Maclaurin's inequality
In mathematics, Maclaurin's inequality, named after Colin Maclaurin, is a refinement of the inequality of arithmetic and geometric means. Let a1, a2, ..., an be positive real numbers, and for k = 1, 2, ..., n define the averages Sk as follows:
- [math]\displaystyle{ S_k = \frac{\displaystyle \sum_{ 1\leq i_1 \lt \cdots \lt i_k \leq n}a_{i_1} a_{i_2} \cdots a_{i_k}}{\displaystyle {n \choose k}}. }[/math]
The numerator of this fraction is the elementary symmetric polynomial of degree k in the n variables a1, a2, ..., an, that is, the sum of all products of k of the numbers a1, a2, ..., an with the indices in increasing order. The denominator is the number of terms in the numerator, the binomial coefficient [math]\displaystyle{ \scriptstyle{n \choose k} }[/math][math]\displaystyle{ . }[/math]
Maclaurin's inequality is the following chain of inequalities:
- [math]\displaystyle{ S_1 \geq \sqrt{S_2} \geq \sqrt[3]{S_3} \geq \cdots \geq \sqrt[n]{S_n} }[/math]
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:
- [math]\displaystyle{ \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} }[/math]
Maclaurin's inequality can be proved using Newton's inequalities or generalised Bernoulli's inequality.
See also
- Newton's inequalities
- Muirhead's inequality
- Generalized mean inequality
- Bernoulli's inequality
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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