Cyclic SOS decomposition

Factorization.
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$

Sum of squares. As shown earlier,
$$a^2+b^2+c^2-ab-bc-ca=\frac12[(a-b)^2+(b-c)^2+(c-a)^2]\ge0.$$

Conclusion. Since $a+b+c\ge0$, the product is $\ge0$. Equality iff $a=b=c$ (or $a=b=c=0$).

Remark. This is the $r=0$ (well, $n=3$) case of Schur's inequality, and also a direct consequence of AM-GM.