Arithmetic Mean/Geometrix Mean Inequality of Degree 3

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!

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!