Các cách chứng minh bất đẳng thức

Buứi Quang Thũnh THC PHNG PHP GII TON BT NG WWW.ToanCapBa.Net Đ1.GII THIU V BT NG THC 1.1.NH NGHA BT NG THC Trong toỏn hc, bt ng thc c nh ngha nh sau: Bt ng thc l mt phỏt biu v quan h th t gia hai i tng ú l kt qu ca phộp so sỏnh hai i tng a v b vi nhau, nú c vit li thnh mt cỏc dng sau: a > b, a < b, a b, a b Trong ú cỏc kớ hiu >, l ln hn; +) < l hn; +) l ln hn hoc bng; +) l hn hoc bng; 1 *Vớ d: < 2; > , x ; L cỏc bt ng thc toỏn hc *Chỳ ý: -Ta cú a > b v ch a b l mt s nguyờn dng Hai i lng a, b cú th l nhng s c th hoc l nhng biu thc cha bin -Nu mt bt ng thc ỳng vi mi giỏ tr ca bin thỡ c gi l Bt ng thc khụng iu kin Cũn nu nú ch ỳng vi mt s giỏ tr ca bin thỡ c gi l Bt ng thc cú iu kin 1.2.CC TNH CHT C BN CA BT NG THC õy l mt phn rt quan trng ca bt ng thc Bt ng thc gm cú nhng tớnh cht c bn nh sau: 1)Cng tng v hai bt ng thc cựng chiu, ta c bt ng thc mi cựng chiu vi bt ng thc ó cho: a > b, c > d a + c > b + d 2)Tr tng v hai bt ng thc ngc chiu, ta c bt ng thc mi cựng chiu vi bt ng thc b tr 3)Tớnh cht n iu ca phộp nhõn: a)Nhõn hai v ca bt ng thc vi cựng mt s dng: a > b, c > ac > bc b)Nhõn hai v ca bt ng thc vi cựng mt s õm v i chiu ca bt ng thc: a > b, c > ac < bc 4)Nhõn tng v hai bt ng thc cựng chiu m hai v khụng õm: a > b 0, c > d ac > bd 5.Nõng lờn ly tha bc nguyờn dng hai v ca bt ng thc: a > b > an > bn; a > b an > bn vi n l |a| > |b| an > bn vi n s chn 6)So sỏnh hai ly tha cựng c s vi s m nguyờn dng Nu m > n > thỡ a > am > an a = am = an ; < a < am < an WWW.ToanCapBa.Net Buứi Quang Thũnh THC PHNG PHP GII TON BT NG 7)Ly nghch o hai v v i chiu bt ng thc nu hai v cựng du: 1 a > b, ab > < a b 8)Cỏc hng bt ng thc: a)Ngoi cỏc hng bt ng thc a2 0, -a2 cũn cú cỏc hng bt ng thc khỏc cú liờn quan n giỏ tr tuyt i: b) |a| Xy ng thc a=0 |a| a Xy ng thc a |a+b| |a| + |b|.Xy ng thc ab |a-b| |a| - |b|.Xy ng thc ab Chng minh bt ng thc |a+b| |a| + |b|nh sau: |a+b| |a| + |b| (1) a + 2ab + b a + | ab | +b (Vỡ hai v ca (1)khụng õm) ab |ab| (2) Bt ng thc (2) ỳng, vy (1)ỳng Chng minh bt ng thc|a-b| |a| - |b| (3) nh sau: Nu |a| b a ( a + b2 ) ( x + y ) ( ax + by ) (Bt ng thc Bu-nhi-a-cp-xki) Đ2.CC PHNG PHP CHNG MINH BT NG THC 1.1 PHNG PHP BIN I TNG NG A)Kin thc cn nh: Gi s cn chng minh bt ng thc : A B (*) Buứi Quang Thũnh THC PHNG PHP GII TON BT NG T tng ca phng phỏp l bin i tng ng bt ng thc (*) v mt bt ng thc ph bin l a (*) v dng: -Cỏc tng bỡnh phng: A B mX + nY + pZ Trong ú m, n, p l cỏc s khụng õm -Tớch cỏc tha s khụng du: A B X Y (X, Y cựng du) -Tớch ca mt s khụng õm v mt biu thc dng(theo iu kin) A B X Y -Xõy dng cỏc bt ng thc t iu kin ca bi toỏn: Nu x, y, z [ a, b ] thỡ ta ngh n cỏc bt ng thc hin nhiờn ỳng: ( x a ) ( x b ) 0, ( x a ) ( y a ) ( z a ) 0, ( x b ) ( y b ) ( z b ) Mt s bt ng thc c bn cn nh: Vi mi s thc a, b, c ta cú: )( ( ) 2 - 4ab ( a + b ) a + b ( a b ) 2 a + b + c ab + bc + ca ( ab + bc + ca ) ( a + b + c ) ( a + b + c ) Hai bt ng thc ny tng ng vi: 1 2 ( a b) + ( b c) + ( c a) 2 2 - ( ab + bc + ca ) 3abc ( a + b + c ) (õy l h qu trc tip ca (x + y + z)2 ( xy + yz + zx ) Ta ch cn cho x = bc, y = ca, z = ab l thu c kt qu nh trờn) B)Cỏc vớ d: Vớ d 1.1.1.Chng minh rng cỏc s thc a, b, c, d, e ta cú: a + b2 + c + d + e2 a ( b + c + d + e ) Li gii: Nhõn c hai v ca bt ng thc cn chng minh cho 4, ta vit c nú li thnh: a 4ab + 4b + a 4ac + 4c + a 4ad + 4d + a 4ae + 4e ( a 2b ) + ( a 2c ) + ( a 2d ) + ( a 2c ) 2 2 Bt ng thc cui cựng ỳng, suy iu phi chng minh ng thc xy v ch a = 2b = 2c = 2d = 2e Vớ d 1.1.2.Chng minh rng vi mi s thc x, y ta cú: x + y + xy + x + y Li gii: 2 x + y + xy + x + y 2 2 Ta cú: x xy + y + x x + + y y + ( x y ) + ( x 1) + ( y 1) 2 Bt ng thc c chng minh ng thc xy v ch x = y = Buứi Quang Thũnh THC PHNG PHP GII TON BT NG Vớ d 1.1.3.Chng minh rng vi mi s thc x, y, z ta cú: x + y + z xy + yz + xz Li gii: 2 x + y + z xy + yz + xz 2 2 2 Ta cú: x xy + y + y yz + z + z xz + x ( x y) + ( y z) + ( z x) 2 Bt ng thc c chng minh ng thc xy v ch x = y = z Vớ d 1.1.4.Cho cỏc s thc a, b, c tha a + b + c = Chng minh ab + 2bc + 3ac Li gii: Theo gi thit thỡ c = -a b nờn bt ng thc ó cho tng ng vi: ab + c ( 2b + 3a ) ab + ( a b ) ( 2b + 3a ) ab 2ab 3a 2b 3ab 3a + 4ab + 2b a2 + ( a + b ) Bt ng thc c chng minh ng thc xy v ch a = b = c = Vớ d 1.1.5.Chng minh rng vi mi s thc x ta cú: x x + x x +1 Li gii: Ta cú: x x3 + x x + ( x x + x ) + ( x x + 1) ( x x ) + ( x 1) 2 Bt ng thc cui ỳng nờn ta cú iu chng minh ng thc xy v ch x = Vớ d 1.1.6.Chng minh rng vi mi a > ta cú: a + 11 a + a2 + 2a Li gii: Ta cú: ( ) Buứi Quang Thũnh THC PHNG PHP GII TON BT NG ( a + 1) 11 a a ( a + 1) + + a2 + 2a a +1 2a ( a 1) ( a 1) ( a 1) + ữ 2a a a2 +1 ( a + 1) ( a 1) 2 2 ( a 1) ì( a 1) + 9(a + 1) 5a a + ì 2a (a + 1) a ( a + 1) 2 Bt ng thc cui ỳng nờn ta cú iu chng minh ng thc xy v ch a = Vớ d 1.1.7.Cho cỏc s thc dng a, b, c Chng minh rng: a b c + + b +c c + a a +b (Bt ng thc Nestbit cho ba s thc dng) Li gii: Hin cú khong 50 cỏch chng minh bt ng thc ny, õy ta chng minh bng phng phỏp bin i tng ng.Ta cú: a b c a b c + + + + b+c c+a a+b b+c c +a a+b a b a c b c b a c a c b + + + + + b+c b+c c+a c+a a+b a+b a b a b bc bc c a c a ữ+ ữ+ ữ b+c c+a c+a a+b a +b b+c ( a b) + ( b c ) ( c a) + ( b + c) ( c + a) ( c + a) ( a + b) ( a + b) ( b + c) 2 Vớ d 1.1.8.Cho cỏc s thc a, b [ 0,1] Chng minh rng a + b ab + 1+ a + b 2 Li gii: Bt ng thc cn chng minh tng ng vi ( a ) (1 b)(a + b) + ab(2 a b) Vỡ a, b [ 0,1] nờn a 0;1 b 0; a b , ú ta cú iu phi chng minh Vớ d 1.1.9.Cho cỏc s thc a, b, c Chng minh rng : (a + b + c ) ( a 3b + b3c + c3a ) Li gii: õy l mt bi toỏn hay v khú, cng cú rt nhiu li gii phc tp, nhng tht bt ng li cú mt li gii ch dựng phng phỏp bin i tng tng C s ca phng phỏp l tỡm cỏch a bt ng thc v dng Buứi Quang Thũnh THC PHNG PHP GII TON BT NG (ma + nb + pc + qab + rbc + sca) + (mb + nc + pa + qbc + rca + sab) + (mc + na + pb + qca + rab + sbc) T hng ng thc (a + b + c ) ( a 3b + b3c + c 3a ) = 2 2 2 1 a b + 2bc ab ac ) + ( b c + 2ca bc ba ) + ( c a + 2ab ca cb ) ( 2 Suy iu phi chng minh Nhn xột Ngoi ra, ta cng cú mt s hng ng thc tng t (m t chỳng cú th suy c nhng bt ng thc rt khú) = Vớ d 1.1.10.Cho a, b, c [ 1, 2] v a + b + c = Chng minh rng a + b + c Li gii: Ta cú a, b, c [ 1, 2] nờn ( a + 1) ( a ) a a Tng t b b; c c nờn a + b + c a + b + c hay a+b+c Vớ d.1.1.11.Cho cỏc s thc x, y, z khỏc v xyz = Chng minh rng: x2 y2 z2 + + 2 ( x 1) ( y 1) ( z 1) ( thi Toỏn quc t nm 2008) Li gii: 1 Vỡ xyz = nờn x, y, z 0, t a = ; b = , c = thỡ ta cú: x y z abc = v a, b, c khỏc 0, khỏc Khi ú, bt ng thc cn chng minh tr thnh 1 1 + + + + ữ 2 a b c ( a) ( b) ( c) 1 + + 1 a b b c c a ( ) ( ) ( ) ( ) ( ) ( ) ( a + b + c ) + ab + bc + ca ( a + b + c) ab + bc + ca a + b + c ab + bc + ca a + b + c ( ) ( ) ( a + b + c) ( a + b + c) + ab + bc + ca ( a + b + c ) ab + bc + ca ( a + b + c ) ( a + b + c) 1+ ab + bc + ca a + b + c ( ) ú chớnh l iu phi chng minh BI TP T GII Bi 1.1.1.Cho cỏc s thc x, y Chng minh rng: Buứi Quang Thũnh THC PHNG PHP GII TON BT NG (a) x + y ( x + y) x + y) (b) x + y ( Bi 1.1.2 Chng minh bt ng thc sau ỳng vi mi s thc a, b a + b ( a + b2 ) Bi 1.1.3 Cho cỏc s thc x, y, z Chng minh rng x + y + z) ( 2 x +y +z Bi 1.1.4.Cho cỏc s thc m, n, p, q Chng minh rng: ( m2 + n2 ) ( p2 + q ) ( mq + np ) Bi 1.1.5 Cho a, b l cỏc s thc khụng õm tựy ý Chng minh rng: a + b a + b 2( a + b) Khi no cú du ng thc ? Bi 1.1.6 Cho a, b, c l cỏc s khụng õm tha a + b + c = Chng minh rng: b + c 19abc Bi 1.1.7.Cho cỏc s thc dng a, b Chng minh rng ab ab a+ b Bi 1.1.8.Chng minh vi mi s thc x, y, z, t ta luụn cú bt ng thc x2 + y + z + t x ( y + z + t ) Bi 1.1.9.Cho cỏc s thc a, b Chng minh rng 3 4 (a) ( a + b ) ( a + b ) ( a + b ) 4 2 (b) ( a + b ) ( a + b + 6a b ) Bi 1.1.10.Cho cỏc s thc dng a, b Chng minh rng: Buứi Quang Thũnh THC PHNG PHP GII TON BT NG 5a b3 2a b 3a + ab a 11b3 b) a 3b 4b + ab a b ab c) + + b a a ab + b a b 9ab 13 d) + + b a a +b a) a 2b ab b ( a + b) a f ) + + e) + 3 2b a + b ( a + 2b ) 3b a + ab + b 2.2.PHNG PHP DNG NH NGHA A)Kin thc cn nh: chng minh A > B, ta xột hiu A B v chng minh rng A B l s dng B)Cỏc vớ d: Vớ d 1.2.1 Chng minh rng ( x 1) ( x ) ( x 3) ( x ) Li gii: Xột hiu: ( x 1) ( x ) ( x 3) ( x ) ( 1) = ( x x + ) ( x x + ) + t x x + = y , biu thc trờn bng ( y 1) ( y + 1) + = y Vy ( x 1) ( x ) ( x 3) ( x ) 2.3.PHNG PHP QUY NP A)Kin thc cn nh: Ni dung ca phng phỏp quy np: Mt bt ng thc ph thuc vo s nguyờn dng n c xem l ỳng nu tha hai iu kin: -Bt ng thc ỳng vi giỏ tr u tiờn ca n -T gi thit bt ng thc ỳng vi n = k ( k Ơ ) suy c bt ng thc ỳng vi n = k + Cỏc bc chng minh bt ng thc bng phng phỏp quy np: Bc 1: Chng minh bt ng thc vi giỏ tr u tiờn ca n Bc 2:Gi s bt ng thc ỳng vi n = k (gi l gi thit quy np), sau ú chng minh bt ng thc ỳng vi n = k + Bc 3: Kt lun bt ng thc ó cho ỳng B)Cỏc vớ d: Vớ d 1.3.1.Chng minh rng vi mi s nguyờn n ta cú: 2n > n Li gii: Vi n = 5, bt ng thc tr thnh 25 > 52 32 > 25 (ỳng) Bt ng thc ỳng vi n = Gi s bt ng thc ỳng vi n = k ( k Ơ , k ) , tc l 2k > k Ta cn chng minh bt ng thc ỳng vi n = k + 1, hay Buứi Quang Thũnh THC 2k +1 > ( k + 1) PHNG PHP GII TON BT NG k +1 k Tht vy, theo gi thit quy np, ta cú: = 2.2 > 2k ( 1) Vỡ k nờn 2k = k + 2k + + k 2k = ( k + 1) + k ( k ) + 3k > ( k + 1) 2 2k > ( k + 1) ( ) T (1) v (2) ta cú bt ng thc ỳng vi n = k + 1, nờn theo nguyờn lớ quy np ta cú iu phi chng minh Vớ d 1.3.2.Cho x l mt s thc cho trc Chng minh rng vi mi s nguyờn n, ta cú n ( + x ) + nx Li gii: Vi n = 1, bt ng thc tr thnh: + x + x (ỳng) Bt ng thc ỳng vi n = Gi s bt ng thc ỳng n n = k ( k Ơ , k 1) , tc l ( 1+ x) k + kx Ta cn chng minh bt ng thc ỳng vi n = k + 1, hay k ( + x ) + ( k + 1) x Tht vy, vỡ x x + nờn theo gi thit quy np, ta cú: k +1 k ( + x ) = ( + x ) ( + x ) ( + x ) ( + kx ) ( + x ) ( + kx ) = + ( k + 1) x + kx + ( k + 1) x k +1 Nờn ( + x ) ( + x ) ( + kx ) + ( k + 1) x Hay bt ng thc ỳng vi n = k + 1, nờn theo nguyờn lý quy np suy iu phi chng minh Vớ d 1.3.3.Chng minh vi mi s thc a, b tha a + b ta cú: n a n + bn a + b ữ Li gii: Vi n = bt ng thc hin nhiờn ỳng Gi s bt ng thc ỳng n n = k (k Ơ , k 1) , tc l k a k + bk a + b ữ Ta chng minh bt ng thc ỳng vi n = k + 1, hay k +1 a k +1 + b k +1 a + b ữ tht vy, vỡ a + b nờn theo gi thit quy np ta cú: k +1 a+b ữ k k k a+b a +b a +b a+b = ì ì ữ 2 a k + b k a + b a k +1 + b k +1 Bt ng thc ỳng vi n = k + nu ta chng minh c ì ( *) 2 Buứi Quang Thũnh THC M (*) tng ng vi (a k PHNG PHP GII TON BT NG + b k ) ( a + b ) ( a k +1 + b k +1 ) a k +1 + b k +1 a k b b k a ( a b ) ( a k b k ) ( **) Vỡ vai trũ ca a, b nh nờn ta cú th gi s a b Khi ú a - b (1) Mt khỏc, t a + b a b , ta cú: (2) a b a k bk b k a k bk T (1) v (2) ta cú (**) ỳng Vy theo nguyờn lớ quy np ta cú iu phi chng minh BI TP T GII 1.3.1.Chng minh rng vi mi s nguyờn n ta cú: n +1 a) > n + 3n, ( n ) n b) > n + 4n + 5, ( n 3) 1.3.2.Chng minh rng vi mi s nguyờn n ta cú: 1 + + + , ( n 3) a) n +1 n + n+n 1 + + + < n , (n 1) b) + n 1.3.3 Chng minh rng vi mi s nguyờn dng n ta cú: 3 a) 13 + 33 + 53 + + ( 2n + 1) > ( n + 1) ( 2n + 1) n ( n + 1) ( 2n + 1) 10 b) 14 + 24 + 34 + + n > 1.3.4 Tỡm tt c cỏc s nguyờn dng n tha bt ng thc 3n > 2n + n 1.3.5 Chng minh rng vi mi s t nhiờn n ta cú n n n > ( n + 1) 1.3.6.Chng minh rng vi mi n , n Ơ , ta cú: 1 + + + < n +1 n + 2n 10 1.3.7 Chng minh bt ng thc sau ỳng vi mi n Ơ * 1 79 + + + + < n 48 1.3.8 Chng minh vi mi n Ơ * , ta cú: 10 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG 1.5.3.Cho a, b, c l cỏc s thc dng tha a + b+ c abc Chng minh rng cú ớt nht hai 6 s cỏc bt ng thc sau ỳng + + 6, + + 6, + + a b c b c a c a b 1.5.4.Cho cỏc s nguyờn dng x, y Chng minh rng cú ớt nht mt hai bt ng thc sai 1 1 1 1 2+ + ữ, xy y x( x + y) 5x x ( x + y ) 1.5.5.Cho a, b, c l cỏc s thc dng tha abc = Chng minh 1 + + + 5x + 5y + 5z 2.6.1.PHNG PHP DNG BT NG THC KINH IN A.Gii thiu bt ng thc CAUCHY Nu a1, a2, , an l cỏc s thc khụng õm thỡ a1 + a2 + + an n a1a2 an n Bt ng thc ny cú tờn gi chớnh xỏc l bt ng thc gia trung bỡnh cng v trung bỡnh nhõn nhiu nc trờn th gii, ngi ta gi bt ng thc ny theo kiu vit tt l bt ng thc AM GM (AM l vit tt ca arithmetic mean v GM l vit tt ca geometric mean) nc ta, bt ng thc ny c gi theo tờn ca nh Toỏn hc ngi Phỏp Augustin Louis Cauchy (1789 1857), tc l bt ng thc Cauchy Tht õy l mt cỏch gi tờn khụng chớnh xỏc vỡ Cauchy khụng phi l ngui xut bt ng thc ny m ch l ngi a mt phộp chng minh c sc cho nú Tuy nhiờn, cho phự hp vi chng trỡnh sỏch giỏo khoa, ti liu ny chỳng ta cng s gi nú l Bt ng thc Cauchy õy l mt bt ng thc c in ni ting v quen thuc i vi phn ln hc sinh nc ta Nú ng dng rt nhiu cỏc bi Toỏn v bt ng thc v cc tr Trong phm vi chng trỡnhToỏn THCS, chỳng ta quan tõm n ba trng hp riờng ca bt ng thc Cauchy l: -Trng hp n = Lỳc ny bt ng thc c vit li rng: Nu a, b l cỏc s thc khụng õm, a+b ab thỡ: Du bng xy v ch a = b Bt ng thc ny cũn c vit hai dng khỏc tng ng l 2 a + b) a+b ( 2 ab ữ v a + b -Trng hp n = Ta cú bt ng thc Cauchy cho ba bin khụng õm: Nu a, b, c l cỏc s thc a+b+c abc khụng õm, thỡ : Du bng xy v ch a = b Trong thc t ỏp dng, ta cũn s dng mt dng khỏc tng ng ca bt ng thc ny l: a+b+c abc ữ 20 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG -Trng hp n = Trong trng hp ny, ta cú bt ng thc Cauchy cho bn bin khụng õm: Nu a, b, c, d l cỏc s thc khụng õm, thỡ a+b+c+d abcd Du bng xy v ch a = b= c =d Tng t nh hai trng hp khỏc, ta cng thng hay s dng bt ng thc ny di dng a+b+c+d abcd ữ a+b ab ta cú th suy cỏc -Ngoi ra, t trng hp bt ng thc Cauchy cú n = cú dng bt ng thc tng ng : a + b 2ab a + b ab ( a + b) 4ab 1 + a b a+b B.Cỏc k thut s dng bt ng thc Cauchy B.1.K thut s dng Cauchy trc tip 1 Vớ d 1.6.1 Cho cỏc s dng a, b tha + = Chng minh a + b a b Li gii: p dng bt ng thc Cauchy kt hp vi gi thit, ta cú 1 2= + ab > a b ab Ti õy, s dng bt ng thc Cauchy mt ln na, ta c a + b ab 2.1 = ng thc xy a = b =1 Vớ d 1.6.2 Cho bn s dng a, b, c, d Chng minh rng ab + cd ( a + d ) ( b + c ) Li gii: Bt ng thc cn chng minh cú th vit li thnh a b c d ì + ì a+d b+c b+c a+d Ti õy, s dng bt ng thc Cauchy dng xy 21 x+ y , ta cú Buứi Quang Thũnh THC PHNG PHP GII TON BT NG a b c d + + a b c d ì + ì a+d b+c + b+c a+d = a+d b+c b+c a+d 2 a+d b+c + a + d b + c =1 = b a a + d = b + c a b = ng thc xy v ch a+d b+c c = d b + c a + d 1 Vớ d 1.6.3.Cho x, y, z l cỏc s thc dng Chng minh rng ( a + b + c ) + + ữ a b c Li gii: 2 S dng bt ng thc Cauchy dng a + b 2ab , d thy x2 y2 z2 x2 y2 z2 P= + + + + =1 x + yz y + zx z + xy x + y + z y + z + x z + x + y ng thc xy v ch x = y = z Vớ d 1.6.4.Cho cỏc s dng a, b, c Chng minh rng 1 ( a + b + c ) + + ữ a b c Li gii: S dng bt ng thc Cauchy b ba s, ta cú: 1 a + b + c 3 abc > 0, + + >0 a b c abc 1 ( a + b + c ) + + ữ 3 abc ì3 =9 abc a b c Du bng xy v ch a = b = c >0 B.2.K thut ghộp i xng Trong nhiu bi toỏn m biu thc hai v tng i phc tp, vic chng minh trc tip tr nờn khú khn thỡ ta cú th s dng k thut Ghộp i xng bi toỏn tr nờn n gin cỏc bi toỏn bt ng thc, thụng thng chỳng ta hay gp phi hai dng toỏn sau: -Dng 1:Chng minh X + Y + Z A + B + C í tng Nu ta chng minh c X + Y 2A Sau ú, tng t húa ch Y + Z 2B v Z + X 2C (Nh tớnh cht i xng ca bi toỏn) Sau ú cng ba bt ng thc trờn li theo v ri rỳt gn cho 2, ta cú iu phi chng minh -Dng 2.Chng minh XYZ ABC vi X, Y, Z í tng: Nu ta chng minh c XY A2 Sau ú tng t húa ch YZ B2 v ZX C2 (nh tớnh cht i xng ca bi toỏn) Sau ú nhõn ba bt ng thc trờn v theo v ri ly cn bc hai , ta cú XYZ A B2C = ABC ABC Vớ d 1.6.5.Chng minh rng vi mi a, b, c dng ta cú 22 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG abc ( b + c a ) ( c + a b ) ( a + b c ) Li gii: Bt ng thc ny cú dng XYZ ABC, vỡ vy s dng k thut ghộp i xng, ta ch cn chng minh: b ( a + b c ) ( b + c a ) Tht vy, s dng bt ng thc Cauchy dng ( x + y) xy , suy ( a + b c ) + ( b + c a ) = b2 ( a + b c) ( b + c a) Bi toỏn c gii quyt xong Du bng xy v ch a = b = c Vớ d 1.6.6.Cho a, b, c l ba s thc dng Chng minh rng: ab bc ca + + a+b+c c a b Li gii: Bi toỏn ny cú dng X + Y + Z A + B + C vi ab bc ca ab bc X= , Y = , Z = , A = a, B = b, C = c ý rng hai biu thc v l i xng vi b c a b c a (tc vai trũ ca a v c nh nhau) Do ú, s dng k thut ghộp i xng, ta s th chng minh ab bc + 2b c a Qu tht, bt ng thc ny l hin nhiờn ỳng vỡ theo bt ng thc Cauchy ab bc ab bc + ì c a c a T ú, bi toỏn c gii quyt hon ton Vớ d 1.6.7.Mt tam giỏc cú di ba cnh l a, b, c tha 3 ( a + b c ) + ( b + c a ) + ( c + a b ) = a3 + b3 + c3 Chng minh tam giỏc ú l tam giỏc u Li gii: x + y) B : Vi mi x, y > 0, ta cú x + y ( 3 2 Chng minh: Do x + y = ( x + y ) ( x xy + y ) nờn bt ng thc tng ng vi: x xy + y 2 ( x + y) S dng bt ng thc Cauchy hai s dng x xy + y = ( x + y ) 3xy ( x + y ) 2 2 ( x + y) xy ( x + y) 3ì ý rng a, b, c l tam giỏc thỡ hin nhiờn ta cú: 2 , ta c ( x + y) = 23 B c chng minh Buứi Quang Thũnh THC PHNG PHP GII TON BT NG a + b c > 0, b + c a > 0, c + a b > p dng b , ta cú: ( a + b c ) + ( b + c a ) = 2b3 ( a + b c) + ( b + c a) ( b + c a ) + ( c + a b ) = 2c3 ( b + c a) + ( c + a b) 3 ( c + a b ) + ( a + b c ) = 2a ( c + a b) + ( a + b c) Cng ba bt ng thc trờn li v theo v v rỳt gn c hai v ca bt ng thc thu c cho 2, ta cú: 3 ( a + b c ) + ( b + c a ) + ( c + a b ) = a3 + b3 + c3 3 Theo gi thit thỡ du bng xy ra, vy ta phi cú: a + b c = b + c a b + c a = c + a b a = b = c c + a b = a + b c iu ny chng t tam giỏc ó cho l tam giỏc u Vớ d 1.6.8.Cho a, b, c l cỏc s thc dng tha abc = Chng minh rng: ab bc ca + 5 + 5 a + b + ab b + c + bc c + a + ca Li gii: 5 2 B Vi mi a, b > 0, ta cú: a + b a b ( a + b ) 5 2 Chng minh: Ta cú a + b = ( a + b ) ( a a b + a b ab + b ) Do ú bt ng thc tng ng a a 3b + a 2b ab3 + b a 2b a + b a 3b + ab S dng bt ng thc Cauchy b bn s, ta cú: a + a + a + b4 a + b4 + b4 + b4 a 3b , ab3 4 Cng hai bt ng thc ny li, ta thu c kt qu nh trờn Bt ng thc c chng minh S dng b kt hp vi gi thit abc = 1, ta cú kt qu ab ab 1 2 = = 5 a + b + ab a b ( a + b ) + ab ab ( a + b ) + ab ( a + b + c ) c = a+b+c 24 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG Tng t: bc a ca b , 5 b + c + bc a + b + c c + a + ca a + b + c Cng ba bt ng thc trờn li vờ theo v, ta thu c ab bc ca a +b+c + 5 + =1 5 a + b + ab b + c + bc c + a + ca a + b + c B.3.K thut t n ph kt hp Cauchy Trong bt ng thc, cú mt quy lut chung, ú l Trong mt dng c th, thỡ nhng bt ng thc cng nhiu bin cng khú iu ny cng ng ngha vi vic khng nh Bi toỏn s tr nờn n gin hn nu ta a c mt bt ng thc nhiu bin v dng ớt bin hn K thut c n ph chớnh l mt cụng c hu ớch thc hin ý tng ny Vớ d 1.6.9.Cho x, y l hai s thc khỏc Chng minh rng: 4x2 y x2 y2 + + 2 2 y x x + y ( ) Li gii: ý rng bt ng thc cn chng minh cú th vit li thnh 4x2 y2 (x t (x t= + y2 ) x y 2 4x2 y2 (x +y ) 2 = + y2 ) (x + + y2 ) x2 y2 t t+ ( t 1) ( t ) t t 5t + t t Ta c bi toỏn v dng mt biu n gin l: Theo bt ng thc Cauchy , ta d thy t Suy t > 0, t ( t 1) ( t ) bi toỏn c gii quyt hon ton T ú ta c t 25 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG ng thc xy v ch x = y x = y 1 + + = Chng minh rng : x y z ( x 2) ( y 2) ( z 2) Vớ d 1.6.10 Cho x, y, z > v Li gii: Vi gi thit x, y, z u ln hn 2, ta ngh n vic t n ph a bi toỏn v dng n gin v quen thuc hn Hóy xột li gii sau: t x = a + 2, y = b + 2, z = c + vi a > 0, b > 0, c > Bi toỏn quy v chng minh abc vi a, b, c > tha 1 a b c + + =1 + + = n õy ta t tip a+2 b+2 c+2 a+2 b+2 c+2 a b c m= ,n = ,p= m+ n + p =1 a+2 b+2 c+2 a+2 2 n+ p 2m = 1+ = = a= Ta cú: = m a a a m m n+ p 2n 2p ,c = Tng t: b = p+m m+n Do ú bt ng thc tr thnh 2m 2n 2p ì ì ( m + n ) ( n + p ) ( p + m ) 8mnp n+ p p+m m+n S dng bt ng thc Cauchy, ta cú ( m + n ) ( n + p ) ( p + m ) mn np pm = 8mnp Bi toỏn c gii quyt xong ng thc xy m = n = p a = b = c = x = y = z = Cỏch khỏc S dng k thut ghộp i xng Ta cú: 1 1 1 = = ữ+ ữ c+2 a+2 b+2 a+2 b+2 = Tng t a b + ( a + 2) ( b + 2) b+2 ab ( b + 2) ( c + 2) ca , ( c + 2) ( a + 2) a+2 bc ( b + 2) ( c + 2) Nhõn ba bt ng thc trờn li v theo v, ta c abc abc < ( a + 2) ( b + 2) ( c + 2) ( a + 2) ( b + 2) ( c + 2) Vớ d 1.6.11.Cho a, b, c, d l cỏc s thc dng tha ab = cd = Chng minh rng ( a + b) ( c + d ) + ( a + b + c + d ) Li gii: 26 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG ý rng nu t x = a + b, y = c + d, thỡ bt ng thc bn bin cn chng minh tng ng ó c quy v dng hai bin n gin hn l: xy + ( x + y ) x ( y ) + ( y ) ( y ) ( x ) Mt khỏc ỏp dng bt ng thc Cauchy kt hp vi gi thit ta cú: a + b ab = 2, c + d cd = Suy x v y T ú ta cú ( y ) ( x ) Bi toỏn c chng minh xong Du bng xy a = b = c = d = Vớ d 1.6.12.Cho a, b, c l cỏc s dng tha abc = Chng minh 1 + + 2a + 2b + 2c + Li gii: x y z t a = , b = , c = vi x, y, z > Bt ng thc cn chng minh tr thnh : y z x 1 y z x + + + + x y z x + y y + z z + x +1 +1 +1 y z x Ti õy, mt cỏch t nhiờn, ta ngh n vic ỏp dng Cauchy lm cho bc ỏnh giỏ, cỏc phõn thc tr nờn cú cựng mu(nh th thỡ bi toỏn s tr nờn n gin) Ta lm nh sau: y ( 2z + y ) y ( 2z + y ) y ( 2z + y ) y = = 2 x + y ( x + y ) ( z + y ) ( x + y ) + ( z + y ) ( x + y + z) z ( 2x + z ) x ( y + x) z x , Tng t : 2x + y ( x + y + z ) 2z + x ( x + y + z ) Cng ba bt ng thc ny li v theo v, ta c: y ( 2z + y ) + z ( 2x + z ) + x ( y + x ) y z x + + 2x + y y + z 2z + x ( x + y + z) = ( xy + yz + zx ) + x + y + z ( x + y + z) =1 Chng minh hon tt x + y = z + y ng thc xy y + z = x + z x = y = z a = b = c = z + x = y + x B.4 K thut Cauchy ngc du Vớ d 1.6.13.Cho ba s dng a, b, c tha a + b + c = Chng minh rng: 27 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG a b c + + 2 1+ b 1+ c 1+ a Li gii: ý rng theo bt ng thc Cauchy thỡ: a ab ab ab = a a =a 2 1+ b 1+ b 2b b bc c ca b , c Hon ton tng t , ta cng cú: 2 1+ c 1+ a Cng ba bt ng thc trờn li v theo v, suy a b c ab + bc + ca ab + bc + ca + + a+b+c = 2 1+ b 1+ c 1+ a 2 Vy chng minh s hon tt nu ta ch c rng: ab + bc + ca iu ny hin nhiờn ỳng vỡ ab + bc + ca ( a + b + c) =3 Bi toỏn c gii quyt xong Du ng thc xy a = b = c = Vớ d 1.6.14.Cho a, b, c l cỏc s thc dng Chng minh rng a3 b3 c3 a +b+c + + 2 2 2 a +b b +c c +a Li gii: S dng bt ng thc Cauchy, ta cú: a3 ab ab b = a a = a 2 2 a +b a +b 2ab 3 b c c a Tng t ta cú: 2 b , c b +c c +a Cng ba bt ng thc ny li theo v, ta cú kt qu cn chng minh ng thc xy v ch a = b = c Vớ d 1.6.15.Cho a, b, c > v a + b + c = Chng minh rng a +1 b +1 c +1 + + b2 + c + a + Li gii: S dng bt ng thc Cauchy, ta cú: b ( a + 1) b ( a + 1) a +1 b + ab = a + a + = a +1 2 b +1 b +1 2b b +1 c + bc c + a + ca b +1 , c +1 Tng t c +1 a +1 Cng ba bt ng thc ny li v theo v, ta c a +1 b +1 c +1 a + b + c ab + bc + ca + + +3 b +1 c +1 a +1 2 28 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG ab + bc + ca ( a + b + c ) = =3 2 ng thc xy v ch a = b = c =1 Vớ d 1.6.14.Cho a, b, c > v a + b + c = Chng minh rng a b c + + b + ab c + bc a + ca Li gii: p dng bt ng thc Cauchy ta cú: a b b 1 11 = = + 1ữ 2 b + ab b a + b b ab b a b 4a b 11 c 11 + 1ữ, + 1ữ Tng t: c + bc c b a + ca a c Cng ba bt ng thc ny li v theo v, ta c a b c 31 1 + + + + ữ b + ab c + bc a + ca a b c V nh th bi toỏn c quy v chng minh 31 1 3 1 + + ữ + + 4a b c a b c + a ữ+ + b ữ+ + c ữ + a + b + c = a b c Bt ng thc cui cựng hin nhiờn ỳng vỡ theo bt ng thc Cauchy ta cú: 1 + a 2, + b 2, + c a b c Bi toỏn c chng minh xong ng thc xy v ch a = b = c = B.5.K thut ỏnh giỏ im biờn Vớ d 1.6.15.Cho x, y, z > v x + y + z = Chng minh rng: xy + z + x + y 1 + xy Li gii: Ta s quy bi toỏn v vic chng minh bt ng thc cựng bc l xy + z ( x + y + z ) + x + y x + y + z + xy ( x + z) ( y + z) + x + y x + y + z + xy S dng bt ng thc Cauchy, ta cú: 2 2 x + y = ( x + y ) xy ( x + y ) ( x + y ) = ( x + y ) 2x2 + y2 x + y 29 Buứi Quang Thũnh THC Do ú ta ch cn chng minh: PHNG PHP GII TON BT NG ( z + x) ( z + y ) z + xy z + xy + z ( x + y ) z + xy + z xy z ( x y ) (ỳng) x = y = Bi toỏn c chng minh hon ton ng thc xy z = Vớ d 1.6.16.Cho cỏc s x, y, z v x + y + z = Chng minh rng x + y + z 4( x) ( y ) ( z ) Li gii: Do x + y + z = nờn bt ng thc cn chng minh cú th vit li thnh: ( x + 2y + z) ( x + y + z) 4( x + y) ( y + z ) ( z + x) Do vai trũ ca x v z bt ng thc trờn l nh nờn ta hon ton cú th gi s x z 2 p dng bt ng thc Cauchy dng ( a + b ) 4ab , ta cú ( x + y + z ) x ( y + z ) S dng ỏnh giỏ ny, d thy chng minh s hon tt nu ta ch c x ( x + y + z ) ( x + y ) ( z + x ) y ( x z ) hin nhiờn ỳng theo gi s x z Bi toỏn c chng minh xong ng thc xy x = z = , y = 2.6.2.S DNG BT NG THC BUNYAKOVSKY A.Gii thiu v bt ng thc Bunyakovsky Trong lnh vc Toỏn s cp, cựng vi bt ng thc Cauchy, bt ng thc Bunyakovzky l hai bt ng thc thụng dng v ph bin nht bi tớnh n gin v hiu qu ca chỳng phn trc, chỳng ta ó tỡm hiu v bt ng thc Cauchy v cỏc k thut s dng nú gii toỏn Trong phn ny, chỳng ta s cựng tỡm hiu thờm v bt ng thc th hai, ú l bt ng thc Bunyakovsky Bt ng thc ny c phỏt biu nh sau: Vi hai b s thc bt kỡ ( a1 , a2 , an ) v ( b1 , b2 , , bn ) ta cú ( a12 + a22 + + an2 ) ( b12 + b22 + + bn2 ) ( a1b1 + a2b2 + + anbn ) (*) Trong ú ng thc xy v ch tn ti s thc k cho a1 = kb1 vi mi i = 1, 2,,n Bt ng thc (*) c nh Toỏn hc Augustin Louis Cauchy xut u tiờn vo nm 1821 Sau ú, vo nm 1859, hc trũ ca ụng l Viktor Yakovlevich Bunyakovesky (1804 1889, nh Toỏn hc ngi Nga) ó m rng c kt qu cho tớch phõn V n nm 1885, nh Toỏn hc ngi c Hermann Amandus Schwarz (1843 1921) ó chng minh c kt qu tng quỏt ca bt ng thc ny trng hp khụng gian tớch Do vy, cng nh bt ng thc Cauchy, cm t Bt ng thc Bunyakovesky m ta thng dựng tht l mt cỏch gi tờn khụng chớnh xỏc m phi gi l bt ng thc Cauchy Bunyakovesky - Schwarz Tuy nhiờn, phự hp vi chng trỡnh sỏch giỏo khoa ca nc ta, xuyờn sut ti liu ny, tụi s gi tờn bt ng thc trờn l bt ng thc Bunyakovesky Ngoi ra, t bt ng thc (*) ta cú th suy c mt h qu khỏc rt hay c s dng cho cỏc bi toỏn bt ng thc dng phõn thc, ú l bt ng thc Bunyakovesky dng phõn thc 30 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG Bt ng thc ny c phỏt biu nh sau :Cho ( a1 , a2 , an ) v ( b1 , b2 , , bn ) l hai dóy s thc vi b1 > 0, i ú a a1 a2 an2 ( a1 + a2 + + an ) a12 a22 = = = n ng thc xy v ch + + + b1 b2 bn b1 b2 bn b1 + b2 + + bn trờn, tụi ó gii thiu vi bn c v bt ng thc Bunyakovesky v h qu ca nú nhng trng hp tng quỏt (s bin khụng xỏc nh) Tuy nhiờn, phm vi chng trỡnh Toỏn THCS chỳng ta ch quan tõm nhiu n hai trng hp c bn l n = v n = -Vi n = 2, ta cú 2 2 _Nu a, b, x, y l cỏc s thc, thỡ: ( a + b ) ( x + y ) ( ax + by ) ng thc xy v ch a b = x y _Nu a, b, x, y l cỏc s thc v x, y > 0, thỡ ng thc xy v ch a b2 ( a + b ) + x2 y x+ y a b = x y -Vi n = 3, ta cú 2 2 2 _Nu a, b, c, x, y, z l cỏc s thc, thỡ: ( a + b + c ) ( x + y + z ) ( ax + by + cz ) ng thc xy v ch a b c = = x y z a b2 c ( a + b + c ) _Nu a, b, c, x, y, z l cỏc s thc v x, y, z > 0, thỡ + + x y z x+ y+z a b c ng thc xy v ch = = x y z B.Cỏc k thut s dng bt ng thc Bunyakovesky Chỳng ta m u phn ny bng mt s vớ d n gin m ú ta cú th thy cỏch s dng bt ng thc Bunyakovesky Vớ d 1.6.17.Cho a, b, c Ă v a + b + c = Chng minh rng a + b2 + c2 Li gii: S dng bt ng thc Bunyakovesky, ta cú: ( + + 1) ( a + b2 + c ) ( a + b + c ) = a + b + c a b c = = ng thc xy v ch 1 a = b = c = a + b + c = Vớ d 1.6.18.Cho a, b, c l cỏc s thc dng Chng minh rng 31 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG a2 b2 c2 a +b+c + + b + 3c c + 3a a + 3a Li gii: S dng bt ng thc Bunyakovesky dng phõn thc, ta cú: a + b + c) ( a2 b2 c2 a +b+c + + = b + 3c c + 3a a + 3b ( b + 3c ) + ( c + 3a ) + ( a + 3b ) Bi toỏn c chng minh xong a b c = = a=b=c ng thc xy v ch b + 3c c + 3a a + 3b Vớ d 1.6.20.Cho x, y, z > Chng minh rng x2 y2 z2 + + x + yz y + zx z + xy Li gii: S dng bt ng thc Bunyakovesky dng phõn thc, ta cú: x + y + z) ( VT =1 ( x + yz ) + ( y + zx ) + ( z + xy ) Bt ng thc c chng minh xong ng thc xy v ch x y z = = x= y=z x + yz y + zx z + xy Vớ d 1.6.21.Cho a, b, c l cỏc s thc dng Chng minh rng 36 + + a b c a+b+c Li gii: S dng bt ng thc Bunyakovesky dng phõn thc ta cú: 22 32 ( + + 3) 36 + + = + + = a b c a b c a+b+c a +b+c = = ng thc xy v ch a b c Cú th thy cỏc vớ d trờn, cỏc bỡnh phng u cú sn v ta cú th d dng nhn cỏch s dng Bunyakovesky nh th no cho hp lớ Tuy nhiờn, phn ln cỏc bi toỏn bt ng thc, nhng bỡnh phng nh th u khụng cú sn Cú v nh lỳc ny vic s dng Bunyakovesky s khụng cũn hiu qu na ? Tht ra, khụng hn nh vy, ta hon ton cú th thờm bt nhng lng thớch hp to bỡnh phng Vớ d 1.6.22.Cho x, y, z > v x + y + z = Chng minh rng x3 + y3 + z Li gii: p dng bt ng thc Bunyakovesky, ta c: 32 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG 2 x4 y4 z ( x + y + z ) Bi toỏn c quy v chng minh x +y +z = + + = x y z x+ y+z x+ y+z x + y + z ( x + y + z ) Kt qu ny ỳng theo bt ng thc x+ y+z 3 Bunyakovesky ( x + y + z) ( + + 1) ( x + y + z ) = ng thc xy v ch x = y = z (x Nhõn xột: Bt ng thc ó cho cú th vit c di dng (x + y2 + z2 ) + y3 + z3 ) 3 BI TP T GII 1.6.23.Cho a, b, c l cỏc s thc dng tha a + b + c = Chng minh rng a2 b2 c2 + = b+c c+a a +b 1.6.24.Chng minh rng vi mi a, b, c, d > 0, ta cú ab + cd 1.6.25.Cho a, b, c, d l cỏc s thc dng Chng minh rng ( a + d ) ( b + c) 1 16 64 + + + a b c d a+b+c+d 1.6.26.Cho x, y, z > v x + y + z = Chng minh rng 1 + + + xy + yz + zx 1.6.27.Cho x, y, z, t > v xy + zt + yz + xt = Chng minh rng xy + zt 1 + + = Chng minh rng x y z 1 + + 2x + y + z y + z + x 2z + x + y 1.6.28.Cho x, y, z l cỏc s dng tha 1.6.29.Cho a, b, c l cỏc s thc dng tha abc = ab + bc + ca Chng minh rng: 1 + + < a + 2b + 3c b + 2c + 3a c + 2a + 3b 16 1.6.30.Cho cỏc s dng a, b, c tha a + b + c Chng minh 2009 + 670 2 a + b + c ab + bc + ca 1.6.31.Chng minh rng vi mi a, b, c > 0, ta cú: 33 Buứi Quang Thũnh THC PHNG PHP GII TON BT NG 1 1 + + + + ữ a b c a + 2b b + 2c c + 2a 1.6.32.Cho a, b, c l cỏc s thc dng Chng minh rng 1 1 1 + + + + ữ b+c c+a a +b 3a + 2b + c 3b + 2c + a 3c + 2a + b MC LC Đ1.Gii thiu v bt ng thc Đ2.Cỏc phng phỏp chng minh bt ng thc 2.1.Phng phỏp bin i tng ng 2.2.Phng phỏp dựng nh ngha 2.3.Phng phỏp quy np 2.4.Phng phỏp lm tri, dựng tng sai phõn11 2.5.Phng phỏp phn chng 18 2.6.Phng phỏp dựng bt ng thc kinh in 20 34 ... iu chng minh ng thc xy v ch a = Vớ d 1.1.7.Cho cỏc s thc dng a, b, c Chng minh rng: a b c + + b +c c + a a +b (Bt ng thc Nestbit cho ba s thc dng) Li gii: Hin cú khong 50 cỏch chng minh bt ng... 0,1] Chng minh rng a + b ab + 1+ a + b 2 Li gii: Bt ng thc cn chng minh tng ng vi ( a ) (1 b)(a + b) + ab(2 a b) Vỡ a, b [ 0,1] nờn a 0;1 b 0; a b , ú ta cú iu phi chng minh Vớ d... chớnh l iu phi chng minh BI TP T GII Bi 1.1.1.Cho cỏc s thc x, y Chng minh rng: Buứi Quang Thũnh THC PHNG PHP GII TON BT NG (a) x + y ( x + y) x + y) (b) x + y ( Bi 1.1.2 Chng minh bt ng thc sau