# Maclaurin's inequality

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

Let a1a2, ..., an be positive real numbers, and for k = 1, 2, ..., n define the averages Sk as follows:

${\displaystyle S_{k}={\frac {\displaystyle \sum _{1\leq i_{1}<\cdots

The numerator of this fraction is the elementary symmetric polynomial of degree k in the n variables a1a2, ..., an, that is, the sum of all products of k of the numbers a1a2, ..., an with the indices in increasing order. The denominator is the number of terms in the numerator, the binomial coefficient ${\displaystyle \scriptstyle {n \choose k}.}$

Maclaurin's inequality is the following chain of inequalities:

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

with equality if and only if all the ai are equal.

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

{\displaystyle {\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 the Newton's inequalities.