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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[{3}]{S_{3}}}\geq \cdots \geq {\sqrt[{n}]{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}}\\[8pt]&{}\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}}}\\[8pt]&{}\geq {\sqrt[{3}]{\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}}}\\[8pt]&{}\geq {\sqrt[{4}]{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 [ edit ]
References [ edit ]
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 .