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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 a 1 , a 2 , … , a n {\displaystyle a_{1},a_{2},\ldots ,a_{n}} be non-negative real numbers, and for k = 1 , 2 , … , n {\displaystyle k=1,2,\ldots ,n} , define the averages S k {\displaystyle S_{k}} as follows: S k = ∑ 1 ≤ i 1 < ⋯ < i k ≤ n a i 1 a i 2 ⋯ a i k ( n k ) . {\displaystyle 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 {\displaystyle k} in the n {\displaystyle n} variables a 1 , a 2 , … , a n {\displaystyle a_{1},a_{2},\ldots ,a_{n}} , that is, the sum of all products of k {\displaystyle k} of the numbers a 1 , a 2 , … , a n {\displaystyle 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 ( n k ) . {\displaystyle {\tbinom {n}{k}}.}

Maclaurin's inequality is the following chain of inequalities: S 1 ≥ S 2 ≥ S 3 3 ≥ ⋯ ≥ S n n {\displaystyle S_{1}\geq {\sqrt {S_{2}}}\geq {\sqrt{S_{3}}}\geq \cdots \geq {\sqrt{S_{n}}}} with equality if and only if all the a i {\displaystyle a_{i}} are equal.

For n = 2 {\displaystyle 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 {\displaystyle n=4} : a 1 + a 2 + a 3 + a 4 4 ≥ 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 ≥ 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 3 ≥ a 1 a 2 a 3 a 4 4 . {\displaystyle {\begin{aligned}&{}\quad {\frac {a_{1}+a_{2}+a_{3}+a_{4}}{4}}\\&{}\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}}}\\&{}\geq {\sqrt{\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}}}\\&{}\geq {\sqrt{a_{1}a_{2}a_{3}a_{4}}}.\end{aligned}}} Maclaurin's inequality can be proved using Newton's inequalities or generalised Bernoulli's inequality.

See also

References

  • Biler, Piotr; Witkowski, Alfred (1990). Problems in mathematical analysis. New York, N.Y.: M. Dekker. ISBN 0-8247-8312-3.

This article incorporates material from MacLaurin's Inequality on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

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