<s>[INST] 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!
 [/INST] 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.
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