Cho x,y,z > 0, xyz = 1 CMR (x+y)(y+z)(x+z) >= 2(1+x+y+z) Em xin hướng giải thôi ạ em cảm ơn.

`(x+y)(y+z)(x+z) >= 2(1+x+y+z)`

`⇔(x+y+z)(xy+yz+xz)-xyz-2(1+x+y+z)>=0`

`⇔(x+y+z)(xy+yz+xz)-2(x+y+z)-3>=0`

Theo AM-GM: `xy+yz+xz>=3\root{3}{xy*yz*xz}=3\root{3}{(xyz)^2}=3`

`x+y+z>=3\root{3}{x*y*z}=3\root{3}{xyz}=3`

`⇒(x+y+z)(xy+yz+xz)-2(x+y+z)-3>=3(x+y+z)-2(x+y+z)-3=(x+y+z)-3>=0`

`⇒(x+y+z)(xy+yz+xz)-2(x+y+z)-3>=0`

`⇔(x+y)(y+z)(x+z) >= 2(1+x+y+z)`

Dấu ''='' xảy ra khi `x=y=z=1`