Arithmetic Mean/Geometrix Mean Inequality of Degree 3 For $a,b$ and $c\geq0$;
$$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$
Is there a simple way to deduce this result from the degree 2 verson? 
That is:
$$\frac{a+b}{2}\geq\sqrt{ab}$$
Some sort of substitution involving $a,b,c \ $ for one of the variables was what I had in mind, but I can't seem to figure it out! 

I think, the previous method is the best (with using AM-GM for two variables),
but we can make also the following proof by the same AM-GM.
Since for non-negatives $x$, $y$ and $z$ by AM-GM we have
$\frac{x^2+y^2}{2}\geq xy$ and $\frac{x^2+z^2}{2}\geq xz$ and $\frac{y^2+z^2}{2}\geq yz$,
after summing we obtain $x^2+y^2+z^2-xy-xz-yz\geq0$.
Now let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, $\frac{a+b+c}{3}-\sqrt[3]{abc}=\frac{x^3+y^3+z^3-3xyz}{3}=\frac{(x+y+z)(x^2+y^2+z^2-xy-xz-yz)}{3}\geq0$
and we are done!
