Question

[Trần Nhã Trúc]

Cho x>=0, y>=0, z>=0. Chứng minh: (x+y)(y+z)(z+x) >=8xyz

Answer 1

[OoO Kún Chảnh OoO]

Xét hiệu: (x+y)(y+z)(z+x)-8xyz=0
(=) (x+y)>=2√xy
(y+z)>=2√yz
(z+x)>=2√zx
(=) (x+y)(y+z)(z+x)>=8√x^2 y^2 z^2
(=) (x+y)(y+z)(x+z)>=8|x| |y| |z|
(=) ( x+y)(y+z)(z+x)>= 8xyz