CHƯƠNG 1.1.1 M sai X – sai A, B “A A > B”, “A < B”, “A ng B B”, A > B” A>B ) ¡ 1.1.2 Đ ¡ “ Lời giải Ta sinAsinB sinAsinB – cosAcosB = cos(A+B) = cosC ABC ta có: sinBsinC cosA, sinAsinC ) cosB (1 sin A)(1 sin B)(1 sin C ) = + sin2A + sin2B + sin2C + sin2A sin2B + sin2B sin2C + sin2C sin2A + sin2A sin2B sin2C + sin2A + sin2B + sin2C + cos2A+ cos2B+ cos2C + sin2A sin2B sin2C >4 Vậy bất đẳng thức chứng minh BsinC cosA điều phải chứng minh - K33D SP rằng: Cho a, b, c b a a2 + b2 + c2 ab bc ca b (ab+bc+ca)2 3abc(a b c) Cho x, y, z 0,1 2( x3 Ch ABC rằng: y3 z ) ( x2 y y2 z z x) a sin2A + sin2B + sin2C sinA sinB sinC b cos A + cos B + cos C > c sin A sin B sin C cos A cos B cos C 3.1.3 ) Cho a1 ,a2 , , an dương cho 1 a1 , , n n , a2 n an a1 a2 an n Lời giải , , , n n ): p1 , N - K33D SP p2 , N p3 N 10 2n 2n 2n 2n 1 99 100 n 1 101 10 99 100 99 100 Suy 99 100 2n 12.32 992 22.42 1002 (2.4)(4.6) (98.100) 22.42 1002 12.(32 1)(52 1) (99 1) 22.42 100 200 225 15 (3) (3) suy điều phải chứng minh 103k n- S= n=1 Lời g 3 m 1 m nên (m 1) 3 (m 1) m rên 3 22 3 32 3 m m(m 1) 3 m m2 (**) : 3 3 12 3 22 - K33D SP 38 3 (103 k ) 103 k 3 103 k 103 k 3 (103 k 1) 103 k 3 (103 k ) 33 S 3 103 k 33 3 103 k 3 3 103 k 33 33 33 1 S 103 k 103 k 103 k 3(10k 1) Lưu điều phải chứng minh VẬN DỤNG n a b 22 32 c 2n 2n (n 1) n n2 Cho n n 2n In Cho n > 1, n : 2n.1.2 n 1.3.5 (2n 3) ¥ - K33D SP (n 1) n n n n 39 CHƯƠNG Ở đây, : hàm số 4.1 (GTLN, GTNN) 4.1.1 , GTNN y f ( x) xác D y M M x D : f ( x) M xo D : f ( xo ) M max f ( x) x D y m M f ( x) D, k f ( x) rên D x D : f ( x) M xo D : f ( xo ) M f ( x) x D ta 4.1.2 , GTNN 4.1.2.1 Cauchy - K33D SP 40 a1a2 an P nn P a1 a1 S n a1 a2 an a2 an a2 an an S a1a2 an P 3, y P S n a1 n a2 S n Cho x , y x c A= (3- x)(4 - y)(2x+ 3y) Lời g A (6 x)(12 y )(2 x y ) (6 x), (12 y) , : A (2 x y) (6 x) (12 y ) (2x y ) 6 x 12 y A = x 0; y x y hay x 36 0; y 2 Cho a 3,b 4,c F= sau ab c- + bc a - + ca b- abc Lời giải F c c a a b b a 3, b 4, c : - K33D SP 41 (c 2).2 c (c 2).2 c c F 2 c a 3 D 4 2 2 b b c c a a (c 2) 2 b a 6, b 8, c a b x+ y+ z 1 P = x+ y+ z+ + + x y z Lời g t suy x t = x + y + z ( x y z) x y z y z 9 hay P t t t =” x = y = z f (t ) t : f (t ) t t t t2 t2 0, - K33D SP 0; t 0; 42 y x y z 0; f ( x) 3 nên P = f 2 15 , D biểu thức sau x+ y+ z= P= x y z + + x+ y+ z+ Lời g P 1 x 1 y ( x y z 1) Suy P 1 x y z 1 z 1 x y z 1 1 x y z z 9 x y P= x - K33D SP y z 43 4.1.2.2 Cho l P = 1+ tan tan + 1+ tan tan + 1+ tan tan Lời g tan tan tan tan tan tan tan tan tan( tan tan tan tan tan tan ) tan cot tan tan tan tan i cho hai tan tan , tan tan P2 : 1 tan tan 1 tan (12 12 12 )(1 tan tan tan tan 1 tan tan tan tan tan ) = 3(3 +1) = 12 P tan tan Dấu “=” xảy tan tan tan tan hay P = - K33D SP 44 2x2 + 3y2 A= 2x+ 3y Lời giải : A2 3 y) ( 2x 2, hai A2 3 y) ( 2x A2 5.5 Suy ra: A = 25 (2 3)(2 x y ) x 2x, y ta y A 1, max A = x y x, y (C elip, hypebol , GTNN (C (C Kh x6 y6 z6 + + x3 + y3 y3 + z3 z3 + x3 x, y, z xy xy+ yz yz + zx zx = a Q x3 , b y3 , c z3 a2 b2 c2 a b b c c a - K33D SP ab bc ca a, b, c 45 a2 a b b2 b c c2 ( a b c) (a b) (b c) (c a) c a (a b c) Q (1) a v nên a b ab b c c c bc a ca (2) (2) suy Q minQ = b a b c x y 3 z N , ví dụ a2 ta có a b b2 b c c2 c a a b b c a b a b b2 b c b c c a ba a2 c2 c a c a a b c xn xn xn x1 (1) x1 , x2 , xn x12 x1 x1 x2 x2 x2 x2 x3 xn xn - K33D SP x1 x2 x3 a a 46 4.1.2.3 x+ y+ z= P= x y z + + x+ y+ z+ Lời giải f ( x) x x ( x 1)3 f ( x) f ( x) Hay P (0, f ( y) ) f ( x) 0, x (0, f ( z) f x y z x y z 3 x f ( x) y z x y z y ( x 1)2 ) x x y z z 3 4 P= Cho ABC TNN T = tanAtanA tanBtanB tanCtanC Lời g - K33D SP 47 x ln x ¡ A B C f ( A) f ( x) f x f ( x) f ( B) x ¡ f (C ) tan A ln(tan A) tan B ln(tan B) tan C ln(tan C ) f tan A tan B tan C 3 (1) ) ln tan Atan A tan B tan B tan C tan C (1) 3 tan Atan A tan B tan B tan C tan C f ( 3) 3 ln ( 3)3 hay T 3 3 T= ABC VẬN DỤNG Cho ba a, b, c P biểu thức abc = bc a b a 2c bc b a b 2c 2 bc c a c 2b 2 Cho ABC P Cho x, y cos A x cos B y P y Trong cos 2C 3x ABC x y y x 32 x S (S > 0) - K33D SP 48 F c2 ( p a ) ( p b) Cho x, y x Cho x, y, z xy x(1 y )(1 z ) K b2 ( p a) ( p c) yz x y zx y(1 z )(1 x ) z(1 x )(1 y ) T y a2 ( p b) ( p c ) cot A cot B cot C tan A tan B tan C 4.2 , Bunhiacopski , sinx+ - sin2 x = Lời g sin x 1, x sin x sin x sin x 0, x ¡ sin x the sin x 2 sin x 2 (1 2 ) sin x sin x sin x sin x x - K33D SP 2 sin x sin x k2 , k sin x sin x ¢ 49 sin x : sin x x x V ¢ k2 , k ¢ k2 , k 2 phương trình 3x2 + 6x+ 19 + 5x2 + 10x+ 14 = - 2x - x2 Lời g 3x x 19 3( x2 x 1) 16 3( x 1)2 16 16 5x2 10 x 14 5( x 1)2 4 x x2 3x ( x2 x 3x x 10 x 14 x 19 ( x 1)2 x 1) x 9 5 x 19 x 10 x 14 x 2x x x 1 x2 + 2x + 2x - = 3x2 + 4x+ ¡ /x : TXĐ: x , Theo x x 2 x x ( x 1) ( x 2 x 1) x 2x x2 x - K33D SP 2x2 x1,2 x 3x 4x x 50 x t V 4 x+ y-1 + 3.42y-1 x+ 3y - log43 xo yo 3.42 yo xo xo , nghĩa ( xo ; yo ) 2 xo yo 3.42 yo yo 3.42 yo xo yo 1 x 3y 3.4 xo yo log 34 3.42 yo 2 log34 3.4 log34 log 34 log 34 xo yo yo yo ( xo ; yo ) log 34 log 34 ; 2 VẬN DỤNG nh: a x b sin x - K33D SP x2 y2 4y 2sin x sin 3x 2 51 B : a x x4 Giải phương trình z y4 z4 xyz y3 2x2 3x b z y2 3y x3 2z2 3z x2 y 3x x2 3x x2 x x x2 (n 1) x n n 4x 2 x2 5x x( x 2)(5x 1) 3(n 2) x n an a > - K33D SP 52 [...]... N 2 pn la n a1 a2 an pn N n n Vy bt ng thc c chng minh a1 a2 = = an Ta a ra 2 Cho a1 ,a2 , ,an l cỏc s dng a1 + a2 + + an 1 1 1 + + + a1 a2 an n2 Li g : a1 + a2 + + an n 1 a1 1 a2 n 1 an na 1 n - K33D SP a2 an , 1 a1a2 an (1) (2) 11 (2) suy ra iu phi chng minh a1 a2 an 1 a1 1 a2 hay a1 1 an khi v ch khi a2 an 3 ) Cho a rng 1+ a 1 Chng minh r r > 1+ ra Li gii Do r m n 1, nờn r m ta m,... daỏ u cn 4a 1 2 iu phi chng minh Vớ d 3 a1 ,a2 , an sao cho 0 a 1 n 1 n 2 m1 = ak ; m2 = ak n k=1 n k=1 ak m2 b = 1,n (a+ b)2 2 m1 4ab Li gii 0 a ak b k 1, n n n (a ak )( a ak ) 0 (*) hay k 1 ngha ak2 n ( a b) k 1 m2 (a b)m1 ab f ( x) f (1) : m2 (a b)m1 2 ak n.ab 0 k 1 0 m2 x 2 ab m1 (a b) 2 (a b)m1 x ab 0 f(x) = 0 luụ 4abm2 0 m2 ( a b) 2 2 m1 4ab ta iu phi chng f x minh VN DNG - K33D SP 34 x,... 2 4a 1 0 2 2 4a 1 : xn o ) a a a a 1 4 4 44 2 4 4 4 43 n = k n daỏ u cn ngha xk xk 2 1 1 4a 1 2 a xk a 1 xk 1 4a 1 2 a xk nờn 2a 1 4a 1 2 2 xk 2 1 4a 2 2 4 a 1 4 1 4a 1 2 xk 1 1 4a 1 2 iu phi chng minh Theo nguyờn lý quy 4 Ơ , n x1 , ,xn , y1 , , yn 0; 1 (1- x1 xn )m + (1- y1m ) (1- ynm ) xi + yi = 1 i = 1,n 1 Li gii n Ơ - K33D SP 21 n=1 o (1 x1 )m (1 y1m ) y1m (1 y1m ) 1 o Gi n1 o n : (1 x1 xn... a(bm 1 bm ) bm (a m a mb m a mb b m a a m 1 ) b(a m 1 a m ) (bm 1 bm )(a a m ) (a m 1 a m )(b bm ) 0 a am 1 a m , b bm 1 bm m (1 x1 xn )m (1 y1m ) (1 yn m ) am bm a mbm Ơ (1 am )(1 bm ) = 1 iu phi chng minh Theo nguyờn lý - K33D SP 22 VN DNG 1 Cho xi 1, i 1, n 1 1 1 x1 1 x2 1 1 xn 1 n 1 x1 x2 xn n k = 1,n 2 k 3 Cho n 4 Cho n h1 Ơ,n 2 2 3 3 2 n k ( 1) k C 2 11 k 2n 1 23 4 1 23 0 1 23 4 3 n 1 5... 2 ,t 3ab c 2 ) b(b2 a(a 2 b3 a3 3ab c 2 ) 3ab c 2 ) b(b 2 3a 2 b 3ab2 c 2 ( a b) ( a b) 3 ( a b) c 2 ( a b) 2 3ab c 2 ) ac 2 b c a b a, b, c 0 0 0 (a b)(c a b)(c a b) a bc 2 0 0 c a b ba 0 iu phi chng minh 0 khỏ nhiu khi ch ú mi - K33D SP 24 VN DNG 1 Cho a, b, c, d 0 : a b c d , (a b)(c d ) 2 Cho x, y, z 0 ab cd v (a b)cd xyz 1 x x, y, z (c d ) ab y 1 x z 1 y 1 z 1 3 2 a 2 b 2 b c ; b2 2 2 c 2 c a... OAB, OBC , OAC a : AB 2 a2 b2 2ab cos 60o 2 2 2 o BC AC 2 b a2 : c c2 2bc cos 60 60o B 60o 2ac cos120o a2 AB O b c ab b 2 , C BC b 2 2 bc c v AC AB a2 ab b 2 BC b2 a 2 ac c 2 AC hay bc c 2 iu phi chng minh a2 ac c 2 (1) A, B, C Tht vy: - K33D SP 27 BD // OA suy ra OBD b BD OA CD CO 1 a 1 c A : b a c b c bc ab ac O 1 b ra khi v ch khi B D 1 a 1 c 1 b C 3 -x+ 2y- 8 0 x+ y+ 2 0 y- 2x - 4 0 16 5 x2... cú phng trỡnh x 2 y2 2)2 ( a c) 2 (b d )2 3 15 5 = 5 y P 2 M(a,b) N(c,d) 1 0 x Q R M (a, b), N (c, d ), P(1,2) a2 b2 c2 d2 5 M , N, P (1) MP MP NP MN 3 15 (2) NP MN MNP 3 15 (2 ( - K33D SP iu phi chng minh 30 yt VN DNG a 1 Cho a, b, c c; b c(a c) c c(b c) ab a2 2 Cho a, b a 10 4a 3b 3 40 b 7b b 2 16 8a 6b 24a x2 x, y, z, t z2 1 3 0 z2 t2 2 xz 2 yt y 2 1 v 3 3.2.2 c hai, : 1 f ( x) ax 2 bx c (a 0)... (a b c) 2 2 2 2 a b c 2(a 2 b 2 ( a b) 2 (1) 3 2 2 (2) 6ab 6ac 6bc c 2 ) 2ab 2ac 2bc 0 (a c) 2 (b c) 2 a * c ( a b) a b c 2ab 2ac 2bc a b c 2ab 2ac 2bc (2) b(c a), b 0 (3) (1), (2) suy ra iu phi chng minh c 3 a b c c a a b b c a b c b c c a a b a, b ,c >0 nờn suy ra s a b c b c c a a b a b c b c c a a b ngha l a b c b c a c b a (b c c a a b) 3(a b c) hoc b c - K33D SP 3 D 2 c a khi v ch khi hoc a ... x, y, z a x y2 z2 xy b x y2 z2 2( x Cho x y z yz : zx y z) x2 y g minh rng z 3.1.2 y2 z x z2 x y x2 y2 z2 X , tng ng vi c chng minh ỳng y x y 1 + + x+ z x z y z 1 + x z x+ z (1) Li gii (1) x... ab 0 iu phi chng minh , b ta luụn a+ b 1+ a+ b a+ b 1+ a + b Li gii - K33D SP a b a b : a a b (1 b a a b b) a b a b a a b a (a b )(1 a b b a b) a b a b a a b b b õy iu phi chng minh C khụ C (1+... + sin2B + sin2C + cos2A+ cos2B+ cos2C + sin2A sin2B sin2C >4 Vy bt ng thc c chng minh BsinC cosA iu phi chng minh - K33D SP rng: Cho a, b, c b a a2 + b2 + c2 ab bc ca b (ab+bc+ca)2 3abc(a b