Schur's inequality via SOS

WLOG. By symmetry, assume $a\ge b\ge c\ge0$.

Factorisation.
$$a(a-b)(a-c)+b(b-c)(b-a)=(a-b)[a(a-c)-b(b-c)]=(a-b)(a-b)(a+b-c)=(a-b)^2(a+b-c).$$
And $c(c-a)(c-b)=c(a-c)(b-c)$ (since both factors are negative in the original, their product is positive).

So the LHS is
$$(a-b)^2(a+b-c)+c(a-c)(b-c).$$

Non-negativity. $a+b-c\ge0$ (since $a\ge c$), $(a-b)^2\ge0$, $c\ge0$, $a-c\ge0$, $b-c\ge0$. All terms are $\ge0$.

Equality. $a=b=c$, or $a=b$ and $c=0$ (and permutations).

Conclusion. Schur's inequality holds.