Maclaurin's inequality

Let ${\displaystyle a_1,a_2,\dots ,a_n}$ be non-negative real numbers, and for ${\displaystyle k=1,2,\dots ,n}$, define the averages ${\displaystyle S_k}$ as follows: $${\displaystyle S_k=\frac{{\displaystyle \sum _{1\le i_1<\cdots <i_k\le n}{a}_{i_1}{a}_{i_2}\cdots a_{i_k}}}{{\displaystyle (\genfrac{}{}{0ex}{}{n}{k})}}.}$$ The numerator of this fraction is the elementary symmetric polynomial of degree ${\displaystyle k}$ in the ${\displaystyle n}$ variables ${\displaystyle a_1,a_2,\dots ,a_n}$, that is, the sum of all products of ${\displaystyle k}$ of the numbers ${\displaystyle a_1,a_2,\dots ,a_n}$ with the indices in increasing order. The denominator is the number of terms in the numerator, the binomial coefficient ${\displaystyle (\genfrac{}{}{0ex}{}{n}{k}).}$

Maclaurin's inequality is the following chain of inequalities: $${\displaystyle S_1\ge \sqrt{S_2}\ge \sqrt[3]{S_3}\ge \cdots \ge \sqrt[n]{S_n}}$$ with equality if and only if all the ${\displaystyle a_i}$ are equal.

For ${\displaystyle n=2}$, this gives the usual inequality of arithmetic and geometric means of two non-negative numbers. Maclaurin's inequality is well illustrated by the case ${\displaystyle n=4}$: $${\displaystyle \begin{array}{cc}& \frac{a_1+a_2+a_3+a_4}{4}\\ & \ge \sqrt{\frac{a_1{a}_2+a_1{a}_3+a_1{a}_4+a_2{a}_3+a_2{a}_4+a_3{a}_4}{6}}\\ & \ge \sqrt[3]{\frac{a_1{a}_2{a}_3+a_1{a}_2{a}_4+a_1{a}_3{a}_4+a_2{a}_3{a}_4}{4}}\\ & \ge \sqrt[4]{a_1{a}_2{a}_3{a}_4}.\end{array}}$$ Maclaurin's inequality can be proved using Newton's inequalities or generalised Bernoulli's inequality.