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!
On
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!
On
By AM-GM (3 variables)
$\displaystyle \frac{a+b}{2}=\frac{\frac{2a}{3}+\frac{2b}3+\frac{a+b}3}{3}*\frac32\geq\frac32({\frac{2a}3*\frac{2b}3*\frac{a+b}3})^\frac13$
put $\displaystyle \frac12({(2a)(2b)(a+b)})^\frac13-(ab)^\frac12=A-B$
$$A^6-B^6$$ $$=\displaystyle \frac{2^4}{2^6}(ab(a+b))^2-(ab)^3$$
$$=\displaystyle (ab)^2\frac14((a+b)^2-4ab)$$
$$=\displaystyle (ab)^2\frac14((a-b)^2\geq0$$
Therefore 2 variables version of AM-GM was proved.
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!