Maclaurin's inequality

Last updated March 05, 2024

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

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

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 $k$ in the $n$ variables $a_{1},a_{2},\ldots ,a_{n}$ , that is, the sum of all products of $k$ of the numbers $a_{1},a_{2},\ldots ,a_{n}$ with the indices in increasing order. The denominator is the number of terms in the numerator, the binomial coefficient ${\tbinom {n}{k}}.$

Maclaurin's inequality is the following chain of inequalities:

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 $a_{i}$ 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{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 generalised Bernoulli's inequality.

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References

Biler, Piotr; Witkowski, Alfred (1990). Problems in mathematical analysis. New York, N.Y.: M. Dekker. ISBN 0-8247-8312-3.

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Maclaurin's inequality

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