—b + c V c + a - J — — - J — b + c V c + a a2(b + c) + b2 (c + a) a _^ b + c v c + a -A2 a2(b + c) + b2(c + a)J>(a + b)^ ụ:> M a t k h a c t a c o : a2(b + c) + b^(c + a) = ab(a + b) + c(a2+ b ^ ) * 3 2 2 u ut u\ 2 u2\ c(a + b) ( a - b ) (a + b - 2 c ) n h u n g ab(a + b) + c a +b^ --^^ '— = -1 L J L < Q ^ V / 4 2 4 Do do F, 2 a J b b + c V c + a >- (a + b f (a + b f _ 4(a + b) \ a + b + 2c ab(a + b) + c(a2+b2) (a + b f c(a + b f
4 ^ 2
Suy ra BDT (1) d i i n g . Dang thuc xay ra k h i va chi k h i a = b . A p d u n g BDT (1) va gia thiet a + b + c = 1 , ta c6: 2 v r T P > 2 . J ^ l i - + 24 24 a + b + 2c 5V5a + 5b N/T+C S V S V T ^ ' v ' ^ u > - ci \^/T^c 24 Xet h a m so f(c) = — p = + —_ , tren - ^ ' y/U^ 5V5Vl-7 Tac6:r(c).-^"-^(^-^)"^^(l:l^ f (c) = 0 « -loVs (l - c) +12(1 + c) VlT^ = 0 o 36c3 - I Z c ^ + 358c - 8 9 = 0 o (4c - l)(9c^ - 2c + 89) = 0 <=> c = - . 1 4 ' Lap bang bien thien ham só f(c) tren
"••-3
78
, suy ra P > —p = r , dang thuc 5N/15
1 3 1 nay xay ra k h i va chi k h i c = —,a = b,a + b + c = 1 o a = b = —,c = —.
4 8 4 78 3 1 78 3 1
V a y m i n P = — d a t d u g c k h i a = b = —,c = —.
5V15 • 8 4
1) Chii y: Bat dang thuc
TAi li'fu Sn thi d0i hQC sang iao v a ^ i a r P I T W n T W F X bar m - Nguyenjnâ^^
Vai cac so thyc duong thoa man a + b > 2c la bat dang thiic chat va rat hay
Ta CO the chiing minh no bang each su d\ing bat d i n g thuc Holder
2) Dc^i vdi bat d i n g thuc: (â + h^)(x^ + y^)(m3 + n^) > (axm + b y n f
Sau khi khai trien va thu g<?n ta c6 bat dang thuc tucmg duong la;
c^b^x^ + a V y 3 + âb^z^ + c^x^y^ + b^x^z^ + ây^z^
>3 a V c ^ x y z + x^y^z^abc) Nhung dieu nay la luon diing vi theo bat dang thuc Cosi ta c6: Nhung dieu nay la luon diing vi theo bat dang thuc Cosi ta c6:
c^b^x^ + âc^y^ + âb^z^ > Sâb^c^xyz;
c\y+bW +a^yV>3x^y^z^abc
V i dv 29. Cho cac so thyc duong a,b,c thoa man a + b + c = 1. Tim gia tri 18 18
nho nhat ciia bieu thuc P =, a + b + 2c
Truoc het ta chung minh: i - i
a
Giii:
b ~ ^ a + b - 1 That vay bat d i n g thuc can chung minh tucmg duong vai That vay bat d i n g thuc can chung minh tucmg duong vai
± _ i _ i > _ _ i _ i -
ab a b ~ ( a + b)^ a + b
1 4 (a + b ) < l
ab (a + bf a b a + b ab(a + b)2 ab(a + b)
dieu nay la luon dung do a + b < a + b + c = 1. Dau bang xay ra khi a = b Tro lai bai toan ta c6: Tro lai bai toan ta c6:
18 2 , 18 2 ^ 18
P= I - - 1 + > : 1 + = + •
a a + b + 2c a + b 18 18
a + b + 2 c 1-c 1+c Xet ham so £(c) = + - 1 tren (0;l) ta c 6 Xet ham so £(c) = + - 1 tren (0;l) ta c 6
1-c 1 + c . 2( - 8 c 2+ 2 0 c- 8 ) 1 . 2( - 8 c 2+ 2 0 c- 8 ) 1 2 . ^ J i , f(c) = O o c = ^ - 1 £'(c) = ( l - c) 2 ( l + c)2 ( l - c) 2( l + c)^
Lap bang bien thien cua ham so f(c) ta suy ra f(c) > f
Dau bang xay ra khi a = b = ^,c = -^ 4 2 4 2
v 2 . = 15
Cty TNHH MTV DWH Khang Vi$t
Doi vai cac bat d i n g thuc 3 bien doi xung nhimg khong cung bac ta c6 the danh gia theo mpt bien dai di$n dya tren quy tic: the danh gia theo mpt bien dai di$n dya tren quy tic:
+ Neu a, b,c > 0 va a + b + c = 3m . Khi can tim mpt chan duoi cho tich be thi ta luon gia su dupe: (b - m)(c - m) > 0 <=> be > m(b + c) - m^ ta luon gia su dupe: (b - m)(c - m) > 0 <=> be > m(b + c) - m^
Khi can tim chan tren va chan duoi cho bien a tac6 2cach: ^ ^- ' • Cach 1: Gia su a = min{a;b;c) => 3a < 3m o a < m => a e [ 0 ; m l * , ], Cach 1: Gia su a = min{a;b;c) => 3a < 3m o a < m => a e [ 0 ; m l * , ], Cach 2: Gia sir a = max{a;b;c} = i > 3 a > 3 m o a > m = > a e [ m ; 3 m ]
Neu CO the nit mpt bien theo hai bien con lai thi ta hoan toan c6 the quy bai
toan can giai quyet thanh bai toan cho 2 bien só. j ^ . ,
Khi chimg minh BDT, hay tim GTLN, GTNN ta can luu y: Phai tan dung toi da cac bát dang thuc trung gian de lam giam bac cua bát d i n g thuc neu c6 da cac bát dang thuc trung gian de lam giam bac cua bát d i n g thuc neu c6 . thẹ Vi^c lam nay se giiip ich rat nhieu trong cac phep toan lien quan dao
ham, tinh gia trj bieu thuc, giai phuong trinh...
V i dy 30: Cho x, y, z la 3 so thvrc thoa man: x + y + z = 0
.2 , ..2 , 2
[ x ^ + y ^ + z ^ = l Tim GTLN cua bieu thuc: A = x^ + y^ + z^ (B. 2012) Tim GTLN cua bieu thuc: A = x^ + y^ + z^ (B. 2012)
Giai:
Tir gia thiet ta suy ra z = -(x + y) => x^ + y^ + (x + y)^ = 1 <=> (x + y)^ - xy = ^
Dat x + y = t=>xy = 2 t 2 - l
V , x y < ( ^ i i y L^ t 2> 4 21^-1
3 3
Ta CO P = x^ + y^ - ( x + y)^ = - ( S x ^ + lOx^y^ + l O x + 5xy^
o P = - 5 x y( x + y) ( x 2 + xy + y^ J . Hay P = -^t^ + | t = f(t)
Xet ham só f(t) = - - t ^ + ^ t tren -:/l<t<:^ "
^ ' 2 4 3 3 Taco f ( t ) = - — 1 2 + - = - ^ ' 2 4 4 Ta b'nh dupe: maxf(t) = f l - 6 f 6 , f ( t ) = 0 < » t = ±- o 5 7 6 . . . S 7 6 k h i X = ; y = z = - — 36 3 ^ 6
Tai Ii$u on thi dai hoc sang tao va gidi PT, bat PT, he rị hn'l l>l \yu i,rn Trung Kien
V i dv 31: Cho x, y, z la 3 so thuc duong thoa man: x + y + z = 3 .
Chung minh rang: 4^x-' + ý' + z^ j + 15xyz > 27
Giai:
Cach 1: Ta y dmh dua ve mpt bién. Gia su x = min(x,y,z) x < 1 ta c6
A{X^ +y^ + z^j + 15xyz = 4(x + y + z) (x + y + z)^ - 3 ( x y + yz + zx) = 108 - 36(xy + yz + zx) + 27xyz = 108 - 36x(y + z) + yz(27x - 36) A = 108 - 36(xy + yz + zx) + 27xyz = 108 - 36x(y + z) + yz(27x - 36) A
N2
+ 27xyz
Vi 27x-36<0,yz:S y + z suy ra
A > 108 - 36x(3 - x) + fc^i(27x - 36) = 108 - 36x(3 - x) + i^-A(27x - 36) Mat khac: 108 - 36x(3- x) + ^^'^^ (27x - 36) > 27 <:> x(x -1)^ > 0. Mat khac: 108 - 36x(3- x) + ^^'^^ (27x - 36) > 27 <:> x(x -1)^ > 0.
Suy ra dieu phai chung minh. Dáu bang xay rakhi x = y = z = l
3 3 (y + z ) ' ( 3 - x ) ' Cach 2: Ta su dung bat dang thuc trung gian: y + z > — = -^^— Cach 2: Ta su dung bat dang thuc trung gian: y + z > — = -^^—
Trong 3 so x,y,z luon ton tai hai so 6 ciing mpt phia so voi 1.
Khong mat tinh tong quat ta gia su hai só do la y,z suy ra
( y - l ) ( z - l ) > 0 < = > y z > y + z - l = 2 - x . Do do 15xyz> 15x(2-x) suy ra 4^x^ + y^ + z^ j + 15xyz > 4x^ + (3 - x)^ + 15x(2 - x). Bay gio ta chi can chung 4^x^ + y^ + z^ j + 15xyz > 4x^ + (3 - x)^ + 15x(2 - x). Bay gio ta chi can chung
minh: 4x3 ^ ^3 _ ^^3 ^ ^5^(2 - x) > 27 o - 6x2 + 3x > 0 » 3^^^ -1)2 > 0.
Nhung bát dang thuc nay luon diing. Dáu bang xay ra khi x = y = z = 1
Nhan xet: Vai vỉc phat hi^n: x = y = z = 1 thi dáu bang cua bát dSng thuc xay rạ Ta c6 dieu hien nhien sau: (x - l)(y - l)(y - l)(z - l)(z - l)(x -1) > 0 xay rạ Ta c6 dieu hien nhien sau: (x - l)(y - l)(y - l)(z - l)(z - l)(x -1) > 0 Vi vay mpt trong ba tich: (x - l)(y - l);(y - l)(z - l);(z - l)(x -1) phai c6 mpt
só khong am.
Gia su (y - l)(z -1) > 0 ta cung suy ra dupe: y z > y + z - l = 2 - x
V i d\f. 32: Cho cac só thuc x, y, z e [-1,2] thoa man x + y + z = 1.
Tim gia trj Ion nhát va GTNN ciia bieu thuc: A = x^ + ý* + z^ Giai: Giai:
GTLN:
Cty TNHH MTV DWH kh,u, , ,
t'.:.j 1
Ta tháy r^ng trong ba só x,y,z phai c6 it nhát hai so cung dáụ Gia sir hai sódo la x, y suy ra xy > 0. sódo la x, y suy ra xy > 0.
Khi do dey rang, x^ +ý* = (x + y)'' - xy(4x2 + 6xy + 4 y 2 ) < (x + y)"* =(l -z)"*.
Xet ham f(z) = ( l - z ) ' * + z \ z G [ - l , 2 ] . , . ,
De dang nhan tháy t'(z) = 0 chi c6 nghiem duy nhát z = ^ .
B Tir do suy ra Maxf(z) = Max|f(-l),f(l),f(2)| = f(-l) = f(2) = 17
P Do dáu dang thuc c6 the xay ra khi (x;y;z) = (-l;2;0)ho|c (2;-l;0) nen ta
ket luan dupe maxA = 17 GTNN: GTNN:
-St