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!
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 $f(x)$.
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.
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 $a=-8, b=-4, c=-2$.
\sqrt[3]{abc}with dollars. – barak manos Jun 27 '16 at 16:27