# Difference between revisions of "Maclaurin's Inequality"

Maclaurin's Inequality is an inequality in symmetric polynomials. For notation and background, we refer to Newton's Inequality.

## Statement

For non-negative $x_1, \ldots, x_n$,

$d_1 \ge d_2^{1/2} \ge \ldots \ge d_n^{1/n}$,

with equality exactly when all the $x_i$ are equal.

## Proof

By the lemma from Newton's Inequality, it suffices to show that for any $n$,

$d_{n-1}^{1/(n-1)} \ge d_{n}^{1/n}$.

Since this is a homogenous inequality, we may normalize so that $d_n = \prod x_i = 1$. We then transform the inequality to

$\frac{\sum 1/x_i}{n} \ge 1^{\frac{n-1}{n}} = 1$.

Since the geometric mean of $1/x_1, \ldots, 1/x_n$ is 1, the inequality is true by AM-GM.