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

$$S_k = \frac\sum_{ 1\leq i_1 < \cdots < i_k \leq n}a_{i_1} a_{i_2} \cdots a_{i_k}}{n \choose k}}.$$

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 $${n\choose k}.$$

Maclaurin's inequality is the following chain of inequalities:

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

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:

\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{\frac{a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4}{4}} \\[8pt] & {} \ge \sqrt{a_1a_2a_3a_4}. \end{align}

Maclaurin's inequality can be proved using the Newton's inequalities.