# Schur’s inequality

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

 $a^{k}(a-b)(a-c)+b^{k}(b-c)(b-a)+c^{k}(c-a)(c-b)\geq 0$
###### Proof.

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

 $\displaystyle(a-b)\left(a^{k}(a-c)-b^{k}(b-c)\right)+c^{k}(a-c)(b-c)\geq 0$

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

 $a^{3}+b^{3}+c^{3}+3abc\geq ab(a+b)+ac(a+c)+bc(b+c)$
• If $c=0$, we get $(a-b)(a^{k+1}-b^{k+1})\geq 0$.

• If $b=c=0$, we get $a^{k+2}\geq 0$.

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

Title Schur’s inequality SchursInequality 2013-03-22 13:19:30 2013-03-22 13:19:30 rspuzio (6075) rspuzio (6075) 11 rspuzio (6075) Theorem msc 26D15