Bất đẳng thức về GTNN và GTLN

NG VIT ANH-BR http://thay-do.net BT NG THC V GI TRI LN NN 1)Cho x, y, z v x + y + z = Chng minh: 2 x3 + y2 + y3 1+ z2 + z3 + x2 2 GII Ta cú: VT + = ( x3 y3 + y )+( 1+ y + z )+( 1+ z z3 1+ x + x2 ) 0.25 VT + =( x3 + + y2 x3 + y2 + + y2 y3 y3 1+ z2 ) +( + + ) 2 + z2 + z2 + x2 +( + + ) + x2 + x2 0.25 z3 z3 x6 y6 z6 3 VT + +3 +3 16 16 16 0.25 3 VT + ( x2 + y2 + z ) = 2 23 2 VT 23 3 2 = 2 2 = = VP (pcm) ( Dõu bng xay va chi x = y = z = 1) 2)Cho x, y, z l cỏc s thc dng ln hn v tho iu kin xy + yz + zx 2xyz Tỡm giỏ tr ln nht ca biu thc A = (x - 1)(y - 1)(z - 1) GII Ta cú xy + yz + xz xyz 1 + + nờn x y z 1 y z ( y 1)( z 1) +1 = + (1) x y z y z yz Tng t ta cú 1 x z ( x 1)( z 1) +1 = + (2) y x z x z xz 1 x y ( x 1)( y 1) +1 = + (3) y x y x y xy Nhõn v vi v ca (1), (2), (3) ta c ( x 1)( y 1)( z 1) NG VIT ANH-BR vy Amax = x = y = z = http://thay-do.net ( ) 2 Vi mi s thc x, y tha iu kin x + y = xy + Tỡm giỏ tr ln nht v giỏ tr nh nht ca biu thc P = G x4 + y xy + 1 ( ) 1 + xy ) xy xy K: t t t = xy Ta cú: xy + = ( x + y ) xy xy xy ( V xy + = ( x y ) (x Suy : P = ) Do ú: P ' = ( + y2 x2 y2 xy + t t ( 2t + 1) = 7t + 2t + ( 2t + 1) ) , P ' = t = 0(th), t = 1(kth) 1 P ữ= P ữ= v P ( ) = 15 1 KL: GTLN l v GTNN l ( HSLT trờn on ; ) 15 4)Vi mi s thc dng x; y; z tha iu kin x + y + z Tỡm giỏ tr nh nht ca biu 1 thc: P = x + y + z + + + ữ x y z G 12 (1) Du bng xóy x = x 2 Tng t: 18 y + 12 (2) v 18 z + 12 (3) y z M: 17 ( x + y + z ) 17 (4) Cng (1),(2),(3),(4), ta cú: P 19 P = 19 x = y = z = KL: GTNN ca P l 19 p dng BT Cụ-si : 18 x + Chng minh a2 b2 c2 + + + a+b b+c c+a ( ) ab + bc + ca a + b + c vi mi s dng a; b; c G Ta cú: a ab ab =a a = a ab (1) a+b a+b 2 ab b2 c2 b bc (2), c ca (3) b+c c+a a2 b2 c2 Cng (1), (2), (3), ta cú: + + + ab + bc + ca a + b + c a+b b+c c+a Tng t: ( ) NG VIT ANH-BR http://thay-do.net 1 1 1 + + = CMR: + + x y z x + y + z x + 2y + z x + y + z 6)Cho x, y, z l cỏc s dng tha 1 1 1 1 1 1 ( + ); ( + ); ( + ) x + y + z x y + z x + y + z 2y x + z x + y + z z y + x 1 1 ( + ); + Li cú : x+y x y 1 1 ( + ); y+z y z 1 1 ( + ); x+z x z +Ta cú : cng cỏc BT ny ta c pcm 7) Cho a, b, c v a + b + c = Tỡm giỏ tr nh nht ca biu thc a3 b3 c3 P= + + + b2 + c2 1+ a2 GII Ta cú: P + = P+ = a3 1+ b a + b2 + 1+ b2 + b3 1+ c a 2 1+ b2 c3 + c2 + + 1+ a 1+ b 2 + + a2 b3 + c2 + b2 + c2 + + c2 1+ a2 a6 b6 c6 3 + + 16 16 16 2 1+ a2 1+ a2 3 9 3 P+ (a + b + c ) = P = = 3 2 2 2 2 2 2 2 + c3 + c2 + PMin a = b = c = Cho cỏc s thc dng a,b,c thay i luụn tho : a+b+c=1.Chng minh rng : a +b2 b +c c + a + + b +c c +a a +b GII 2 a b c b c a Ta cú :VT = ( + + )+( + + ) = A+ B b+c c+a a+b b+c c+a a+b A+3 = 1 1 + + [ (a + b) + (b + c) + (c + a)] a + b b + c c + a 1 1 3 (a + b)(b + c )(c + a )3 = a+b b+c c+a A NG VIT ANH-BR http://thay-do.net 2 a b c 12 = (a + b + c) ( + + )(a + b + b + c + c + a) a+b b+c c+a B.2 B T ú tacú VT + = = VP 2 Du ng thc xy a=b=c=1/3 Cho s dng x, y, z tha : x +3y+5z Chng minh rng: xy 625 z + + 15 yz x + + zx 81 y + 45 xyz GII Bt ng thc 4 + y + + 25 z + 9y x 25 z x2 + 45 36 2 2 + ) 9(.3 x.3 y.5 z ) + VT ( x + y + z ) + ( + x y 5z ( x.3 y.5 z ) t t = ( x.3 y.5 z ) ta cú x + y + 5z ( x.3 y.5 z ) = ú t iu kin < t Xé hm s f(t)= 9t + Du bng xy khi: t=1 hay x=1; y= 36 36 36 = 36t + 27t 36t 27 =45 t t t 1 ; z= 10 Cho x, y, z l s thc thuc (0;1] Chng minh rng 1 + + xy + yz + zx + x + y + z ý rng ( xy + 1) ( x + y ) = ( x ) ( y ) ; yz + y + z v tng t ta cng cú zx + z + x Vỡ vy ta cú: GII NG VIT ANH-BR http://thay-do.net 1 x y z + + + + +1+1+1 ( x + y + z) ữ xy + yz + zx + yz + zx + xy + x y z + + +3 yz + zx+y xy + z z y = x ữ+ yz + zx + y xy + z z y x ữ+ z+ y y+z =5 11.Cho a, b, c l ba cnh tam giỏc Chng minh b c a + + + c Vỡ a, b, c l ba cnh tam giỏc nờn: b + c > a c + a > b a+b c+a = x, = y , a = z ( x, y , z > ) x + y > z , y + z > x, z + x > y 2 V trỏi vit li: a+b a+c 2a VT = + + 3a + c 3a + b 2a + b + c x y z = + + y+z z+x x+ y 2z z > Ta cú: x + y > z z ( x + y + z ) < z ( x + y ) x+ y+z x+ y x 2x y 2y < ; < Tng t: y+z x+ y+z z+x x+ y+z 2( x + y + z) x y z + + < = Do ú: y+z z+x x+ y x+ y+z b c + + + Tỡm giỏ tr nh nht ca biu thc P = x + y + 16 z ( x + y + z) GII Trc ht ta cú: x + y ( x + y) (bin i tng ng) ( x y ) ( x + y ) t x + y + z = a Khi ú P ( (vi t = x + y ) + 64 z ( a z ) + 64 z 3 = = ( t ) + 64t 3 a a 3 z , t 1) a Xột hm s f(t) = (1 t)3 + 64t3 vi t [ 0;1] Cú f '(t ) = 64t ( t ) , f '(t ) = t = [ 0;1] Lp bng bin thiờn 64 Minf ( t ) = GTNN ca P l 16 t c x = y = 4z > 81 t[ 0;1] 81 1 14 Chng minh: ( x + y + z ) + + ữ 12 vi mi s thc x , y , z thuc on [ 1;3] x y z GII Ta cú: t ( t 1) ( t ) t 4t + t + t 3 Suy : x + ; y + ; z + x y z 1 Q = ( x + y + z ) + + + ữ 12 x y z 1 Q 1 ( x + y + z ) + + ữ ( x + y + z ) + + ữ 12 x y z x y z 15.Tỡm giỏ tr nh nht ca hm s y = ( x 1) ln x GII x x x y= x = ; y(1) = vỡ y = ln x + l HSB x Khi < x < y ' < ; x > y ' > TX: D = ( 0; + ) ; y ' = ln x + NG VIT ANH-BR KL: miny = x = http://thay-do.net 16 Cho x, y, z l s thc thuc (0;1] Chng minh rng 1 + + xy + yz + zx + x + y + z GII ý rng ( xy + 1) ( x + y ) = ( x ) ( y ) ; yz + y + z v tng t ta cng cú zx + z + x Vỡ vy ta cú: 1 x y z + + + + +1+1+1 ( x + y + z) ữ xy + yz + zx + yz + zx + xy + x y z + + +3 yz + zx+y xy + z z y vv = x ữ+ yz + zx + y xy + z z y x ữ+ z+ y y+z =5 17 Cho x, y, z l cỏc s thc dng tha món: x2 + y2 + z2 Tỡm giỏ tr nh nht ca biu thc: P= 1 + + + xy + yz + zx Gii 1 + + ữ + xy + yz + zx 2/ Ta cú: [ (1 + xy ) + (1 + yz ) + (1 + zx) ] P 9 + xy + yz + zx + x + y + z Vy GTNN l Pmin = P x = y = z = 18 Cho a, b, c l cỏc s thc tho a + b + c = Tỡm giỏ tr nh nht ca biu thc M = 4a + 9b + 16c + 9a + 16b + 4c + 16a + 4b + 9c GII Theo cụ si cú 22 + 2b + 2c 33 2a + b + c = Tng t r r uur r r uur a b c c a b b c a t u = ;3 ; , v = ;3 ; , w = ;3 ; M = u + v + w ( ) r r uur M u+v+w = ( ( 2a + 2b + 2c ) ( ) + 3a + 3b + 3c ) ( ) ( 2 + 4a + 4b + 4c ) NG VIT ANH-BR http://thay-do.net Vy M 29 Du bng xy a = b = c = 19 Cho x, y, z tho x+y+z > Tỡm giỏ tr nh nht ca biu thc P = x + y + 16 z ( x + y + z) GII Trc ht ta cú: x + y ( t x + y + z = a Khi ú P ( (vi t = x + y) ( x y ) ( x + y ) x + y ) + 64 z ( a z ) + 64 z 3 = = ( t ) + 64t 3 a a 3 z , t 1) a Xột hm s f(t) = (1 t)3 + 64t3 vi t [ 0;1] Cú f '(t ) = 64t ( t ) , f '(t ) = t = [ 0;1] Lp bng bin thiờn 64 Minf ( t ) = GTNN ca P l 16 t c x = y = 4z > 81 t[ 0;1] 81 20.Xột cỏc s thc dng x, y, z tha iu kin x + y + z = Tỡm giỏ tr nh nht ca biu thc: P= x (y + z) y (z + x) z (x + y) + + yz zx xz GII 2 2 2 x x y y z z + + + + + (*) y z z x x y Nhn thy : x2 + y2 xy xy x, y Ă Ta cú : P = Do ú : x + y xy(x + y) x, y > 3 x y2 + x + y x, y > hay y x y2 z2 + y + z y, z > z y Tng t, ta cú : z2 x + z + x x, z > x z Cng tng v ba bt ng thc va nhn c trờn, kt hp vi (*), ta c: P 2(x + y + z) = x, y, z > v x + y + z = 1 Hn na, ta li cú P = x = y = z = Vỡ vy, minP = 21 Cho x, y, z tho x+y+z > Tỡm giỏ tr nh nht ca biu thc P = Trc ht ta cú: x + y 3 ( x + y) t x + y + z = a Khi ú P ( (bin i tng ng) ( x y ) ( x + y + z) ( x + y) x + y ) + 64 z ( a z ) + 64 z 3 = = ( t ) + 64t 3 a a x + y + 16 z 3 NG VIT ANH-BR z (vi t = , t ) a http://thay-do.net Xột hm s f(t) = (1 t)3 + 64t3 vi t [ 0;1] Cú f '(t ) = 64t ( t ) , f '(t ) = t = [ 0;1] Lp bng bin thiờn 64 Minf ( t ) = GTNN ca P l 16 t c x = y = 4z > 81 t[ 0;1] 81 ( ) a1 + b1 + c1 ữ 32 b +a c + c +b a + a +c b ữ 3 22.Cho a,b,c l ba s thc dng Chng minh: a + b + c * Ta cm vi a, b > cú a3 + b3 a2b + ab2 (*) Tht vy: (*) (a + b)(a2 -ab + b2) - ab(a + b) (a + b)(a - b)2 ỳng ng thc xy a = b * T (*) a3 + b3 ab(a + b) b3 + c3 bc(b + c) c3 + a3 ca(c + a) 2(a3 + b3 + c3 ) ab(a + b) + bc(b + c) + ca(c + a) * p dng BT co si cho s dng ta cú: 1 1 + + 33 3 = a a a abc a b c * Nhõn v vi v ca (1) v (2) ta c BT cn cm ng thc xy a = b = c 3 (1) (2) 23 Cho x, y, z l ba s thc dng thay i v tha món: x + y + z xyz Hóy tỡm giỏ tr ln nht ca biu P= thc: P= x y z + + x + yz y + zx z + xy x y z + + x + xy y + zx z + xy Vỡ x; y; z > , p dng BT Cụsi ta cú: P = x x yz + y y zx + z z xy = 2 + + yz zx xy 1 1 1 yz + zx + xy x + y + z + + + + + = y z z x x y xyz xyz xyz = xyz NG VIT ANH-BR http://thay-do.net Du bng xy x = y = z = Vy MaxP = (x 24 Cho x,y R v x, y > Tỡm giỏ tr nh nht ca P = t t = x + y ; t > p dng BT 4xy (x + y)2 ta cú xy P= + y3 ) ( x2 + y ) ( x 1)( y 1) t t t xy (3t 2) t2 Do 3t - > v xy nờn ta cú xy t + t (3t 2) t2 P = t2 t2 t +1 t2 t 4t ; f '(t ) = ; f(t) = t = v t = Xột hm s f (t ) = t2 (t 2) t f(t) + f(t) + t3 t2 + + x+ y=4 x = f (t ) = f(4) = t c Do ú P = (2; + ) xy = y = 25.Cho x > 0, y > 0, x + y = Tỡm giỏ tr nh nht ca biu thc T= x y + x y ữ ú cos a sin a cos3 a + sin a ( sin a + cos a ) ( sin a.cos a ) T= + = = sin a cos a sina.cos a sin a.cos a t2 t t = sin a + cos a = sin a + ữ sin a.cos a = Vi < a < < t 2 t 3t Khi ú T = = f (t) ; t t f '( t ) = 2 f ( t ) f = 2 < t 1; t t x = cos a; y = sin a a 0; ( ) ( ( ) NG VIT ANH-BR f ( t) = f Vy tmin ( 1; http://thay-do.net ( 2) = x = y = Hay T = x = y = 2 t 2t + 1, t f '(t) = t > t 2 f (t) f ( ) = 16 x = y = Vy : A = 16 f (t) = 27.Chng minh rng vi mi s thc dng x, y, z tho x(x + y + z) = 3yz, ta cú: 3 ( x + y) + ( x + z) + 3( x + y) ( x + z) ( y + z) 5( y + z) Gii: T gi thit ta cú: x2 + xy + xz = 3yz (x + y)(x + z) = 4yz t a = x + y v b = x + z Ta cú: (a b)2 = (y z)2 v ab = 4yz Mt khỏc a3 + b3 = (a + b) (a2 ab + b)2 2 2(a + b ) ( a b ) + ab = 2 (a b) + 2ab ( a b ) + ab = (y z) + 2yz ( y z ) + 4yz = (y + z) + 4yz ( y + z ) 2 4(y + z) ( y + z ) = 2(y + z) (1) Ta li cú: 3(x + y)(y +z)(z + x) = 12yz(y + z) 3(y + z)2 (y + z) = 3(y + z)3 (2) Cng tng v (1) v (2) ta cú iu phi chng minh 30 Cho x, y, z số thực dơng thỏa mãn xyz=1 Chứng minh 1 + + x + y +1 y + z +1 z + x +1 Đặt x=a3 y=b3 z=c3 x, y, z >0 abc=1.Ta có a3 + b3=(a+b)(a2+b2-ab) (a+b)ab, a+b>0 a2+b2-ab ab NG VIT ANH-BR http://thay-do.net a3 + b3+1 (a+b)ab+abc=ab(a+b+c)>0 1 a + b + ab ( a + b + c ) Tơng tự ta có 1 , b + c + bc ( a + b + c ) 1 c + a + ca ( a + b + c ) Cộng theo vế ta có 1 1 1 + + = + + 3 x + y + y + z + z + x + a + b +1 b + c + c + a3 +1 1 1 + + ữ= a + b + c ( c + a + b ) = ) ( a + b + c ) ab bc ca ( Dấu xảy x=y=z=1 [...]... v (1) v (2) ta cú iu phi chng minh 30 Cho x, y, z là 3 số thực dơng thỏa mãn xyz=1 Chứng minh rằng 1 1 1 + + 1 x + y +1 y + z +1 z + x +1 Đặt x=a3 y=b3 z=c3 thì x, y, z >0 và abc=1.Ta có a3 + b3=(a+b)(a2+b2-ab) (a+b)ab, do a+b>0 và a2+b2-ab ab NG VIT ANH-BR http://thay-do.net a3 + b3+1 (a+b)ab+abc=ab(a+b+c)>0 1 1 3 a + b + 1 ab ( a + b + c ) 3 Tơng tự ta có 1 1 , 3 b + c + 1 bc ( a + b + c )