Maclaurin's inequality

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

Let $a\_1,\; a\_2,\backslash ldots,a\_n$ be non-negative real numbers, and for $k=1,2,\backslash ldots,n$, define the averages $S\_k$ as follows:

The numerator of this fraction is the elementary symmetric polynomial of degree $k$ in the $n$ variables $a\_1,\; a\_2,\backslash ldots,a\_n$, that is, the sum of all products of $k$ of the numbers $a\_1,\; a\_2,\backslash ldots,a\_n$ with the indices in increasing order. The denominator is the number of terms in the numerator, the binomial coefficient $\backslash tbinom\; n\; k.$ Maclaurin's inequality is the following chain of inequalities:

$S\_1\; \backslash geq\; \backslash sqrt\{S\_2\}\; \backslash geq\; \backslash sqrt[3]\{S\_3\}\; \backslash geq\; \backslash cdots\; \backslash geq\; \backslash sqrt[n]\{S\_n\}$,

with equality if and only if all the $a\_i$ are equal.

For $n=2$, this gives the usual inequality of arithmetic and geometric means of two non-negative numbers. Maclaurin's inequality is well illustrated by the case $n=4$:

\backslash begin\{align\}

&\quad \frac{a_1+a_2+a_3+a_4}{4} \\[8pt]

&\ge \sqrt{\frac{a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4}{6}} \\[8pt]

&\ge \sqrt[3]{\frac{a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4}{4}} \\[8pt]

&\ge \sqrt[4]{a_1a_2a_3a_4}.

\end{align}

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