Schur’s inequality

If $a$, $b$, and $c$ are non-negative real numbers and $k\ge 1$ is real, then the following inequality holds:

$$a^k(a-b)(a-c)+b^k(b-c)(b-a)+c^k(c-a)(c-b)\ge 0$$

Proof.

We can assume without loss of generality that $c\le b\le a$ via a permutation of the variables (as both sides are symmetric in those variables). Then collecting terms, we wish to show that

$(a-b)\left(a^k(a-c)-b^k(b-c)\right)+c^k(a-c)(b-c)\ge 0$

which is clearly true as every term on the left is positive.∎

There are a couple of special cases worth noting:

• Taking $k=1$, we get the well-known

  $$a^3+b^3+c^3+3abc\ge ab(a+b)+ac(a+c)+bc(b+c)$$

• If $c=0$, we get $(a-b)(a^{k+1}-b^{k+1})\ge 0$.

• If $b=c=0$, we get $a^{k+2}\ge 0$.

• If $b=c$, we get $a^k{(a-c)}^2\ge 0$.