(x + y) x2 + xy + y 2 (3) x2 + y 2 3 2

Transcription

1 A Nice Sequence of Inequlities Co Minh Qung In this ppe, we would like to intoduce to the edes some pplictions of nice sequence of inequlities fo solving inequlities. Theoem. Fo ny x > 0 nd y > 0, we get xy (x y) () x y () (x y) x xy y () x y (4) x y (x y) Equlity (t (), (), (), (4)) holds if nd only if x = y Poof. x y () = (x y) (x y) 4xy (x y) xy (x y) = 8 8 () (x y) x xy y = (x y) 4x 4xy 4y 4 (x y) (x y) (x y) (x y) (x y) x y = = 4 4 () x y = (x y) x xy y (x y) (x y) x xy y = (x y) x xy y (4) x y = (x y) (x y) 4x 4xy 4y 8 (x y) (x y) (x y) 4x 4xy 4y 8 (x y) = x y (x y) The bove sequence of inequlities is vey simple. But we will show some inteesting inequlities cn lso be poved by using it. Hee e some exmples. Exmple. (Poposed by Pnos E. Tsoussoglou, M, Myhem Poblem, Cux Mthemticoum Vol8_008) Let ; b; c be positive el numbes. Pove tht b c bc ( b) (b c) (c ). Solution. Fist, we will show the following well - know inequlity b c b c (5) Indeed, fo ; b; c; d > 0, fom () nd (), we hve b b ; c d c d! b c d Hence, b c d b b b c d ( b c d) c d o c d

2 Fo d = ( b c), we obtin b c 7 ( b c) b c ( b c) o b c 7 ( b c) 4 7 ( b c) o b c b c By using (5) nd the AM - GM inequlity, we get b c b c ( b) (b c) (c ) = p ( b) (b c) (c ) ( b) (b c) (c ) = 8 Theefoe, by the well - know inequlity b b, fo ; b > 0; we hve b c bc ( b) (b c) (c ) ( b) (b c) (c ) ( b) (b c) (c ). Exmple. (Poposed by Jose Luis Diz - Beo, J4, Mthemticl Re ections 5_007) Let ; b; c be positive el numbes. Pove tht b c p 4 (b c ) c b p 4 (c ) b c p 4 ( b ). Solution. Since b c p bc o 4 (b c ) b c Hence, p 4 (b c ) bc. Theefoe Similly, we obtin c b p 4 (c ) Theefoe b c p 4 (b c ) Exmple. tht c b c ; b c p 4 ( b ) b c p 4 (b c ) b b c. b c b c c b p 4 (c ) b ( b c) c p = 4 ( b ) b c (USAMO 007) Let ; b; c be positive el numbes. Pove b bc b c bc c bc bc. Solution. Fom (), (), (), we hve b b ( b) o b bc b ( b c). Hence, b bc b ( b c) Similly, we obtin b c bc bc ( b c) ; c bc c ( b c) Adding thee bove inequlities, we get b bc b c bc c bc b c b bc = c bc

3 Exmple 4. (Romni 997) Let x; y; z be positive el numbes stisfying xyz =. Pove tht x 9 y 9 x x y y y 9 z 9 y 9 y z z z 9 x 9 z z x x. Solution. By the substitutions = x ; b = y ; c = z, then bc =. The inequlity becomes b b b b c b bc c c c c. x y Recll (), we hve x xy y x y. Theefoe b b b b c b bc c c c c b b c c = ( b c) p bc = Exmple 5. (Polnd 007) Let ; b; c; d be positive el numbes stisfying b c d = 4. Pove tht b b c c d d ( b c d) 4 Solution. By the substitutions = x ; b = y ; c = z, then bc =. The inequlity becomes q Recll (4), we hve x y q q q b b c q. Theefoe, x y xy c d It su ces to pove tht b b b c b c c d c d d d Note tht u v d b b b c bc c d cd d d ( b c d) 4 () u v u v = uv u v = u, fo ll u; v = 0 So v y z z t, whee t z (), x y x = ; y = b ; z = c ; t = d,x; y; z; t > 0 nd x y z t = 4 By the well-know inequlity ( n ) n n, fo ll i > 0; i = ; ; ; n We obtin xy yz zt tz 4 (xy)(yz)(zt)(tx) = This completes the poof. Exmple. (Poposed by Geoge Apostopoulos, Messolonghi, Geece, Poblem 78, Cux Mthemticoum). Let ; b; c be positive el numbes such tht b bc c =. Pove tht b b b b b c b bc c c c c c 0bc. Solution.

4 x y Recll (), we hve x xy y x y, fo ny x > 0; y > 0 Theefoe, by using the AM - GM nd the given condition, we hve b b b ( b) = b ( b) = b ( b) b 9 p b c p b c 9 p bc p b c = 0bc Equlity holds if nd only if = b = c =. Exmple 7. (Poposed by Pngiote Ligous, Leondo d Vinci High School, Noci, Itly, Poblem 79, Cux Mthemticoum). Let ; b; nd c be the sides, the indius nd R the cicumdius of tigle ABC. Pove tht c b b bc c c b c c R. b b Solution. We note tht b c b b = ( b) c = 0. It su ces to pove tht b c b b R x y Recll (), we hve x xy y x y, fo ny x > 0; y > 0 Theefoe b c b b ( b) c Lst, we need to pove tht ( b) c R o b 8R. On the othe hnd, since 8R = b 9bc. The lst inequlity is tue since b p b c 9bc bc, if su ces to pove tht p bc = 9bc. We e done. Exmple 8. (Poposed by Pedo Henique O. Pntoj, student, UFRN, Bzil, poblem, Cux Mthemticoum). Let ; b; c be positive el numbes. Pove tht 4 4b c b 4b 4c c 4c 4 b < Solution. 4

5 x y Recll (), we hve x y o p 4 (x y ) p x p y. The given inequlity is equivlent to b 4c b b c 4 c c 4b < We hve b 4c = p p 4 ((4 4b) c) p p 4 ( b) c p p b p c. Hence, b 4c p p p b p c =. If equlity holds, then we must hve 44b = c; 4b4c = ; nd 4c4 = b which imply tht 8 ( b c) = b c, o b c = 0; contdiction. Ou poof is complete. Some execises fo the edes. Execise. Let ; b; c be positive el numbes. Find the minimum vlue of the expession p 4 ( b ) p 4 (b c ) p 4 (c ) b b c c Execise. (Poposed by Tn Tun Anh) Let ; b; c be positive el numbes nd k. Pove tht k k b c b c c b k Execise. (Poposed by Thi Nht Phuong) Let ; b; c be positive el numbes stisfying bc =. Pove tht b b c b b c c c c b Execise 4. (Poposed by Nguyen B Nm) Let ; b; c be positive el numbes. Pove tht b c b c b c c b b c Execise 5. (JBMO 00, Shotlist) Let ; b; c be positive el numbes. Pove tht b b c c b b c c Execise. (Poposed by Mildof) Let ; b; c be positive el numbes. Pove tht p 4 4b p 4b 4c p 4c 4 4 b 4b b c 4c c Execise 7. (Poposed by Co Minh Qung) Let ; b; c be positive el numbes. Setting A = b c; B = p p b p b c p c ; C = b b bc c c ; D = ( b c ) bc ; nd E = b b c c Pove tht A B C D E. Refeences 5

6 [] Phm Kim Hung, Secets in Inequlities (in Vietnmese), 00. [] Cux Mthemticoum, 0, 0, 04. [] Mthemticl Re ections, 0, 0, 04. Contct infomtion Co Minh Qung, Nguyen Binh Khiem High School, Vinh Long, Vietnm. E-mil ddess