Schur's Inequality ($r=1$)

Step 1: Reduce by symmetry. The expression $a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)$ is symmetric in $a,b,c$, so without loss of generality assume $a \ge b \ge c \ge 0$.

Step 2: Group the first two terms. Group $a(a-b)(a-c) + b(b-a)(b-c) = (a-b)\big[a(a-c) - b(b-c)\big]$, factoring out the common factor $(a-b)$ from both terms (note $b(b-a)(b-c) = -(a-b)\cdot b(b-c)$).

Step 3: Simplify the bracket. Expand $a(a-c) - b(b-c) = a^2 - ac - b^2 + bc = (a^2-b^2) - c(a-b) = (a-b)(a+b) - c(a-b) = (a-b)(a+b-c)$.

Step 4: Combine. Substituting back, the grouped terms equal $(a-b)\cdot(a-b)(a+b-c) = (a-b)^2(a+b-c)$. So the full expression equals $(a-b)^2(a+b-c) + c(a-c)(b-c)$ (the last term is exactly the third original term, unchanged).

Step 5: Sign-check each piece. Since $a \ge b \ge c \ge 0$: $(a-b)^2 \ge 0$ always. For the sign of $a+b-c$: since $a \ge c$, we have $a - c \ge 0$, and $b \ge 0$, so $a+b-c = (a-c)+b \ge 0$. Hence $(a-b)^2(a+b-c) \ge 0$. For the second piece: $c \ge 0$, $a-c \ge 0$ (since $a\ge c$), and $b - c \ge 0$ (since $b \ge c$), so $c(a-c)(b-c) \ge 0$ as a product of three nonnegative numbers.

Step 6: Conclude. The full expression is a sum of two nonnegative terms, $(a-b)^2(a+b-c) + c(a-c)(b-c) \ge 0$, proving Schur's inequality for $r=1$. Equality requires both pieces to vanish: either $a=b$ (making the first term $0$) together with $c(a-c)(b-c)=0$ (forced if additionally $a=c$, giving $a=b=c$, or $c=0$ giving $a=b, c=0$), matching the stated equality condition.