Schur's Inequality

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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.