Maclaurin's inequality

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

Let $a_{1}, a_{2}, \dots, a_{n}$ be non-negative real numbers, and for $k = 1, 2, \dots, n$ , define the averages $S_{k}$ as follows: $S_{k} = \frac{\sum_{1 \leq i_{1} < \dots < i_{k} \leq n} a_{i_{1}} a_{i_{2}} \dots a_{i_{k}}}{(\binom{n}{k})} .$ The numerator of this fraction is the elementary symmetric polynomial of degree $k$ in the $n$ variables $a_{1}, a_{2}, \dots, a_{n}$ , that is, the sum of all products of $k$ of the numbers $a_{1}, a_{2}, \dots, a_{n}$ with the indices in increasing order. The denominator is the number of terms in the numerator, the binomial coefficient $(\binom{n}{k}) .$

Maclaurin's inequality is the following chain of inequalities: $S_{1} \geq \sqrt{S_{2}} \geq \sqrt[3]{S_{3}} \geq \dots \geq \sqrt[n]{S_{n}}$ 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 $n = 4$ : $\begin{matrix} \frac{a_{1} + a_{2} + a_{3} + a_{4}}{4} \\ \geq \sqrt{\frac{a_{1} a_{2} + a_{1} a_{3} + a_{1} a_{4} + a_{2} a_{3} + a_{2} a_{4} + a_{3} a_{4}}{6}} \\ \geq \sqrt[3]{\frac{a_{1} a_{2} a_{3} + a_{1} a_{2} a_{4} + a_{1} a_{3} a_{4} + a_{2} a_{3} a_{4}}{4}} \\ \geq \sqrt[4]{a_{1} a_{2} a_{3} a_{4}} . \end{matrix}$ Maclaurin's inequality can be proved using Newton's inequalities or generalised Bernoulli's inequality.

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Maclaurin's inequality

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Maclaurin's inequality

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