Từ gt`⇔(x-y)^2+z=xy`

`⇔z=xy-(x-y)^2`

Ta có `(x-y)^2>=0`

`⇒z=xy-(x-y)^2<=xy`

Ta có:

`x/(y+z)+y/(x+z)+(x^3+y^3)/(16z)`

`>=x/(y+xy)+y/(x+xy)+((x+y)(x^2-xy+y^2))/(16xy)`(`z<=xy`)

`>=(x^2)/(xy+x^2y)+(y^2)/(xy+xy^2)+(xy(x+y))/(16xy)`(`x^2-xy+y^2>=xy`)

`>=((x+y)^2)/(2xy+x^2y+xy^2)+(x+y)/16`(Cauchy-Schwarz)

`=((x+y)^2)/(xy(x+y+2))+(x+y)/16`

`>=(4xy)/(xy(x+y+2))+(x+y)/16`(`(x+y)^2>=4xy`)

`=4/(x+y+2)+(x+y)/16`

`=4/(x+y+2)+(x+y+2)/16-1/8`

`>= 2\sqrt{4/(x+y+2)*(x+y+2)/16}-1/8`(Cauchy)

`=1-1/8`

`=7/8(đpcm)`

Dấu ''='' xảy ra khi `x=y=3⇒z=9`