## Journal of Inequalities and Applications

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# Some Equivalent Forms of the Arithematic-Geometric Mean Inequality in Probability: A Survey

Journal of Inequalities and Applications20082008:386715

DOI: 10.1155/2008/386715

Accepted: 24 June 2008

Published: 6 July 2008

## Abstract

We link some equivalent forms of the arithmetic-geometric mean inequality in probability and mathematical statistics.

## 1. Introduction

The arithmetic-geometric mean inequality (in short, AG inequality) has been widely used in mathematics and in its applications. A large number of its equivalent forms have also been developed in several areas of mathematics. For probability and mathematical statistics, the equivalent forms of the AG inequality have not been linked together in a formal way. The purpose of this paper is to prove that the AG inequality is equivalent to some other renowned inequalities by using probabilistic arguments. Among such inequalities are those of Jensen, Hölder, Cauchy, Minkowski, and Lyapunov, to name just a few.

## 2. The Equivalent Forms

Let be a random variable, we define
(2.1)

where denotes the expected value of .

Throughout this paper, let be a positive integer and we consider only the random variables which have finite expected values.

In order to establish our main results, we need the following lemma which is due to Infantozzi [1, 2], Marshall and Olkin [3, Page 457], and Maligranda [4, 5]. For other related results, we refer to [619].

Lemma 2.1.

The following inequalities are equivalent.

AG inequality: , where is a nonnegative random variable.
if and for with . The arithmetic-geometric mean inequality is usually applied in a simple version of with for each .
if and , and the opposite inequality holds if or .
if and , and the opposite inequality holds if or and .
for if with , and the opposite inequality holds if with .
if and for , and the opposite inequality holds if .
if for and .
Let be a measure space. If is finitely , and let . Then is finitely integrable and .
If and is -integrable, where is a probability space, then .
Artin's theorem. Let be an open convex subset of and satisfy

(a) is Borel-measurable in for each fixed

(b) is convex in for each fixed .

If is a measure on the Borel subsets of such that is integrable for each then is a convex function on .

Jensen's inequality. Let be a probability space and be a random variable taking values in the open convex set with finite expectation . If is convex, then .

Proof.

The proof of the equivalent relations of can be found in [1, 2, 4, 5].

The proof of the equivalent relations of , and can be found in [3].

Theorem 2.2.

The following inequalities are equivalent.

if are random variables and with and .
if are random variables and with and .
if are random variables and with .
if are random variables and with , that is, if with .
if are random variables and with and .
if are random variables and with .
if are random variables and .
if are random variables and .
if is a random variable and , that is, is nondecreasing on .
(see [10, 18]) , where is a random variable if or , and the opposite inequality holds if .
if are random variables and with .
if are random variables and with .
if are random variables and with .
if are random variables and .
if are random variables and .
if are random variables and .
Cauchy-Bunyakovski and Schwarz's (CBS) inequality: if are random variables.
if are random variables.
if are random variables and (the inequality is reversed if or ).
for any if are random variables and .
if are random variables and either or .
if are random variables and .
if are random variables and either or .
if are random variables and either or .
increases with if are random variables and .
Minkowski's inequality: if are random variables, , and the opposite inequality holds if .
Triangle inequality: if are random variables, and the opposite inequality holds if .
if is a random variable, and are two continuous and strictly increasing functions such that is convex.
for any if is a random variable.

The above listed inequalities are also equivalent to the inequalities in Lemma 2.1.

Proof.

The sketch of the proof of this theorem is illustrated by the following maps of equivalent circles:

(1) ;

(2) ;

(3) ;

(4) ;

(5) ;

(6) ;

(7) ;

(8) ;

(9) .

Now, we are in a position to give the proof of this theorem as follows.

, see Casella and Berger [7, page 187].
is clear.
: If and , then and . This and imply
(2.2)

Replacing by in the above inequality, we obtain .

Similarly, we can prove the case that and .

is proved similarly.
is clear.
. Letting , and be replaced by and in , respectively, we obtain .
is similarly proved.
: Let . Then . It follows from that
(2.3)

That is, holds.

is similarly proved.
. Taking in , we see that
(2.4)
which implies
(2.5)
Replacing by ,
(2.6)
Thus
(2.7)

This proves .

Next, let . Then and . This and imply
(2.8)
Replacing and by and in the above inequality, respectively, and using , we obtain
(2.9)

This proves holds.

is proved similarly.
. (a) Taking and in ,
(2.10)
which implies
(2.11)
1. (b)
Taking and in ,
(2.12)

which implies
(2.13)
1. (c)
Taking and in ,
(2.14)

which implies
(2.15)
1. (d)

It follows from (a), (b), and (c) that holds.

Thus, we complete the proof.

is similarly proved.
is clear.
by using the technique of .
. Letting and in , we obtain .
. It follows from and that . This and imply
(2.16)

Replacing and by and in the above inequality, respectively, we obtain .

and are similarly proved.
and follow by taking and .
and follow by taking in and , respectively.
Casella and Berger [7, page 188].
(see [5]): Let with and . It follows from Benoulli's inequality that
(2.17)
This and imply
(2.18)
Hence
(2.19)

This proves holds.

follows by replacing and by and in , respectively.
follows by replacing and in with and , respectively.
. Let for . Then, it follows from that
(2.20)
Thus, is midconvex on , and hence is convex on . Hence
(2.21)
Therefore,
(2.22)
Letting in the both sides of the above inequality,
(2.23)

This shows (see [13]).

. First note that, as shown above, and are equivalent. It follows from that, for ,
(2.24)

These imply for the case .

Similarly, we can prove the case for or by using .

follows by replacing in by if or if , respectively.
follows by replacing in by , respectively.
follows by replacing by or in , respectively.
follows by taking with in .
follows by taking and in .
is clear.
follows by letting and in .
follows by replacing and in by and , respectively.
. Replacing by and by in and changing appropriately the notation for the exponents, we obtain .
is clear.

To complete our proof of equivalence of all inequalities in this theorem and in Lemma 2.1, it suffices to show further the following implications.

follows by taking in .
: Let , where (hence or ). Then it follows from that . Setting , we obtain , see [14, page 162].
follows by taking in .
follows by taking and replacing by in .

Remark 2.3.

Letting and with and in , we obtain the inequality (5) of [18]:
(2.25)
That is,
(2.26)

is a decreasing function of Sclove et al. [18] proved this property by means of the convexity of , see [14]. Clearly, our method is simpler than theirs.

Remark 2.4.

Each (or ) is called Hölder's inequality, each (or ) is called CBS inequality, each is called Lyapunov's inequality, each is called Radon's inequality, each is related to Jensen's inequality.

## Declarations

### Acknowledgments

The authors wish to thank three reviewers for their valuable suggestions that lead to substantial improvement of this paper. This work is dedicated to Professor Haruo Murakami on his 80th birthday.

## Authors’ Affiliations

(1)
Department of Information Management, Lunghwa University of Science and Technology
(2)
Department of Biostatistics, University of Kansas
(3)
Division of Biostatistics, University of Texas-Health Science Center at Houston

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