# Schur's Inequality

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Schur's Inequality is an inequality that holds for positive numbers. It is named for Issai Schur.

## Theorem

Schur's inequality states that for all non-negative $a,b,c \in \mathbb{R}$ and $r>0$: $$a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0$$

The four equality cases occur when $a=b=c$ or when two of $a,b,c$ are equal and the third is ${0}$.

### Common Cases

The $r=1$ case yields the well-known inequality: $$a^3+b^3+c^3+3abc \ge a^2 b+a^2 c+b^2 a+b^2 c+c^2 a+c^2 b$$

When $r=2$, an equivalent form is: $$a^4+b^4+c^4+abc(a+b+c) \ge a^3 b+a^3 c+b^3 a+b^3 c+c^3 a+c^3 b$$

### Proof

Without loss of Generality, let ${a\ge b\ge c}$. Note that $a^r(a-b)(a-c)+b^r(b-a)(b-c)$ $= a^r(a-b)(a-c)-b^r(a-b)(b-c) = (a-b)(a^r(a-c)-b^r(b-c))$. Clearly, $a^r\ge b^r \ge 0$, and $a-c \geq b-c \geq 0$. Thus, $(a-b)(a^r(a-c)-b^r(b-c)) \geq 0 \implies a^r(a-b)(a-c)+b^r(b-a)(b-c) \geq 0$. However, $c^r(c-a)(c-b) \geq 0$, and thus the proof is complete.

### Generalized Form

It has been shown by Valentin Vornicu that a more general form of Schur's Inequality exists. Consider $a,b,c,x,y,z \in \mathbb{R}$, where ${a \geq b \geq c}$, and either $x \geq y \geq z$ or $z \geq y \geq x$. Let $k \in \mathbb{Z}^{+}$, and let $f:\mathbb{R} \rightarrow \mathbb{R}_{0}^{+}$ be either convex or monotonic. Then, $$f(x)(a-b)^k(a-c)^k+f(y)(b-a)^k(b-c)^k+f(z)(c-a)^k(c-b)^k \geq 0.$$

The standard form of Schur's is the case of this inequality where $x=a,\ y=b,\ z=c,\ k=1,\ f(m)=m^r$.