# Schur's Inequality

## Theorem

Let $x, y, z \in \R_{\ge 0}$ be positive real numbers.

Let $t \in \R, t > 0$ be a (strictly) positive real number.

Then:

- $x^t \left({x-y}\right) \left({x-z}\right) + y^t \left({y-z}\right) \left({y-x}\right) + z^t \left({z-x}\right) \left({z-y}\right) \ge 0$

The equality holds if and only if either:

- $x = y = z$
- Two of them are equal and the other is zero.

When $t$ is a positive even integer, the inequality holds for *all* real numbers $x, y, z$.

## Proof

We note that the inequality, as stated, is symmetrical in $x, y$ and $z$.

Without loss of generality, we can assume that $ x \ge y \ge z$.

Consider the expression:

- $\left({x-y}\right) \left({x^t \left({x-z}\right) - y^t \left({y-z}\right)}\right) + z^t \left({x-z}\right) \left({y-z}\right)$

We see that every term in the above is non-negative. So, directly:

- $(1): \quad \left({x-y}\right) \left({x^t \left({x-z}\right) - y^t \left({y-z}\right)}\right) + z^t \left({x-z}\right) \left({y-z}\right) \ge 0$

If $x = y = z = 0$, it clearly evaluates to $0$.

Inspection on a case-by-case basis provides evidence for the other conditions for equality.

$(1)$ can then be rearranged to **Schur's inequality**.

$\Box$

Now, let $t$ be a positive even integer.

Then from Even Power is Non-Negative:

- $\forall w \in \R: w^t = \left({-w}\right)^t$

## Source of Name

This entry was named for Issai Schur.