cho ba số không âm x, y, z thỏa mãn điều kiện x+y+z=1 tìm GTLN,GTNN của biểu thức F=xy+yz+zx2xyz
Dễ chứng minh BĐT (x+yz)(y+zx)(x+zy)xyz,a,b,c0
Suy ra 
      (12x)(12y)(12z)xyz
12(x+y+z)+4(xy+yz+zx)8xyzxyz
1+4(xy+yz+zx)8xyzxyz
xy+yz+zx2xyz14(1+xyz)
P14(1+(x+y+z3)3)=727
Vậy max.
ấn xác nhận giúp mình nhé :) –  Trần Nhật Tân 12-10-13 08:37 PM
tks bạn nhìu –  hoan96vip 12-10-13 07:56 PM
Ta có do 0 \leq x,y,z\leq1

\Rightarrow xy+yz+zx\geq 3xyz\geq xyz

\Rightarrow xy+yz+z-xyz\geq 0

Vậy \min F = 0. Dấu = xảy ra khi (x;\ y;\ z) = (0;\ 0;\ 1) và các hoán vị
b nghĩ ra thì..đăng lên mình xem với nhé :)) tks –  hoan96vip 12-10-13 11:12 AM
mình mới nghĩ min còn cái kia để rảnh mình nghĩ sau nha –  Dép Lê Con Nhà Quê 12-10-13 07:02 AM
thế còn GTLN thì như nào b?? –  Tiến Thực 11-10-13 11:46 PM

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