---
tag: 958TMJZ51Y
kind: statement
role: theorem
name: Schur's inequality
msc: [26D05]
uses: [08J2M1JYAG]
about: [0ZMY8441GW]
sources:
  - url: https://en.wikipedia.org/wiki/Schur%27s_inequality
same_as:
  wikidata: Q1317813
  wikipedia: Schur's inequality
---

## Statement

Let $x, y, z \ge 0$ and $t > 0$ be real numbers.
Then
$$x^t (x - y)(x - z) + y^t (y - z)(y - x) + z^t (z - x)(z - y) \ge 0.$$
Equality holds if and only if $x = y = z$, or two of $x, y, z$ are equal and the third is $0$.

## Remarks

Proof idea: the expression is symmetric, so assume $x \ge y \ge z$.
Group the first two terms as $(x - y)\big(x^t (x - z) - y^t (y - z)\big)$, which is $\ge 0$ because $x^t \ge y^t \ge 0$ and $x - z \ge y - z \ge 0$.
The last term $z^t (x - z)(y - z)$ is also $\ge 0$.

Example: for $t = 1$, expanding gives
$$x^3 + y^3 + z^3 + 3xyz \ge xy(x + y) + yz(y + z) + zx(z + x).$$
In terms of the [[elementary symmetric polynomials|0ZMY8441GW]] $p = x + y + z$, $q = xy + yz + zx$ and $r = xyz$, this reads $p^3 - 4pq + 9r \ge 0$.

## Explanations

### L1 human

Schur's inequality is a sum of three terms, one for each of three nonnegative numbers.
Some of the terms can be negative, but the total never is.
The case $t = 1$ is a cubic inequality that is often the missing step in symmetric problems.

### L2 human

Schur's inequality is a degree $t + 2$ symmetric inequality that does not follow from [[Muirhead's inequality|PBPFVY20WT]] alone, because it has negative coefficients.
It is a standard companion of Muirhead's inequality in proofs of symmetric inequalities in three variables.
Its equality cases, which include $(1, 1, 0)$, show that it is sharp at the boundary as well as at the centre.