MacLaurin's inequality

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

Let a₁, a₂, ..., a_n be positive real numbers, and for k = 1, 2, ..., 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 a₁, a₂, ..., a_n, that is, the sum of all products of k of the numbers a₁, a₂, ..., a_n with the indices in increasing order. The denominator is the number of terms in the numerator, the binomial coefficient $\scriptstyle {n\choose k}.$

MacLaurin's inequality states that the following chain of inequalities is true:

$S_1 \geq \sqrt{S_2} \geq \sqrt[3]{S_3} \geq \cdots \geq \sqrt[n]{S_n}$

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 numbers. MacLaurin's inequality is well illustrated by the case n = 4:

$\begin{align} & {} \quad \frac{a_1+a_2+a_3+a_4}{4} \\ \\ & {} \ge \sqrt{\frac{a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4}{6}} \\ \\ & {} \ge \sqrt[3]{\frac{a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4}{4}} \\ \\ & {} \ge \sqrt[4]{a_1a_2a_3a_4}. \end{align}$