Ta có:$x^{3}+y^{3} \geq xy(x+y)$(1) $\Leftrightarrow(x+y)(x-y)^{2}\geq 0 $(đúng mọi x,y>0)
(1)$\Leftrightarrow$$x^{3}+y^{3}+1\geq xy(x+y)+xyz$(vì xyz=1)
$\Leftrightarrow$$x^{3}+y^{3}+1\geq xy(x+y+z)$
$\Leftrightarrow $$\frac{1}{x^{3}+y^{3}+1}$$\leq\frac{z}{xyz(x+y+z)}$(2)
Tương tự:$\frac{1}{y^{3}+z^{3}+1}$$\leq$$\frac{x}{xyz(x+y+z)}$(3)
$\frac{1}{z^{3}+x^{3}+1}$$\leq$$\frac{y}{xyz(x+y+z)}$(4)
Cộng (2),(3)và(4)$\Rightarrow$đpcm(Vì xyz=1)