0, x > 0 pplying Jensen s inequlity f f b c 4 b c b c 4 5 b c b c 4 Now, (Proved) () is true Solution by Sptk Bhttchry Kolkt-Indi 66

68 6; Let x, x 6 x x 8 x x 8 x (AM HM) x x [ x x; then Titu s lemm] x 4 (Proved) x x x 4 x Solution 4 by Soumitr Mndl Chndr Ngore Indi 4 b c b c b c we know, 7 b c 7 7 b c Similrly, 74 4 b c b c

69 48. If x, y, z > 0 then: (Proved) 4 x y z 4xyz Solution by Mihlce Andrei Stefn-Romni xy x y Solution by Soumitr Mndl-Chndr Ngore-Indi Solution by Redwne El Mellss-Morroco 7xyz Proposed by Dniel Sitru Romni Solution by Mihlce Andrei Stefn-Romni 4 x y z 4xyz 4 x y y x 4 xy x y 7xyz : xyz > 0 x y y 7 x We ll prove: 4 x y 4 x 9 y x. x y y x not α 4α α 4 0 α y.y x Solution by Soumitr Mndl-Chndr Ngore-Indi 4 z x y 4xyz 4α 7 0, true (α ) xy x y 7xyz 4 z x y 6xyz xyz 4xy x y 68

70 4 x y z xyz x y x y 4 x y xy x y x y x y 4 x.y 7xy xy x.y 0, which is true 4 z 4 x y xy 5xy > 0 x y Hence, 4xyz (proved) Solution by Redwne El Mellss-Morroco The inequlity 4 Let f x 4 x x xy x y 7xyz x y xy y x xuy 7 for x > 0. x. x Since f x x x. x with x x we get x > 0 : f x f 9. LHS f x y 7 with equlity if 49. If, b, c, then: x y y x z x y z > 0 x log b e log bc e log c e log e log b e log c e Solution by Mygmrsuren Ydmsuren Drkhn Mongoli Proposed by Dniel Sitru Romni 69

71 Solution by Rvi Prksh - New Delhi Indi Solution by Mygmrsuren Ydmsuren Drkhn Mongoli log b e log bc e log c e log e log b e log c e ln b ln ln b 4 ln ln b ln bc ln c ln b ln c 4 ln b ln c 4 ln ln b ln c Solution by Rvi Prksh - New Delhi Indi ln ln b ln c ln c ln 4 ln c ln ln ln ln b ln c Cuchy Let x log e, y log e b, z log e c. As, b, c >, x, y, z > 0 Now, log b e log e log.log b x.y We hve:, etc. [chnge of bse] xy x y x y x y 4 x y Thus, x.y 4 x y x log b e log e 50. If, b, c > 0 then: b b 6 b b 4 Proposed by Dniel Sitru Romni Solution by Leonrd Giugiuc Romni 70

72 Solution by Soumv Chkrborty-Kolkt-Indi Solution by Leonrd Giugiuc Romni By Cuchy, b b b 7b.b.b.b b b 6 b b b b 6 b b. By AM-QM, b b b b 4. Solution by Soumv Chkrborty-Kolkt-Indi b b 6 b b x x Let s prove:.b 7b.b 4 ().b b b b b 4 b 4 b b b 4 b 4 b b b b b 0 true Similrly, b.c b 7bc.b b.c 4 () nd 7

73 c. c 7c. c. 4 () () () ().b 7b.b 4 (4).b b b 6 b b b b 4 (using (4)) (proved) 5. For the positive rel numbers, b, c, n, where n N, prove tht n n n n 7 bc b c n Proposed by Abhy Chndr-Indi Solution by Leonrd Giugiuc-Romni: It s equivlent to n ln n ln n ln ln b ln c n ln b c 0 Let the function f: n, n R, f n ln n ln n ln ln b ln c n ln b c. fʹ n Then But n n 7 n 7n.b.c.b.c.b.c > n b c,, n, hence f is strictly decresing n min f f n n ln n ln n ln n ln b ln c n ln n b c Apply twice more this procedure nd get the desired result. Equlity if. 7

74 5. If, b, c, d > 0 then: b c n. 6 b c d > b b c b c d Solution by Ans Adlny - El Jdid Morroco Solution by Henry Ricrdo - New York USA Solution by Mygmrsuren Ydmsuren Drkhn Mongoli Solution by Ans Adlny - El Jdid Morroco 6 b b c b c d 6 b b c b c d Proposed by Dniel Sitru Romni b b c b c d 6 b 4c 6 d 6 (true) Solution by Henry Ricrdo - New York USA b 4c 6 d 6 The AM GM inequlity gives us < b c d 6 b b c b c d lic b b c b c d < 6 lic b c d 6 lic b c d lic lic lic 9 b c d b c d. Solution by Mygmrsuren Ydmsuren Drkhn Mongoli ; b; c; d > 0 7

75 6 > b b c b c d b c d b c d 6 9 b c d 6 b b c b c d 6 Cuchy 6 b b c b c d 5. If, b, c, d 0,,..b.c.d then: b c d 6 Solution by Mygmrsuren Ydmsuren Mongoli ) n i±n n i± x i x i x i n n i± n x i x i i± x x x n x x x n x.x x.x n x i i±n n n i±n x i x i x n.x n x i x i Proposed by Dniel Sitru Romni n x i Chebyshev x i n n i±n Assume x n i n i±n x i x i n i±n x i x i Cuchy n n x i x i n x i x i n 74

76 ; n i±n x i x i n i±n x i x i n x i x i n x i i±n n i±n x i x i n n x i i±n ) n i±n n i±n x i x i x i n Cuchy n x i i±n n n i±n n i±n x i x i x i x i n n n x i i±n n 4 x x x x 4 6 b c d If, b, c, d > 0, b c d then: b c b c d c d d b 8 Proposed by George Apostolopoulos - Messolonghi - Greece Solution by Kevin Soto Plcios Hurmey Peru Si:, b, c, d > 0. b c d. Probr que: b c b c d c d d b 8 Por l desiguldd de Holder: b c b c d c d d b b c c d d b b c d 75

77 b c 55. If x, y, z > 0 then: b c d c d d b (LQQD) b c d 8 b c d 8 x x z y 4 y y x z 4 z z y x 4 > xy yz yx zx zy xy Solution by Mygmrsuren Ydmsuren Mongoli Solution by Nguyen Minh Triet - Qung Ngi Vietnm Solution by Mygmrsuren Ydmsuren Mongoli Proposed by Dniel Sitru Romni x x z y 4 x 4 z y 4 x x z y x x z y x y x z y z y x z x z y x y x z y z y x z xy xz xy yz xz zy > xy xz yx zx zy xy Solution by Nguyen Minh Triet - Qung Ngi Vietnm By AM GM inequlity, we hve: x z y 4 xzy x x z y 4 xy xz () Similrly, we get: y y x z 4 yz xy () z x y x 4 xz xy () () () () LHS RHS > RHS q.e.d. 56. If, b, c > 0 then the following reltionship holds: 76

78 b c b c c b c c b c Proposed by Dniel Sitru Romni Solution by Kevin Soto Plcios Hurmey Peru, Solution by Ans Adlny-El Jdid-Morroco, Solution by Sk Rejun-West Bengl-Indi, Solution 4 by Nirpd Pl-Indi, Solution 5 by Mygmrsuren Ydmsuren-Drkhn-Mongoli Solution by Kevin Soto Plcios Hurmey Peru Si:, b, c > 0. Probr l siguiente desiguldd: b c b c c b c c b c Desde que:, b, c > 0. Por: MA MG: c b c 4 c b c.b.c.c b.c c..b.c. b.c 4 Solution by Ans Adlny-El Jdid-Morroco We hve A B b c (LQQD) 4AB, so if we tke A c, B b c, we get b c c b c c b c A B AB A B A B 4 Solution by Sk Rejun-West Bengl-Indi We hve to prove 77

79 b c c b c c b c b c LHS.b.c.c b.c.c. b.c () Now, c b c b c c b c b c b c c b c b c c b c c b c b c Solution 4 by Nirpd Pl-Indi c b c b c 4 [by G.M A.M] b c.b.c.c b.c.c. b.c s b.b.c.c b.c.b.c b.c b.c.b.c uc. buc.c. b.c s HM AM 4 b c b c Solution 5 by Mygmrsuren Ydmsuren-Drkhn-Mongoli c x c y b c b z x y z x y z x.y xy x.y x y z x y xy x y (ASSURE) 78

80 x y 57. If, b, c, then: y z z x x.y x.y xy x.y (ASSURE) xy x.y x y xy (TRUE) y z z x x y z Solution by Redwne El Mellss-Morroco Similrly y z zy y z z x zx z x x y xy y x b 4 c( b) bc b c x y xy x y b b b Proposed by Dniel Sitru Romni b b 4 b c b 4 c b 4 bc c 0 b b 0 b 4 c b bc. with equlity if nd only if b c. b b c c 58. If, b, c (0, ) then: 79

81 be 7 ce b7 e c7 e Solution by Redwne El Mellss-Morrocco c b c b Let f x xe xf, 0 < x <. Since: Proposed by Dniel Sitru Romni f x x x ex7 x, 59. If, b, c > 0 then: b b we get f x f e. So: b f e b. b > 6 b bc c b Proposed by Dniel Sitru Romni Solution by Mihlce Andrei Ștefn Romni, Solution by Rvi Prksh - New Delhi Indi, Solution by Seyrn Ibrhimov Msilli Azerbidjin Solution 4 by Soumitr Mndl - Chndr Ngore Indi Solution 5 by Abdllh El Frissi Bechr Algerie Solution by Mihlce Andrei Ștefn Romni Let s prove:.b.b b 6 b.b b not s 80

82 b not p s 6s p 9p > 9 s p 0 if b b b c bc b c c c b ± 5 c ± 5 ± 5 Anywy we will hve the signs in ± 5, it will result tht rtionl number is equl to n irrtionl number. Contrdiction LHS > RHS Solution by Rvi Prksh - New Delhi Indi Consider b b b 6 b b b b b 6 b b 9b b b 6 b b b b b b 0 b b b 6 b b Similrly for other two terms. Solution by Seyrn Ibrhimov Msilli Azerbidjin b b b b b b 9b AM7GM 6 b b 8

83 Solution 4 by Soumitr Mndl - Chndr Ngore Indi b b > 6 b b b b 9b 6 b b > 0 b b b > 0 b which is true b b > 6 b b b (Proved) Solution 5 by Abdllh El Frissi Bechr Algerie b b b b 9 b 6 b b b nd equlity if nd only if 7b b 0, b b nd equlity if nd only if b c b 5.7 b 6 b bc c b 7b b 0 b 7bc c 0 c 7c 0 c this is contrdiction then b then it follow tht 8

84 60. If, b, c then: b b b > 6 b bc c b log log b log log bc log Proposed by Dniel Sitru Romni Solution by Mihlce Andrei Ștefn Romni Solution by Soumv Chkrborty Kolkt Indi Solution by Abdllh El Frissi Bechr Algerie Solution by Mihlce Andrei Ștefn Romni x ln 0 Tke y ln b 0 z ln c 0 The inequlity xy x xy x g p g 4q p g 4gp p 4 8g gp 0 g p 0 true Solution by Soumv Chkrborty Kolkt Indi Let log u, log b v, log c w u, v, w 0 Given inequlity uv u u u uv u uv u Let uv x nd u y 8

85 given inequlity x y x y x y 0 which is true (Proved) N.B.: Inequlity is true, b, c > 0 Solution by Abdllh El Frissi Bechr Algerie log log b log 0 then log log b log log log b log 6. If, b, c > 0 then: b b log log bc log b c b b c Solution by Redwne El Mellss Morroco Let f t t.t 6 c c c b b c c Proposed by Dniel Sitru Romni t, t > 0..t6 Since f t 7t t7 t 4.t.t.t..t 6 0 f b 0 b b b b with equlity if nd only if b c c b c > 0. b 6. If x, y, z > 0 then: 84

86 x xy y y z Solution by Mihlce Andrei Ștefn Romni xy y 4 y z y z yz z 4 Solution by Mygmrsuren Ydmsuren-Drkhn-Mongoli Solution by Soumitr Mndl - Chndr Ngore Indi Solution by Mihlce Andrei Ștefn Romni LHS We ll prove: xy y z Proposed by Dniel Sitru Romni xy y z xy y 4.z 4 7y z7yz.y z y 4 y z z 4 yz 0 y z y yz z Equlity for b c Solution by Mygmrsuren Ydmsuren-Drkhn-Mongoli 0 true. x xy y x y xy xy xy xy Cuchy. y z 0 0 y4 z 4 y z y 4 z 4 y4 z 4 y z y 4 z 4 y z y 4 z 4 y z y 4 z 4 zy y z y 4 z 4 y z yz 0 y 4 z 4 y z y z yz y z y z Cuchy. x 7xy.y y z ; xy y 4 7y z.y z 7yz.z 4 85

87 Solution by Soumitr Mndl - Chndr Ngore Indi x xy y y z x y y z x y xy y 4 y z y z yz z 4 xy y z y 5 z 5 x.y y 5.z 5 7xy z y.z x.y 0 () y z x.y y 5.z 5 Now, x y y 5 z 5 xy z x y y z 4 x y 6 y z 5 xy z x y Hence, () is estblished. y z 0 x xy y y z xy y 4 y z y z yz z 4 (proved) 6. If, b, c then: log b b c c 6 Solution, by Leonrd Giugiuc Romni Solution by Soumitr Mndl-Chndr Ngore-Indi Solution by Leonrd Giugiuc Romni b b c 4 Proposed by Dniel Sitru Romni Lemm: b ln 6b..4. b,, b 86

88 Lemm s proof: For ll, let f b b ln f b ln 6b b on, ; Now let g 4 f b 4 ln on,. We hve: g ln 5 4 nd g ln 4 > 0 g 7 0, g 0, g 0, g() 0 ¹ f is strictly incresing. Since f g().4. 0 the conclusion follows. Similrly, the other two inequlities. Adding up, we re done. Solution by Leonrd Giugiuc Romni Lemm: The function f x ln x 6.x x.4x. Lemm s proof: f x x x.x. x.4x.. We hve: is strictly incresing on,. f x 0 x x x x x x, which is true by AM GM for u x x nd v x. Bck to the problem. By the lemm, f x f x. We get: bf cf b f c b c ln b b c c 6b. b c..4. Solution by Soumitr Mndl-Chndr Ngore-Indi 87

89 Let f x ln x f x x x.x.x x7.4x.x 6.x.4x.x for ll x, x 4x x 4 x.4x.x 4x x x 0 for ll x, now f is continuous on, nd f x 0 for ll x, f x f 0 for ll, b, c, then f c bf cf b 0 ln b b c c 64. If, b, c 0, then: 6 b 4 (Proved) b c b b b 4 4 b b bc c Solution by Seyrn Ibrhimov Msilli Azerbidjin b b b 4 4 b b b b b b 4 4 b b 4 4 b b b b Proposed by Dniel Sitru Romni 6 b 4 b 4 b 6 4 b b 4 5 b b 5 6 b 4 b 6 4 b 4 5 b b 6 b b 4 b 6 4 b 4 b 5 proved(am-gm) 88

90 65. If, b, c, d > 0, bcd then: 6 Solution by Mihlce Andrei Ștefn Romni Solution by Seyrn Ibrhimov-Msilli-Azerbidjin Solution by Mihlce Andrei Ștefn Romni Proposed by Dniel Sitru Romni Inequlity () true becuse 4 bcd Solution by Seyrn Ibrhimov-Msilli-Azerbidjin x b c d 4 x x x 6 x x 5x x 5x 6x x 5x 6 6 x 6x x 0 x x 4 x x 4 x 4 0 x 4 x x 0 ( ) 0 therefore x 4 Equlity x x x x 66. If, b, c R, bc then: c b b b c c c b < 4 Proposed by Dniel Sitru Romni 89

91 Solution by Redwne El Mellss-Morroco Let Δ, b, c R c b. Δ, b, c c c b b Δ, b, c c b b c () Δ b, c, b c c Δ, b, c b c c () b Δ c,, b b c b Δ, b, c c b Now lets study some cses bout, b nd c : If b c : c c > 0 b () c then () Δ, b, c c c < c c. c By the sme ide, if b c we use () nd if c b we use (). If c b : c > 0 b b > 0 c If b : () Δ, b, c b If b : () Δ, b, c c b b < b b c < c b b b b c c b By the sme ide, if c b we use () nd () nd if b c we use () nd (). Finlly Δ, b, c R bc < < Let x, y, z be positive rel numbers such tht: xyz. Prove tht: x x 4.y 4.4xy y (y 4.z 4 ).4yz z z 4.x 4.4zx x.y.z 5 () Proposed by Hong Le Nht Tung Hnoi Vietnm 90

92 Solution by Hong Le Nht Tung Hnoi Vietnm * Since Inequlity Cuchy Schwrz. We hve: x 4 y 4 xy x 4 y 4 4x y 4 x 4 x y y 4 4 x y x 4 y 4 xy x y x 4 y 4 4xy x xy y x 4 y 4 4xy x xy y x x 4.y 4.4xy x x.xy.y () y x 4.y 4.4yz - Similr: y ; z y.yz.z z 4.x 4.4zx - Since (), (): z z.zx.x () x x 4 y 4 4xy y y 4 z 4 4yz z z 4 x 4 x y z (4) x.xy.y y.yz.z z.zx.x - Other, Since inequlity Cuchy Schwrz: 4zx x x xy y x x x y xy y y yz z y y y z yz z z zx x z z z x zx x y z x x y xy y y z yz z z x zx x x xy y y y yz z z z zx x 9

93 x y z x x y z y x y z z x y z x x.xy.y y y.yz.z z z.zx.x x.y.z x.y.z x.y.z x.y.z x.y.z (5) - Since (4), (5): x x 4.y 4.4xy y y 4.z 4.4yz z z 4.x 4.4zx x.y.z x.y.z x (x 4.y 4 ).4xy y y 4.z 4.4yz z z 4.x 4.4zx x.y.z x.y.z x.y.z x.y.z (6) - Since Inequlity AM-GM for 5 positive rel numbers we hve: x y z x y z x y z x y z x y z x y z x y z x y z x y z x.y.z x.y.z x.y.z 6 x.y.z 6 x.y.z 6 x.y.z 6 x y z x y z x y z 5 5 x y z x y z - Since inequlity: xy yz zx xyz x y z nd supposed: xyz. We hve: x y z x y z x y z x y z xyz x y z x y z xy yz zx x y z x y z xy yz zx xy yz zx 9

94 x y z x y z x.y.z xy.yz.zx xy.yz.zx 4 - Other, Inequlity AM-GM for positive rel numbers: (8) x y z x y z xy yz zx xy yz zx x y z xy yz zx xy yz zx x y z 6 7 x y z xy yz zx xy yz zx x y z xy yz zx xy yz zx - Since (8), (9): x y z x y z x y z x.y.z 5 8 x.y.z 5 x.y.z x.y.z 6 7 x.y.z 6 7 (9) x.y.z (0) - Since (7), (0): x.y.z x.y.z x.y.z () - Since (6), (): x (x 4.y 4 ).4xy y y 4.z 4.4yz z z 4.x 4.4zx x.y.z Inequlity () True nd we get the desired result. Equlity occurs if: x, y, z > 0; xyz (x 4 y 4 ) xy; y 4 z 4 yz; z 4 x 4 zx x xy y y yz z z xz x x y z x y z x y z 6 xy yz zx x y z xy yz zx x y z 68. If x, y, z > 0 then: x 7x 5 y z 7, 9 x 5x 7 y z 5 9

95 Solution by Abdllh El Frissi-Bechr-Algerie Solution by Rvi Prksh-New Delhi-Indi Solution by Imd Zk-Sid-Lebnon Solution by Abdllh El Frissi-Bechr-Algerie Proposed by Dniel Sitru Romni Let f.5 x.y.z, > 0 we hve f 70 x.y.z.5 x.y.z < 0 then f is concve function, x 7x 5 y z f x f y f z f x y z 7 Let g, < 7 7 x.y.z 7 x y z, we hve g 8 x.y.z 7 x.y.z 7 0, then g is convex function, x 5x 7 y z f x f y f z f x y z 9 Solution by Rvi Prksh-New Delhi-Indi Consider E x 5 y z 7x 7x 7x 5 y z 7 7 7x 5 y z Similrly, 5 x y 5 x z 7 7x 5y 5z E 5 y7x.5 y7z 7 7y.5x.5z Now, 7 5 nd E E E E 5 z7x.5 z7y 7 7z.5x.5y 94

96 x y 7x 5y 5z 7y 5x 5z y z 7y 5x 5z 7z 5x 5y z x 7z 5x 5y 7x 5y 5z x y 7x 5y 5z 5x 7y 5z y z 5x 7y 5z 5x 5y 7z z x 5x 5y 7z 7x 5y 5z 0 x 7x 5 y z 7 Next, consider F 9 x 5x 7 y z 7 y z 5x 9x 9 5x 7y 7z 7 y x 7 z x 9 5x 7y 7z Similrly, y x F 7 z y 7 x y 9 5y 7x 7z 9 7 5x 7y 7z 5y 7x 7z, F F F F x z 7 x z 7 y z 9 5z 7x 7y 5z 7x 7y 5x 7y 7z z y 5y 7x 7z 5z 7x 7y y x 5x 7y 7z 5y 7x 7z x z 5z 7x 7y 5x 7y 7z z y 5y 7x 7z 5z 7x 7y 0 Solution by Imd Zk-Sid-Lebnon Let f x x 5x 7 y z 9 x x.5 for x 0; we hve: f x 5x Let g x 70 x7 89 x.5 x 7x 5x 0 f x for x 0; we hve: () 95

97 g x x x7 6 7x x 0 g x () 6 6 Both inqulities re homogeneous, so let x y z f x Ineq () f x? 5x x.y. 89 cc. to () we my write! Q.E.D. Ineq () g x?! g x x cc. to () we my ffirm Q.E.D. when x y z or x y z in generl N.B.: y 5x 89 8 is the tngent, t x to G f y x 6 6 // // // // // // to G g 69. Let, b, c be positive rel numbers such tht: b c. b b b bc c Solution by proposer Prove tht: c c c c Solution by Anh Ti Trn Hnoi Vietnm Solution by Leonrd Giugiuc Romni Solution by proposer * We hve inequlity: b b Proposed by Hong Le Nht Tung Hnoi Vietnm 4 b 4 c 4 bc b c b b bc b c c c () 96

98 - () 4 b 4 c 4 bc b c b b bc b c c c 4 b 4 c 4 bc b c b b bc b c c c 0 b c bc b b bc b c c c c cb b 0 b c b b b c c c c b 0 () * b c > 0. We hve: c c 0 c b c b 0 c c b 0 c c c b 0 (4) We hve: b c b b b c b c b b c b c b b b c b b c b b b b b c b b b b b b c bc b b b c bc (5) b c > 0 c 0; b c 0 Therefore: b b c bc c b b c b b > 0. b 0;, b R b b b c bc 0 (5): b c b b b c 0 (6) - Since (4), (6): b c b b b c c c c b 0 Inequlity () True True. - Since inequlity AM-GM for positive rel numbers: b b bc b c c c b b bc bc c c b b c c (7) * Since (7): 4 b 4 c 4 bc b c b b c c 4 b 4 c 4 b b c c 4 b b c c bc b c b c 4 b b c c bc b c.b.c 4 b.b c.c 7bc.b.c (8) 97

99 - Since inequlity Cuchy Schwrz. We hve: b b b bc c b bc c b c c c c b c c bc b Fbcuc. b c Fcu. c Fbub b.bc.c b b c b b c (9) - Other, since inqulity Cuchy Schwrz: b bc c b c c c b b 4 b bc c b 4 b c b c b b c c 4 c bc b c b bc c b c b c b c bc b c - Since (8), (9): - Since (9), (): b bc c b c c.b.c 4 b.b c.c 7bc.b.c b c b 7bc.c c 7c. 7b.b c b b (0) () b c b b 7bc.c c c 7c. 7b.b b.bc.c b b bcc b c c c c b.bc.c bb () b c.b - Since inequlity AM-GM we hve: b.c c. b bc c b bc c b c b bc c b c b bc c b bc c b c b bc c b bc c (becuse b c ) () b c 98

100 - Since (), (): Equlity occurs if: b c b b 7bc.c c c 7c. 7b.b Inequlity () true nd we get the desired result. b Fbcuc b b c b c > 0 b 7bc.c b c Fcu bc c 7c. Solution by Anh Ti Trn Hnoi Vietnm c Fbub c 7b.b b c. b b bc c b bc c b b bc c b b bc c b c We will prove: T Solution by Leonrd Giugiuc Romni T b bc c T 4 bc bc 4 bc b It s true by Shur nd AM-GM LHS b b b By Cuchy, 99

101 b b bc c b bc c b bc c. Suffice it to show tht b bc c b bc c Apply Jensen nd get b bc c b c 4 b b c c bc. We ll show tht.b.c 4 b.b c.c 7bc t 4 t p, where b bc c t, 0 t < nd bc p. Cse : p. Then 4 t t t 4 t p nd t 4 t t t t t 0 Cse : p <. Then t 4 t 4 t p. Becuse b.bc.c, done. 70. If, b, c > 0 then: b c b c b c b bc c 6 bc bc Proposed by Dniel Sitru Romni 00

102 Solution by Soumv Chkrborty Kolkt Indi Solution by proposer Solution by Soumv Chkrborty Kolkt Indi LHS Solution by proposer b bc c b c b 4 b 4 c 4 b b b c b c b bc c b c b c A7G b c 4 b 4 c 4 4 b 4 c b 8 c 8 4 b 4 c b 4 c b 4 c 4 (Proved) 7 4 b 4 c 4 bc bc 6 bc b c 9 4 b 4 c 4 b c b c 4 b c b c b bc c b bc c b bc c b c b c b bc c b c b bc c 0

103 b c b c b bc c b c b bc c b c b c b c b c b bc c b c b bc c 6 6 b c b bc c 6 4 b 4 c 4 6 bc bc 7. Let, b, c be positive rel numbers such tht b c. b 4 4 Prove tht: Solution by Anish Ry-Sntrgchi-Indi Solution by Soumitr Mndl-Chndr Ngore-Indi Solution by Henry Ricrdo - New York USA Solution by Anish Ry-Sntrgchi-Indi b > b 4 4 c b c > 5 Proposed by Anish Ry-Sntrgchi-Indi Given tht, b, c > 0 We cn write, b 4 4 > b b 4 4 > b 0

104 So, > b 4 4 b Now, By Bergstorm s Lemm, we get tht b b c b c 6 Now, for, b, c > 0 b c > b c b c 6 > b c 6 b c 6 > b c b c 6 > so, b c 6 5 which implies, b c b c therefore, b 5 Thus, > b Solution by Soumitr Mndl-Chndr Ngore-Indi Let, b, c 0 nd b c then > b

105 b 4 4 b b b b b 4 b b c b we need to prove, 5 > b 6 5 > b b c 5 > 9 b 5 4 b > 4 b 4 7 b > 4 b we need to prove (). Now we hve 4 7 b 4 b b bc b 4 b bc 04

106 since, 4 b bc b c 4 b bc 4 b 54bc > 4 b hence sttement () is true. > b 4 5 (proved) Solution by Henry Ricrdo - New York USA Without loss of generlity we my ssume tht b c. It follows tht b c nd.4 b.4 c.4. Now the Rerrngement Inequlity give us b 4 b c 4 c b b 4 c c 4 It cn be seen grphiclly (nd proved with some tedious lgebr/nlysis) tht the curve given by y x x.4 lies on or bove the tngent line to the curve t, x, y lic 5 x, on the intervl 0,. Thus we hve 5 b 4 lic

107 lic > Prove tht: b c d 5.b 5.c 5.d 5,, b, c, d > 0 bcd Proposed by Dniel Sitru Romni Solution by Kunihiko Chiky Tokyo Jpn 7. If x, y, z > 0 then: 5 5 b 5 c 5 d b 5 b 5 c 5 d b 5 c 5 c 5 d b 5 c 5 d 5 d 5 5 x y z y z 5( 5.b 5.c 5.d 5 ) 5 x 8 > x y b 5 c 5 d 5 bcd 5 b 5 b 5 c 5 d 5 b cd 5 b 5 c 5 c 5 d 5 bc d 5 b 5 c 5 d 5 d 5 bcd bcd( b c d) y z z x x y y z z x Proposed by Dniel Sitru Romni Solution by Soumv Chkrborty Kolkt Indi Solution by Mygmrsuren Ydmsuren Drkhn Mongoli Solution by Adity Nryn Shrm Knchrpr Indi 06

108 Solution 4 by Seyrn Ibrhimov Msilli Azerbidin Solution by Soumv Chkrborty Kolkt Indi x, y, z > 0 x x 8 > y x y Let f t t t t 6 t > 0 y Now, t t 4 5 A7G t 4 5 > 5t t 4 5 t t 6 5 > t t 6 5 t t 5 > t 9 t 6 t > 0 t t 6 > 0 t > 0 4 t 6 > t t t > 0 x x 6 > x y y y y y 6 > y z z z z z 6 > z x x x () (putting t x y ) () (putting t y z ) () (putting t z x ) () () () x 8 > x y y x Solution by Mygmrsuren Ydmsuren Drkhn Mongoli If x, y, z > 0 y x x 8 > y x y t 6 > t t (ASSURE) y 07

109 t 6 t t 4 t Cuchy t t t i i± t x y ; t y z ; t z x 8 > t i t i i± i± Solution by Adity Nryn Shrm Knchrpr Indi x y z y z x 8 > x y z y z x x y y z y x Let x y, y z b, z x c bc with, b, c > 0 To prove, b c 9 b c b c > 0 Now define, f, b, c b c z b c b c Evluting first prtil derivtives, f () f b 6b 6b 0 () f c 6c 6c 0 () All of the equtions re identicl, thus we hve sme solution for, b, c. b c Now to justify tht b c is minimum we consider the hessin mtrix, 08

110 b 0 0 H f, b, c 0 b c b b c Now, from the solutions of (), (), () we know, > 0, b > 0, c > 0 becuse we neglect the negtive roots becuse, b, c > 0 Solution 4 by Seyrn Ibrhimov Msilli Azerbidin Equivlent to: x y ; y z c; z x b 6 > 6 > < 6 f f ; ; Ã7 nd Ã. ; 09