Maclaurin's inequality

In mathematics, Maclaurin's inequality, named after Colin Maclaurin, is a refinement of the inequality of arithmetic and geometric means.

Let ${\displaystyle a_{1},a_{2},\ldots ,a_{n}}$ be non-negative real numbers, and for ${\displaystyle k=1,2,\ldots ,n}$, define the averages ${\displaystyle S_{k}}$ as follows:

${\displaystyle S_{k}={\frac {\displaystyle \sum _{1\leq i_{1}<\cdots <i_{k}\leq n}a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}}{\displaystyle {n \choose k}}}.}$

The numerator of this fraction is the elementary symmetric polynomial of degree ${\displaystyle k}$ in the ${\displaystyle n}$ variables ${\displaystyle a_{1},a_{2},\ldots ,a_{n}}$, that is, the sum of all products of ${\displaystyle k}$ of the numbers ${\displaystyle a_{1},a_{2},\ldots ,a_{n}}$ with the indices in increasing order. The denominator is the number of terms in the numerator, the binomial coefficient ${\displaystyle {\tbinom {n}{k}}.}$ Maclaurin's inequality is the following chain of inequalities:

${\textstyle S_{1}\geq {\sqrt {S_{2}}}\geq {\sqrt[{3}]{S_{3}}}\geq \cdots \geq {\sqrt[{n}]{S_{n}}}}$,

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 ${\displaystyle n=4}$:

${\begin{aligned}&\quad {\frac {a_{1}+a_{2}+a_{3}+a_{4}}{4}}\\[8pt]&\geq {\sqrt {\frac {a_{1}a_{2}+a_{1}a_{3}+a_{1}a_{4}+a_{2}a_{3}+a_{2}a_{4}+a_{3}a_{4}}{6}}}\\[8pt]&\geq {\sqrt[{3}]{\frac {a_{1}a_{2}a_{3}+a_{1}a_{2}a_{4}+a_{1}a_{3}a_{4}+a_{2}a_{3}a_{4}}{4}}}\\[8pt]&\geq {\sqrt[{4}]{a_{1}a_{2}a_{3}a_{4}}}.\end{aligned}}$

Maclaurin's inequality can be proved using Newton's inequalities or a generalised version of Bernoulli's inequality.