Elementary proof for $\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$ I am searching for an elementary proof of the AM-GM inequality in three variables:
$\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$
The inequality of the geometric mean vs the arithmetic mean of two variables can be proven elementarily via 
$x - 2 \sqrt{xy} + y \geq 0$
whenever $x,y > 0$. I am searching for a proof that uses a similar technique based on elementary arithmetic.

An equivalent form is
$$x^3+y^3+z^3\ge 3xyz\tag{*}$$
for $x$, $y$, $z\ge0$. Without loss assume that $0\le x\le y\le z$.
Then $y=x+a$ and $z=x+a+b$ with $a$, $b$, $x\ge0$.
Also
$$x^3+y^3+z^3-3xyz=x^3+(x+a)^3+(x+a+b)^3-3x(x+a)(x+a+b)
=3xa^2+3xab+3xb^2+2a^3+3a^2b+3ab^2+b^3\ge0.$$
Then (*) follows.
