Schur's inequality

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

Factorisation. For $r=1$, the expression is
$$a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b).$$
Note $b(b-c)(b-a)=-b(b-c)(a-b)$, so the first two terms group as $(a-b)[a(a-c)-b(b-c)]=(a-b)(a-b)(a+b-c)=(a-b)^2(a+b-c)$.
The third term is $c(c-a)(c-b)=c(a-c)(b-c)\ge0$.
So the total is $(a-b)^2(a+b-c)+c(a-c)(b-c)\ge0$ (since $a+b-c\ge0$ and the rest is $\ge0$).

General $r>0$. For $r\ge1$, one uses the substitution $x=a^r$ etc. and the same factorization. For $0<r<1$, the inequality also holds; it is proved by a similar argument after ordering the variables.

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