Proof of AM-GM inequality for $n=3$: $\frac{a+b+c}{3}\geq\sqrt[3]{abc}$ Sorry for bad formatting, I couldn't mark the 3rd root on the right hand side...
I've figured this out into the point where
(and yeah, the problem is to prove that this applies to all non-negative real numbers)

$(a+b+c)/3\geq (abc)^{1/3}$

$$(a+b+c)^3\geq27abc\\
a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2+b^3+3b^2c+3bc^2+c^3\geq27abc$$
I'm not sure how to proceed. Any advice is appreciated - a full answer would of course be better. Thanks in advance!
 A: If $a,b$ or $c$ equals zero the inequality is trivial. Hence we may assume:
$$ (a,b,c)=\left(e^{A},e^{B},e^{C}\right) $$
without loss of generality, and by taking $f(x)=e^{x}$ the inequality can be written as:
$$ \frac{f(A)+f(B)+f(C)}{3}\geq f\left(\frac{A+B+C}{3}\right) $$
that holds as a consequence of the convexity of $f(x)$.
A: The many online proofs for general $n$ and for $n=2$ leave little to do for $n=3$ except maybe to look for an algebraic identity proving the inequality. Let $(a,b,c) = (x^3,y^3,z^3)$, then:
$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - yz - zx) = (1/2) (x+y+z)(\sum (x-y)^2)$ 
The conclusion is slightly more precise than the inequality limited to  non-negative variables:

the Arithmetic mean minus Geometric mean of $a,b,c$ has the same sign as $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c}$, except for $a=b=c$ where the difference is $0$.

The polynomial in the question, $(a+b+c)^3 - 27abc$, is not factorizable.
A: For non-negative $a,b,c$ this is provable by elementary methods. For example, for a fixed $c$ and a fixed average of $a$ and $b$ ($\mu = \frac{a+b}{2}$) the triplet is
$$(\mu + \delta,\mu-\delta,c)$$ and the largest this can be comes when $\delta = 0$, so the values that make the inequality closest to being false come when $a=b$ and applying that argument again also $a=c$... at which point the equality holds.
But the premise you asked to prove is not restrilcted to non-negative $a,b,c$ and in fact is false:  Try $a=-8, b=-4, c=-2$.
