A Brief Introduction to Olympiad Inequalities

Transcription

1 Ev Che Aprl 0, 04 The gol of ths documet s to provde eser troducto to olympd equltes th the stdrd exposto Olympd Iequltes, by Thoms Mldorf I ws motvted to wrte t by feelg gulty for gettg free 7 s o problems by smply regurgttg few trcks I hppeed to kow, whle other studets were uble to solve the problem Wrg: These re otes, ot full hdout Lots of the exposto s very mml, d my thgs re left to the reder I problem wth vrbles, these respectvely me to le through the vrbles, d to go through ll! permuttos To provde exmple, three-vrble problem we mght wrte = + b + c b = b + b c + c = + + b + b + c + c sym b = b + c + b c + b + c + c b sym Polyoml Iequltes AM-GM d Murhed Cosder the followg theorem Theorem (AM-GM) For oegtve rels,,, we hve Equlty holds f d oly f = = = For exmple, ths mples + b b, + b + c bc Addg such equltes c gve us some bsc propostos

2 Ev Che (Aprl 0, 04) Exmple Prove tht + b + c b + bc + c 4 + b 4 + c 4 bc + b c + c b Proof By AM-GM, Smlrly, + b b + c b d 4 + b 4 + c 4 4 bc d b4 + c bc b c c + c d c b 4 c b 4 Summg the bove sttemets gves + b + c b + bc + c d 4 + b 4 + c 4 bc + b c + c b Exercse Prove tht + b + c b + b c + c Exercse 4 Prove tht 5 + b 5 + c 5 bc + b c + c b bc(b + bc + c) The fudmetl tuto s beg ble to decde whch symmetrc polyomls of gve degree re bgger For exmple, for degree, the polyoml + b + c s bggest d bc s the smllest Roughly, the more mxed polyomls re the smller From ths, for exmple, oe c mmedtely see tht the equlty ( + b + c) + b + c + 4bc must be true, sce upo expdg the LHS d ccellg + b + c, we fd tht the RHS cots oly the pddlg term 4bc Tht mes strght AM-GM wll suffce A useful formlzto of ths s Murhed s Iequlty Suppose we hve two sequeces x x x d y y y such tht d for k =,,, x + x + + x = y + y + + y, x + x + + x k y + y + + y k, The we sy tht (x ) mjorzes (y ), wrtte (x ) (y ) Usg the bove, we hve the followg theorem Theorem 5 (Murhed s Iequlty) If,,, re postve rels, d (x ) mjorzes (y ) the we hve the equlty sym x x x y y y sym

3 Ev Che (Aprl 0, 04) Exmple 6 Sce (5, 0, 0) (,, ) (,, ), b 5 + b 5 + c 5 + c 5 bc + bc + b c + b c + c b + c b b c + b c + b c + b c + c b + c b From ths we derve 5 + b 5 + c 5 bc + b c + c b bc(b + bc + c) Notce tht Murhed s symmetrc, ot lc For exmple, eve though (, 0, 0) (,, 0), Murhed s equlty oly gves tht ( + b + c ) b + c + b c + b + c + c b d prtculr ths does ot mply tht + b + c b + b c + c These stutos must stll be resolved by AM-GM No-homogeeous equltes Cosder the followg exmple Exmple 7 Prove tht f bc = the + b + c + b + c Proof AM-GM loe s hopeless here, becuse wheever we pply AM-GM, the left d rght hd sdes of the equlty ll hve the sme degree So we wt to use the codto bc = to force the problem to hve the sme degree The trck s to otce tht the gve equlty c be rewrtte s + b + c / b / c / ( + b + c) Now the equlty s homogeeous Observe tht f we multply, b, c by y rel umber k > 0, ll tht hppes s tht both sdes of the equlty re multpled by k, whch does t chge ythg Tht mes the codto bc = c be gored ow Sce (, 0, 0) ( 4,, ), pplyg Murhed s Iequlty solves the problem The mportce of ths problem s tht t shows us how to elmte gve codto by homogezg the equlty; ths s very mportt (I fct, we wll soo see tht we c use ths reverse we c mpose rbtrry codto o homogeeous equlty) Prctce Problems 7 + b 7 + c 7 4 b + b 4 c + c 4 If + b + c =, the + b + c + ( +b +c ) bc bc + b c + c b + b + c 4 If + b + c =, the ( + )(b + )(c + ) 64 5 (USA 0) If + b + c + ( + b + c) 4, the b + ( + b) + bc + (b + c) + c + (c + ) 6 If bcd =, the 4 b + b 4 c + c 4 d + d 4 + b + c + d

4 Ev Che (Aprl 0, 04) Iequltes Arbtrry Fuctos Let f : (u, v) R be fucto d let,,, (u, v) Suppose tht we fx = (f the equlty s homogeeous, we wll ofte sert such codto) d we wt to prove tht f( ) + f( ) + + f( ) s t lest (or t most) f() I ths secto we wll provde three methods for dog so We sy tht fucto f s covex f f (x) 0 for ll x; we sy t s cocve f f (x) 0 for ll x Note tht f s covex f d oly f f s cocve Jese / Krmt Theorem (Jese s Iequlty) If f s covex, the f( ) + + f( ) ( + + f ) The reverse equlty holds whe f s cocve Theorem (Krmt s Iequlty) If f s covex, d (x ) mjorzes (y ) the f(x ) + + f(x ) f(y ) + + f(y ) The reverse equlty holds whe f s cocve Exmple (Shortlst 009) Gve + b + c = + b + c, prove tht ( + b + c) + ( + b + c) + ( + b + c) 6 Proof Frst, we wt to elmte the codto The orgl problem s equvlet to ( + b + c) + ( + b + c) + ( + b + c) 6 + b + c + b + c Now the equlty s homogeeous, so we c ssume tht + b + c = Now our orgl problem c be rewrtte s 6 ( + ) 0 Set f(x) = 6x (x+) We c check tht f over (0, ) s covex so Jese completes the problem 4

5 Ev Che (Aprl 0, 04) Exmple 4 Prove tht + b + ( c + b + b + c + ) c b + c Proof The problem s equvlet to + b + c +b + b+c + c+ +b+c + +b+c + +b+c Assume WLOG tht b c Let f(x) = /x Sce ( + b (, b, c), + c, b + c ) ( + b + c, + b + c, + b + c ) the cocluso follows by Krmt Exmple 5 (APMO 996) If, b, c re the three sdes of trgle, prove tht + b c + b + c + c + b + b + c Proof Ag ssume WLOG tht b c d otce tht (+b c, c+ b, b+c ) (, b, c) Apply Krmt o f(x) = x Tget Le Trck Ag fx = + + If f s ot covex, we c sometmes stll prove the equlty f(x) f() + f () (x ) If ths equlty mges to hold for ll x, the smply summg the equlty wll gve us the desred cocluso Ths method s clled the tget le trck Exmple 6 (Dvd Stoer) If + b + c =, prove tht 8 + (b + bc + c) 5 ( c)(4 c) Proof We c rewrte the gve equlty s ( ) 8 ( c)(4 c) c 6 Usg the tget le trck lets us obt the mgcl equlty 8 ( c)(4 c) c c + d the cocluso follows by summg c(c ) (c 9) 0 5

6 Ev Che (Aprl 0, 04) Exmple 7 (Jp) Prove (b+c ) +(b+c) 5 Proof Sce the equlty s homogeeous, we my ssume WLOG tht + b + c = So the equlty we wsh to prove s ( ) + ( ) 5 Wth some computto, the tget le trck gves wy the mgcl equlty: EV ( ) ( ) ( ) 5 5 ( + ) The lst such techque s EV Ths s brute force method volvg much clculus, but t s oetheless useful wepo Theorem 8 ( EV) Let,,, be rel umbers, d suppose s fxed Let f : R R be fucto wth exctly oe flecto pot If f( ) + f( ) + + f( ) cheves mxml or mml vlue, the of the re equl to ech other Proof See pge 5 of Olympd Iequltes, by Thoms Mldorf The m de s to use Krmt to push the together Exmple 9 (IMO 00 / APMOC 04) Let, b, c be postve rels Prove +8bc < Proof Set e x = bc, e y = c, e z = b We hve the codto x + y + z = 0 d wt to b c prove f(x) + f(y) + f(z) < where f(x) = +8e x You c compute f (x) = 4ex (4e x ) (8e x + ) 5 so by EV, we oly eed to cosder the cse x = y Let t = e x ; tht mes we wt to show tht + < + 8t + 8/t Sce ths fucto of oe vrble, we c just use stdrd Clculus BC methods 6

7 Ev Che (Aprl 0, 04) Exmple 0 (Vetm 998) Let x, x,, x be postve rels stsfyg = 998+x = 998 Prove x x x 998 Proof Let y = x Sce y + y + + y =, the problem becomes = ( ) ( ) y Set f(x) = l ( x ), so the equlty becomes f(y ) + + f(y ) f ( ) We c prove tht f (y) = y (y y) So f hs oe flecto pot, we c ssume WLOG tht y = y = y Let ths commo vlue be t; we oly eed to prove ( ) ( ) ( ) l t + l ( )t l( ) Ag, sce ths s oe-vrble equlty, clculus methods suffce 4 Prctce Problems Use Jese to prove AM-GM If + b + c = the + + b + + c + 6b+c + 6bc+ + 6c+b If + b + c = the (MOP 0) If + b + c + d = 4, the + b + c + d + b + c + d Elmtg Rdcls d Frctos Weghted Power Me AM-GM hs the followg turl geerlzto Theorem (Weghted Power Me) Let,,, d w, w,, w be postve rels wth w + w + + w = For y rel umber r, we defe (w r + w r + + w r ) /r r 0 P(r) = w w w r = 0 If r > s, the P(r) P(s) equlty occurs f d oly f = = = 7

8 Ev Che (Aprl 0, 04) I prtculr, f w = w = = w =, the bove P(r) s just ( r + r + + ) /r r r 0 P(r) = r = 0 By settg r =,, 0, we derve whch s QM-AM-GM-HM Moreover, AM-GM lets us dd roots, lke + b + c + b + c Exmple (Tw TST Quz) Prove ( + b + c) 8 bc + +b +c Proof By Power Me wth r =, s =, w = 9, w = 8 9, we fd tht ( ) + b + c + 8 bc ( + b + c ) (bc) so we wt to prove + b + c + 4bc ( + b + c), whch s cler Cuchy d Hölder Theorem (Hölder s Iequlty) Let λ, λ b,, λ z be postve rels wth λ + λ b + + λ z = Let,,,, b, b,, b,, z, z,, z be postve rels The ( + + ) λ (b + + b ) λ b (z + + z ) λz = λ b λ b z λz Equlty holds f : : : b : b : : b z : z : : z Proof WLOG + + = b + + b = = (ote tht the degree of the o ether sde s λ ) I tht cse, the LHS of the equlty s, d we just ote = λ b λ b z λz (λ + λ b b + ) = If we set λ = λ b =, we derve wht s clled the Cuchy-Schwrz equlty = ( ) (b + b + + b ) ( b + b + + b ) 8

9 Ev Che (Aprl 0, 04) Cuchy c be rewrtte s x y + x y + + x y (x + x + + x ) y + + y Ths form t s ofte clled Ttu s Lemm the Uted Sttes Cuchy d Hölder hve t lest two uses: elmtg rdcls, elmtg frctos Let us look t some exmples Exmple 4 (IMO 00) Prove + 8bc Proof By Holder ( ( + 8bc) ) ( ) + 8bc ( + b + c) So t suffces to prove ( + b + c) ( + 8bc) = + b + c + 4bc Does ths look fmlr? I ths problem, we used Hölder to cler the squre roots the deomtor Exmple 5 (Blk) Prove (b+c) + b(c+) + c(+b) 7 (+b+c) Proof Ag by Holder, ( ) ( b + c ) ( ) (b + c) + + = Exmple 6 (JMO 0) Prove +5b +b ( + b + c ) Proof We use Cuchy (Ttu) to obt + b = ( ) + b ( + b + c ) + b We c esly prove ths s t lest 4 ( +b +c ) (recll +b +c s the bggest sum, so we kew dvce ths method would work) Smlrly 5b +b 5 4 ( + b + c ) 9

10 Ev Che (Aprl 0, 04) Exmple 7 (USA TST 00) If bc =, prove 5 (b+c) + b 5 (c+) + c 5 (+b) Proof We c use Hölder to elmte the squre roots the deomtor: ( ) ( b + c Prctce Problems ) ( 5 (b + c) ) (b + bc + c) If + b + c =, the b + c + bc + + c + b + b + bc + c If + b + c =, the b + c + b c + + c + b (ISL 004) If b + bc + c =, prove + 6b + b + 6c + c (MOP 0) b + b + b bc + c + c c bc 4( + b + c) 5 (Ev Che) If + b + c + bc = 4, prove (5 + bc) ( + b)( + c) + (5b + c) (b + c)(b + ) + (5c + b) (0 bc) (c + )(c + b) + b + c Whe does equlty hold? 4 Problems (MOP 0) If + b + c =, the + b + b + b + bc + c + c + c + bc (IMO 995) If bc =, the (b+c) + b (c+) + c (+b) (USA 00) Prove (+b+c) +(b+c) 8 4 (Rom) Let x, x,, x be postve rels wth x x x = Prove tht = +x 5 (USA 004) Let, b, c be postve rels Prove tht ( 5 + ) ( b 5 b + ) ( c 5 c + ) ( + b + c) 6 (Ev Che) Let, b, c be postve rels stsfyg + b + c = b + 7 c Prove b b c c 0