Prove $x^3+y^3+z^3\geq 3xyz$ Prove that, $\forall x,y,z\in \mathbb{R}_+\cup \{0\}$,
$$x^3+y^3+z^3\geq 3xyz\text{.}$$

I've been trying for a while now whithout making any significant progress.

hint
$x\mapsto \ln(x) \; $ is concave at $ (0,+\infty) \;$ since its second derivative $\; x \mapsto \frac{-1}{x^2}\; $  is negative.
$$\frac 13 + \frac 13 + \frac 13 = 1$$
thus
$$\ln\Bigl(\frac 13(x^3+y^3+z^3)\Bigr)\ge$$
$$ \frac 13\Bigl(\ln(x^3)+\ln(y^3)+\ln(z^3)\Bigr)$$
$$(=\ln(xyz))$$
