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

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

If $b=c$, we get $a^{{k}}(ac)^{2}\geq 0$.
Keywords:
sum,inequality,Schur
Type of Math Object:
Theorem
Major Section:
Reference
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Corrections
typo by kshum ✓
the lemma? by matte ✓
type by Mathprof ✓
Proven by Andrea Ambrosio ✓
contains own proof by CWoo ✓
the lemma? by matte ✓
type by Mathprof ✓
Proven by Andrea Ambrosio ✓
contains own proof by CWoo ✓