Maclaurin's inequality

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

Let $$a_1, a_2,\ldots,a_n$$ 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 $$n=4$$: $$\begin{align} & {} \quad \frac{a_1+a_2+a_3+a_4}{4} \\[8pt] & {} \ge \sqrt{\frac{a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4}{6}} \\[8pt] & {} \ge \sqrt[3]{\frac{a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4}{4}} \\[8pt] & {} \ge \sqrt[4]{a_1a_2a_3a_4}. \end{align}$$ Maclaurin's inequality can be proved using Newton's inequalities or generalised Bernoulli's inequality.