Abstract
We link some equivalent forms of the arithmetic-geometric mean inequality in probability and mathematical statistics.
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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
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 [6–19].
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
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
That is, 
holds.
is similarly proved.
. Taking 
in 
, we see that
which implies
Replacing 
by 
,
Thus
This proves 
.
Next, let 
. Then 
and 
. This and 
imply
Replacing 
and 
by 
and 
in the above inequality, respectively, and using 
, we obtain
This proves 
holds.
is proved similarly.
. (a) Taking 
and 
in 
,
which implies
-
(b)
Taking
and 
in 
, 
(2.12)
and 
in 
, 
which implies
-
(c)
Taking
and 
in 
, 
(2.14)
and 
in 
, 
which implies
-
(d)
It follows from (a), (b), and (c) that
holds.
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
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
This and 
imply
Hence
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
Thus, 
is midconvex on 
, and hence 
is convex on 
. Hence
Therefore,
Letting 
in the both sides of the above inequality,
This shows 
(see [13]).
. First note that, as shown above, 
and 
are equivalent. It follows from 
that, for 
,
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]:
That is,
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.
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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.
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Yeh, CC., Yeh, HW. & Chan, W. Some Equivalent Forms of the Arithematic-Geometric Mean Inequality in Probability: A Survey. J Inequal Appl 2008, 386715 (2008). https://doi.org/10.1155/2008/386715
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DOI: https://doi.org/10.1155/2008/386715