Proving inequality using root of unity Let $\omega$ be a complex cube root of unity. It can be shown that if $a,b,c \in \mathbb{R}$, then $$(a+b+c)(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)= a^3+b^3+c^3-3abc$$
I was wondering if this could be extended to prove the AM/GM inequality for $n=3$. All that needs to be done is to prove that the left expression is non-negative, but I cannot find a way.

Proof of AM-GM:
Using $\omega^3=1, \omega^2+\omega+1=0$, We know that $(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)=a^2+b^2+c^2-ab-bc-ca$, then
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\ge 0$
If $a,b,c>0$, then  $a+b+c>0$ and $\frac{a^2+b^2}{2}\ge ab$ etc give $a^2+b^2+c^2\ge ab+bc+ca$
Hence $\frac{a^3+b^3+c^3}{3}\ge abc \implies \frac{x+y+z}{3} \ge (xyx)^{1/3}.$
