Solution by Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania

Transcription

1 Revista Virtuala Ifo MateTehic ISSN ISSN-L Probleme rouse sre rezolvare Nicusor Zlota, Focsai 08.Prove that C, j N,where the fiboacci, F F F 0 F F, F 0, F + = + + = = = 0 + j + j 09.Let a,b,c be umbers real ad,, <.Prove that a b c a + b + c + + a + b b + c c + a 0.Let a>0 ad sequece ( x) 0, x x,. Evaluate lim!( x a ) Solutio Mathematical Reflectios 5/04, 6/04 Solutio Mathematical Excalibur 460, School Sciece ad Mathematics Associatio (ssma) Solutio Recreatii Matematice /04 J5 Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia By settig a= x, b= y, c= z,whe iequality is equivalet to : x+ y+ z x+ y+ z x+ y+ z 4a+ + ab+ + ac+ 5+ 5x+ y+ z+ x+ 5y+ z+ x+ y+ 5 z ( 5+ ) x+ y+ z By squarig, we obtai

2 Revista Virtuala Ifo MateTehic ISSN ISSN-L x+ y+ z+ x+ 5y+ z+ x+ y+ 5z+ 5x+ y+ z x+ 5 y+ z ( x+ y+ z)(9+ 4 5),(*) 5x+ y+ z x+ 5 y+ z (+ 5)( x+ y+ z) We show that 5x+ y+ z x+ 5y+ z 5( x+ y) + z 8 xy+ ( 5) yz+ ( 5) zx> 0, x, y, z> 0, is true similarly we get x+ 5y+ z x+ y+ 5z 5( y+ z) + x x+ y+ 5z 5x+ y+ z 5( x+ z) + y,that by gatherig obtai iequality (*) J9. Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia Alyig Cauchy-Schwarz-Buiaovsi, we a a a a ( a+ a+.. + a) = aa aa aa aa aa aa aa a a aa + aa a J. Let x,y,z be ositive real umbers such that xyz(x+y+z)=.prove that 54 9 x + y + z + ( x+ y+ z) Proosed by Marius Staea, Zalau, Romaia Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia Let

3 Revista Virtuala Ifo MateTehic ISSN ISSN-L x+ y+ z= u xy+ yz+ zx= v xyz= w, the we have xyz( x+ y+ z) = uw = uw =, () This iequality is equivalet to 54 x y z ( x+ y+ z) ( x+ y+ z) ( x y + y z + z x ) + 54( xyz) 9( x+ y+ z) ( xyz) u (9v 6 uw ) 54w 9(9 u )( w ) u (v ) + w Hece, our iequality is equivalet to, where is a icreasig fuctio. J. I triagle ABC sia+ sib+ sic= 5 Proosed by Titu Adreescu, Uiversity of Texas at Dallas, USA Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia I triagle ABC, we have : a< b+ c sia< sib+ sic Therefore 5 5 sia+ sia< sia+ sib+ sic= sia< = si6 4 0 A> 6

4 Revista Virtuala Ifo MateTehic ISSN ISSN-L S Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia Let f ( abc,, ) = ( a+ b+ )( c+ ) we show that : f ( abc,, ) 0 We have f ( abc,, ) = ( a+ b+ )( c+ ) = ( a+ b+ )( c+ ) 9 ( a+ b+ )( c+ ) + ac+ a+ bc+ b+ ( a+ b+ c) f ( abc, ) = = ( a+ b+ )( c+ ) + ( a b) + ( bc ) + ( ca ) ( a+ b+ )( c+ ) + 0, abc,, 0 S4 Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia we show that : P= ( x y )( y z )( z x ) = q We have : ( x y)( y z)( z x) = q,() ( x + xy+ y )( y + yz+ z )( z + zx+ x ) = q + q+,() We wor i C. Let + i α =.Tthe, α = ad thus

5 Revista Virtuala Ifo MateTehic ISSN ISSN-L ( x αy)( y αz)( z αx) ( α ) xyz α( x y y z z x) α ( xy yz zx ) = = α+ α q= α( + αq) The asume comutatio with α relaced by α = ) roves : α every where (ad usig ( ) = istead of α ( x+ y)( y+ z)( z+ x) = ( + q) α α α α α But ay two comlex u ad v satisfy ( u+ αv)( u+ v) = u + uv+ v α Hece, ( x + xy+ y )( y + yz+ z )( z + zx+ x ) = ( x+ αy)( y+ αz)( z+ αx)( x+ y)( y+ z)( z+ x) = α α α α( + αq) ( + q) = + q+ q α α From () ad (), obtai P= ( x y )( y z )( z x ) = q S9.Let a,b,c be ositive real umbers such that a+b+c=.prove that for ay ositive real umber t, ( at + bt+ c)( bt + ct+ a)( ct + at+ b) t Proosed by Titu Adreescu, Uiversity of Texas at Dallas, USA Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia S9.Let a,b,c be ositive real umbers such that a+b+c=.prove that for ay ositive real umber t, ( at + bt+ c)( bt + ct+ a)( ct + at+ b) t

6 Revista Virtuala Ifo MateTehic ISSN ISSN-L Proosed by Titu Adreescu, Uiversity of Texas at Dallas, USA Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia Usig atural logarithm, we obtai l[( at + bt+ c)( bt + ct+ a)( ct + at+ b)] lt,() l( at + bt+ c) + l( bt + ct+ a) + l( ct + at+ b) lt Cosider the fuctio f :[0, ) R, f ( t) = l( at + bt+ c) + l( bt + ct+ a) + l( ct + at+ b) lt We shall rove that this fuctio has o-egative derivative f ' ( t) 0,(*) We have at+ b bt+ c ct+ a ( ) = + + ' f t at + bt+ c bt + ct+ a ct + at+ b t at+ b bt+ c ct+ a ( at+ b) (t+ ) + + = at + bt+ c bt + ct+ a ct + at+ b ( at+ b)( at + bt+ c) ( t + t) a + (t + t+ ) ab t,() t(t+ ) ( t + t) a + (t + t+ ) ab For x y z a=, b=, c=,=x+y+z,q=xy+yz+zx,r=xyz, the iequality () becomes x+ y+ z x+ y+ z x+ y+ z (4 4 )( ) ( ) ( ) t + t + t x t t + x + t + t+ xy ( t )((8t + 7t+ ) xy t( t ) x ) 0 t 0, or t (4 xy x ) + t(7 xy+ x ) + xy 0 = xy+ x xy xy x = q + q= q + q q + q t 4 4 (7 ) 4 (4 ) (4 ) 9 0 We have roved (*), therefore the fuctio f is icreasig. It follows that f ( t) 0, t, q.e.d

7 Revista Virtuala Ifo MateTehic ISSN ISSN-L U4 Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia Deote by t 0 =, for, we have : t t l= lim ( ) = lim t 0( ) Alyig l'hosital's rule, we : t t l t t l l.. l ex lim l= t 0 = ex limt 0 = t t t l + l l ex =... =! We show by iductio that: t! >,(), or =, the iequality is true. e Suose that () is true ad rove ( + )! > ( )! ( + > ( + )( ) > ( ) e> ( ) e e e e e> ( + ) S07 Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia

8 Revista Virtuala Ifo MateTehic ISSN ISSN-L Tae D BC, the i triagle ADB, we have succesively : ADB AD= AB sib a= 4csiB From the theorem of siuses we get : sia= 4siC sib si(40 B) = 4si( B 60) sib si(0 B) cos(0 B) = 4siB si(60 B) ( cosb+ si B)( sib cos B) = 4si B(siB cos B) ( 4) si B+ (+ 4 ) sib cosb cos B= 0 Deotig tgb=t, the equatio becomes : ( 4) t ( 4 ) t 0 t + + = = +, whece tgb= + B= 75 0 O4.Let a,b,c,d be oegative real umbers such that a b c d abcd = 5.Prove that abc+ bcd+ cda+ dab abcd Proosed by A Zhe-ig, Xiayag Normal Uiversity, Chia Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia We fix m= a+ b ad = c+ d.let x=ab ad y=cd, the we have m mx y xy + + = 5,(),ad abc+ bcd+ cda+ dab abcd = x+ my xy = f ( x, y) Is a liear (covex) fuctio i both x ad y. It oly reaches the maximum at boud ary values, amely : m max f ( x, y) = f ( α, β ); α {0, }, β {0, } 4 4 If m α = ad 4 β =, we have : a=b, c=d. I this case, the roblem becomes : 4

9 Revista Virtuala Ifo MateTehic ISSN ISSN-L a c ac + + = 5, ac ac ac + a= c The equality holds for a=b=c=d= m Otherwise, if α # ad 4 β # 4, we must have m=0 or abcd=0, assume that d=0, the iequality becomes abc if a + b + c = 5 This follows immediately from AM-GM iequality ad attais equality for a= b= c= 5 We are doe. 460.If x,y,z >0 ad x+y+z+=xyz, the rove that x+ y+ z+ 6 ( yz+ zx+ xy) Solutio () Solutio by Nicuşor ZLOTA, Traia Vuia Techical College, Focşai, Romaia Iegalitatea di eut se mai oate scrie astfel yz zx xy x+ y+ z ,(*) yz zx xy Notam cu : a=, b=, c=, atuci yz zx xy a b c x=, y=, z=, deci coditia di eut devie bc ca ab a b c abc = 4, () π Petru a,b,c umere reale ozitive exista ABC,, (0, ) astfel icat

10 Revista Virtuala Ifo MateTehic ISSN ISSN-L ,a=cosA,b=cosB,c=cosC, atuci () devie este adevarata cos A cos B cos C cosa cosb cosc =, care Ilociud i (*), obtiem iegalitatea a b c abc ab bc ca ( + + ) cos A cos B cos C 6 cosa cosb cosc (cosa cosb cosb cosc cosc cos A) + 4 cosa cosb cosc (cosa cosb+ cosb cosc+ cosc cos A),() Utilizad formulele s ( R+ r) 4R cosa=, cosacosb= s + r 4R 4R, atuci iegalitatea (), devie : s ( R+ r) s + r 4R 4 8 4R 4R + s R + Rr+ r,() Petru a demostra iegalitatea (), avem urmatorul rezultat Itr-u triughi eobtuzughic exista iegalitatea Waler a + b + c 4( R+ r) s r 8Rr 4( R+ r) 0 s R + 8Rr+ r, adica () Demostratie(iegalitatea lui Waler) Avem succesiv Solutio () Notam cu x=a+ab,y=b+bc,z=c+ca, astfel icat abc=, care verifica coditia di eut. Atuci iegalitatea devie a+ ab+ b+ bc+ c+ ca+ 6 ( ( b+ bc)( c+ ca) + ( c+ ca)( a+ ab) + ( a+ ab)( b+ bc)) = ( ( bc+ )( + c) + ( ca+ )( + a) + ( ab+ )( + b), ude am folosit iegalitatea α+ β αβ si ( b+ bc)( c+ ca) = bc+ abc+ bc + abc = bc+ + bc + c= ( bc+ )( + c)

11 Revista Virtuala Ifo MateTehic ISSN ISSN-L If lim ( + ) = a> 0 =, the comute + = lim ( ) a Proosed by D.M. Ba tietu-giurgiu, Matei Basarab Natioal College, Bucharest, Romaia ad Neculai Staciu George Emil Palade School, Buzau, Romaia Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia Avem cazul Notam cu a = + =, atuci limita se oate scrie astfel : a a a a a lim a a a a a a e lim ( ) lim [( ) ] a l= + = + = a a Fie a a a a l = lim = lim, si alicad lema lui Cesaro-Stolz, avem succesiv : a a l a a lim lim + a a = = ( + ) + + ( ( )) (9 )( ( ) ( ) ) lim + + lim l = = = a + a (+ ) + (9+ ) ( + ) + 9 ( + ) a Deci limita este l= e a

12 Revista Virtuala Ifo MateTehic ISSN ISSN-L Geeralizare : If, the comute im ( + ) = a> 0 = + = lim ( ), N, a I mod similar se rocedeaza la fel, deci ri urmare, limita este egala cu : ( ) a e 57: Show that i ay triagle ABC, with the usual otatios, that ( ab ) + ( bc ) + ( ca ) 9r a+ b b+ c c+ a Proosed by D.M. Ba tietu-giurgiu, Matei Basarab Natioal College, Bucharest, Romaia ad Neculai Staciu, George Emil Palade School, Buzau, Romaia Solutio () by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia Notam cu x=, y=, z=, atuci iegalitatea devie a b c ( x y) ( y z) ( z x) 4( xy yz zx),care rerezita iegalitatea data la Olimiada Ira 996 Folosid otatiile de mai sus, obtiem 9abc 6Rrs 9r 9r R r, care rerezita iegalitatea lui Euler,ude 4( a+ b+ c) 8s,()

13 Revista Virtuala Ifo MateTehic ISSN ISSN-L abc= 4 Rrs, a+ b+ c= s Petru mai multe detalii rivid demostrarea iegalitatii (), uteti vedea urmatoarele : [.] Yu-Dog Wu, Chag-Jia Zhao- Buildig triagle to rove algebraic iequalities, Octogo Mathematical Magazie, vol, o..a./004, October 004. []. [4] Cezar Luu, Asura iegalitatii lui Gerretse, RMT, vol XI ( seria a IV-a), ag. -0, o.4/006., htt:// []. Titu Adreescu, Old ew iequalities, Solutio () by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia Folosid iegalitatiile x + y + z xy+ yz+ zx 9xyz xy+ yz+ zx x + y + z, si uad : x= ab, y= bc, z= ca a+ b b+ c c+ a, obtiem 9xyz ab c 9( abc) xy+ yz+ zx 9r x+ y+ z ( a+ b)( b+ c) ab ( ) ( a+ b) a+ b 8R s + r + Rr Alicad iegalitatea lui Gerretse, avem s R Rr r R r Rr R r R r ( )( + ) 0, care este evideta, deoarece R r, ude am utilizat urmatoarele formule : a+b+c=s,

14 Revista Virtuala Ifo MateTehic ISSN ISSN-L abc= Rrs ab+ bc+ ca= s + r + Rr 4, 4, ( a+ b) = a ab abc= s( s + r + Rr), ab a+ b+ c = s a+ b 59: Proosed by Arady Alt, Sa Jose, CA Fid the smallest value of x y z + + x + y y + z z + x where real x; y; z > 0 ad xy + yz + zx = Solutio by Nicuşor Zlota, Traia Vuia Techical College, Focşai, Romaia Avem x y z x xy xy E= + + = = ( x ) ( x ) = x + y y + z z + x x + y x + y xy y x+ y+ z = ( x ) =, ude am utilizat x + y xy Deci x+ y+ z ( xy+ yz+ zx) E =

15 Revista Virtuala Ifo MateTehic ISSN ISSN-L Geeralizare : Fid the smallest value of x y z + + x + y y + z z + x, where real x; y; z > 0 ad xy + yz + zx =, < Demostratie Vom arata ca : x y z x + +,() x + y y + z z + x Folosim iegalitatea lui Cauchy-Buiaovschi-Schwarz : x x ( )( ( ) ( ( ) ( ) x + y x + y + + x x + y x x + y = x Este suficiet acum sa demostram ca : ( x ) x x ( x + y ),ceea ce este echivalet cu : x x ( x + y ) x x y,() Folosim acum iegalitatea oderata a mediilor, α β γ αa+ βb+ γc a b c, α+ β + γ =, si obtiem iegalitatii de forma

16 Revista Virtuala Ifo MateTehic ISSN ISSN-L x + y ( x ) ( y ) y + z ( y ) ( z ) z + x ( z ) ( x ) De aici, ri isumare, obtiem iegalitatea () si astfel iegalitatea () este demostrata Petru a rezolva cerita dorita, vom utiliza iegalitatea x + y + z ( x+ y+ z) ( ( xy+ yz+ zx)) = Deci, x y z x E= + + = x y y z z x Petru = si =, obtiem ceea ce trebuia demostrat. icuzlota@yahoo.com Recreatii Matematice /04 IX-5 Solutie : Nicusor Zlota Colegiul Tehic Auto Traia Vuia, Focsai Folosid iegalitatea lui Cauchy-Buiaowschi-Schwartz, avem : a ( a+ b+ c) 4 = = 4 rr rr + rr + rr a b a b b c c a, ude : rr + rr + rr = a b b c c a

17 Revista Virtuala Ifo MateTehic ISSN ISSN-L X-54 Solutie : Nicusor Zlota Colegiul Tehic Auto Traia Vuia, Focsai Fie A,B,C,O uctele de afixe ale lui z, z, z si 0, atuci iegalitatea di eut se scrie astfel : max(,, ) AB AC BC + max( OA, OB, OC ),() Fie AB cea mai mica latura a triughiului ABC si = max( OA, OB, OC ).Iegalitatea() devie : + AB, () Folosid relatia lui Leibiz,avem : AB + BC + CA OA + OB + OC = OG + AB, deci AB. Atuci + + AB si este suficiet sa aratam ca : + AB AB ( AB ) 0 L68 Solutie : Nicusor Zlota- Colegiul Tehic Auto Traia Vuia, Focsai Notam cu = a+ b+ c, q= ab+ bc+ ca, r = abc, atuci iegalitatea di eut devie a ab a q a ( a) a b c a + = ( q) ( a) 4 a ( b)( c) ( a) q+ r 6q 6qr 0,(),ude: ( + a) = + q+ r, 4 ( ) a + b+ c+ bc = + q+ r

18 Revista Virtuala Ifo MateTehic ISSN ISSN-L Iegalitatea () se oate scrie astfel : ( 4q 9 r) q( 4q 9 r) 6( q r qr) 5 r( q) ,, care este evideta deoarece , q r q r qr q 0 0,care rerezita iegalitatile lui Schur. L7 Solutie Nicusor Zlota, Colegiul Tehic Auto Traia Vuia, Focsai Notam cu a=y+z,b=z+x,c=x+y, atuci iegalitatea di eut este echivaleta cu : bc ( y+ z)( z+ x) x y z + ( xy+ yz+ zx)( x y + y z + z x ) 0R 4 ( a) x x y z r = = x y z 4xyz x y z ( xy yz zx)( x y y z z x ) 0 ( x y)( y z)( z x) 4,(), ude R abc ( x+ y)( y+ z)( z+ x) = = r 4S S 4xyz Daca x,y,z sut umere reale ozitive si otam cu : = x+ y+ z, q= xy+ yz+ zxr, = xyz, atuci iegalitatea () devie : r + q( q r) 5( q r) 4 r r, care este evideta, deoarece : q 4qr+ 9r 0 q 9r 0, care rerezita iegalitatile lui Schur. 0 q qr 8r 0 ( q 4qr 9 r ) r( q 9 r) 0

19 Revista Virtuala Ifo MateTehic ISSN ISSN-L