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! 

This is a nice method:
We know that $\frac{a+b}{2}\geq\sqrt{ab}$
and similarly $\frac{c+d}{2}\geq\sqrt{cd}$
so $\frac{a+b+c+d}{2}\geq\sqrt{ab}+\sqrt{cd}\geq2\sqrt[4]{abcd}$
$$\Rightarrow\frac{a+b+c+d}{4}\geq\sqrt[4]{abcd}$$
and by making the substitution $d=\frac{a+b+c}{3} \rightarrow$
$$\frac{a+b+c+\frac{a+b+c}{3}}{4}\geq \sqrt[4]{ \frac{a+b+c}{3} abc }$$
$$\frac{a+b+c}{3}\geq\sqrt[4]{ \frac{a+b+c}{3} abc }$$
Raising both sides the the power of $4$
$$(\frac{a+b+c}{3})^4\geq(\frac{a+b+c}{3})abc$$
which gives us:
$$\frac{a+b+c}{3}\geq\sqrt[3]{abc}$$
as required. 
I would love to see some more methods if you have any! 
