Schur’s inequality
If $a$, $b$, and $c$ are nonnegative real numbers and $k\ge 1$ is real, then the following inequality holds:
$${a}^{k}(ab)(ac)+{b}^{k}(bc)(ba)+{c}^{k}(ca)(cb)\ge 0$$ 
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
$(ab)\left({a}^{k}(ac){b}^{k}(bc)\right)+{c}^{k}(ac)(bc)\ge 0$ 
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 wellknown
$${a}^{3}+{b}^{3}+{c}^{3}+3abc\ge ab(a+b)+ac(a+c)+bc(b+c)$$ 
•
If $c=0$, we get $(ab)({a}^{k+1}{b}^{k+1})\ge 0$.

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

•
If $b=c$, we get ${a}^{k}{(ac)}^{2}\ge 0$.
Title  Schur’s inequality 

Canonical name  SchursInequality 
Date of creation  20130322 13:19:30 
Last modified on  20130322 13:19:30 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  11 
Author  rspuzio (6075) 
Entry type  Theorem 
Classification  msc 26D15 