Open Access

Jensen Type Inequalities Involving Homogeneous Polynomials

Journal of Inequalities and Applications20102010:850215

https://doi.org/10.1155/2010/850215

Received: 4 November 2009

Accepted: 8 February 2010

Published: 7 April 2010

Abstract

By means of algebraic, analytical and majorization theories, and under the proper hypotheses, we establish several Jensen type inequalities involving th homogeneous polynomials as follows: , , and , and display their applications.

1. Introduction

The following notation and hypotheses in [14] will be used throughout the paper:
(1.1)
Also let
(1.2)
where
(1.3)
is a nonempty and finite subset of
(1.4)

and the permanent of matrix is given by (see [2, 4])

(1.5)

here, the sum extends over all elements of the symmetric group .

If , then is called homogeneous polynomial; if , then is called homogeneous symmetric polynomial (see [3]).

The famous Jensen inequality can be stated as follows: if is a convex function, then for any we have

(1.6)

A large number of generalizations and applications of the inequality (1.6) had been obtained in [1] and [58]. An interesting generalization of (1.6) was given by Chen et al., in [8]: Let and . If with and , then we have the following Jensen type inequality:

(1.7)

In this paper, by means of algebraic, analytical, and majorization theories, and under the proper hypotheses, we will establish several Jensen type inequalities involving homogeneous polynomials and display their applications.

2. Jensen Type Inequalities Involving Homogeneous Polynomials

In this section, we will use the following notation (see [1, 4, 9]):
(2.1)

2.1. A Jensen Type Inequality Involving Homogeneous Polynomials

We begin a Jensen type inequality involving homogeneous polynomials as follows.

Theorem 2.1.

Let . If with and , then
(2.2)

The equality holds in (2.2) if there exists , such that .

Lemma 2.2.

(Hölder's inequality, see [1, 10]). Let with and . If , then
(2.3)

The equality in (2.3) holds if for

Lemma 2.3.

(Power mean inequality, see [1, 10–11]). Let and . If , then
(2.4)

The inequality is reversed for . The equality in (2.4) holds if and only if , or

Lemma 2.4.

Let and . If and with , then
(2.5)

The equality in (2.5) holds if , or there exists , such that .

Proof.

According to , and Lemmas 2.2-2.3, we get that
(2.6)
From
(2.7)

we deduce to the inequality (2.5). Lemma 2.4 is proved.

Proof of Theorem 2.1.

First of all, we assume that . According to , and Lemmas 2.3-2.4, we find that
(2.8)

That is, the inequality (2.2) holds.

Secondly, for some of with satisfing , we have the following cases.

(1)If , then the inequality (2.2) holds from the above proof.

(2)If , then there exists that satisfies . By the result in (1), we obtain that
(2.9)

which implies that inequality (2.2) is also true.

(3)If , then there exist sequences , such that

(2.10)
We get by the case in (2) that
(2.11)

and taking in (2.11), we can get the inequality (2.2). The proof of Theorem 2.1 is thus completed.

2.2. Jensen Type Inequalities Involving Difference Substitution

Exchange the row and row in unit matrix , then this matrix, written , is called exchange matrix. If are exchange matrixes, then the matrix is called difference matrix, and the substitution is difference substitution, where , and

(2.12)

Let . If is true for any difference matrix , then (see [11]), and the homogeneous polynomial is called positive semidefinite with difference substitution.

If we let

(2.13)

then is a finite set and the count of elements of is , and .

We have the following Jensen type inequality involving homogeneous polynomials and difference substitution.

Theorem 2.5.

Let . If , and with , then
(2.14)

The equality holds in (2.14) if there exists , such that , and .

Lemma 2.6.

(Jensen's inequality, see [12]). For any and , we have
(2.15)

The equality in (2.33) holds if and only if , or at least numbers equal zero among the set

Lemma 2.7.

If and , then for the difference substitution , one has the following double inequality:
(2.16)

The equality holds if and only if , or , or

Proof.

From , it is easy to know that . By and Lemma 2.6, we find that
(2.17)

This shows that the double inequality (2.16) holds.

Proof of Theorem 2.5.

Consider the difference substitution . Since , . From , we have that , for all . Hence,
(2.18)
According to Theorem 2.1, we obtain that
(2.19)
In view of and with Lemma 2.7, we have
(2.20)
By noting that , it implies that is increasing with respect to . Thus,
(2.21)
Therefore,
(2.22)

This evidently completes the proof of Theorem 2.5.

As an application of Theorem 2.5, we have the following.

Theorem 2.8.

Let If with then the inequality (2.14) holds. The equality holds in (2.14) if there exists , such that .

Proof.

First of all, we prove that . If the function satisfies the condition that is continuous, then we have the following identity:
(2.23)
where
(2.24)
In fact,
(2.25)
and
(2.26)

That is, the identity (2.23) holds.

Setting
(2.27)
in (2.23), we have that
(2.28)
Since , if and only if Consider the difference substitution . From
(2.29)
for arbitrary , it is easy to see that if . , then
(2.30)

for arbitrary . Therefore, we get that . It follows that the inequality (2.14) holds by using Theorem 2.5. Since , the equality holds in (2.14) if there exists , such that .

The proof of Theorem 2.8 is thus completed.

Remark 2.9.

Theorem 2.8 has significance in the theory of matrices. Let   be an positive definite Hermitian matrix and its eigenvalues, let be the diagonal matrix with the components of as its diagonal elements, and also let Then for some unitary matrix (where is the conjugate transpose of and see [9, 13]). If , then
(2.31)
Write
(2.32)
then Theorem 2.8 can be rewritten as follows, let and . If are positive definite Hermitian matrix, , with , then
(2.33)
In fact, if are positive definite Hermitian matrix and , there exists a unitary matrix such that (see [13])
(2.34)
Thus,
(2.35)
From with , we get that
(2.36)

According to Theorem 2.8, the inequality (2.33) holds.

Remark 2.10.

Theorem 2.8 has also significance in statistics. By using the same proving method of Theorems 2.1–2.8, we can prove the following: under the hypotheses of the Theorem 2.8, if , then and the inequality (2.14) also holds, where
(2.37)
Let be a random variable, let be the probability of random events with . If , then
(2.38)
is the variance of random variable . The is called variance of random variable and for arbitrary , where
(2.39)
Let be also a random variable, with , and let the function be increasing with . Then the inequality (2.14) can be rewritten as follows:
(2.40)

where , .

2.3. Applications of Jensen Type Inequalities

By (1.7) and the same proving method of Theorem 2.1, we can obtain the following result.

Corollary 2.11.

Let . If with and , then
(2.41)

One gives several integral analogues of (2.2) and (2.41) as follows.

Corollary 2.12.

Let be bounded closed region in , and let the functions and be continuous, and . If and , then
(2.42)
If , and is an ordered set, that is,
(2.43)
for arbitrary and , then
(2.44)

As an application of the proof of Theorem 2.8, one has the following.

Corollary 2.13.

Let If with then one has the following Jensen type inequality:
(2.45)

where

Proof.

We can suppose that , and
(2.46)
Since , from Lemma 2.3, we get that
(2.47)
By using (2.28), we find that
(2.48)

The proof of Corollary 2.13 is thus completed.

Corollary 2.14.

If with , then
(2.49)

Proof.

The Vandermonde determinant is wellknown (see [14]):
(2.50)
By Theorem 2.1, for arbitrary , we get that
(2.51)
Letting
(2.52)

in inequality (2.51), it implies that the inequality (2.49) holds. The proof is completed.

Example 2.15.

Given -inscribed-polygon with . Defining the summation of them is an -inscribed-polygon , and its sides lengths are given by with . Also defining , and with .

Wen and Zhang in [15] raised a conjecture: prove that
(2.53)

where is the area of the -inscribed-polygon .

Now, we prove that the inequality (2.53) holds for by using Theorem 2.1.

Denote
(2.54)
If , we have that
(2.55)
Setting
(2.56)

in Theorem 2.1, then inequality (2.2) is just (2.53).

For , we get that
(2.57)
Taking
(2.58)

in Theorem 2.1, it is clear to see that inequality (2.2) deduces to (2.53).

Remark 2.16.

The following result was obtained in [15]. Let with and all be -inscribed-polygons. If , then for , we have
(2.59)

This inequality can also be deduced from inequality (1.7).

3. Jensen Type Inequalities Involving Homogeneous SymmetricPolynomials

In this section, we will also use the following notation (see [4, 16]):

(3.1)

Definition 3.1.

(see [17, 18]). is called the control ordered set if
(3.2)

for arbitrary and

The well-known Chebyshev inequality states: let , and . If and , then
(3.3)

The inequality is reversed for .

We remark here that Wen and Wang generalized the inequality (3.3) in [4]: if , and , then we have the following Chebyshev type inequality:
(3.4)

3.1. Jensen Type Inequalities Involving Homogeneous Symmetric Polynomials

In this subsection, we first present a Jensen type inequality involving homogeneous symmetric polynomials as follows.

Theorem 3.2.

Let let be a control ordered set. If with , then
(3.5)

where

Proof.

By using the same proving method of Theorem 2.1, we can suppose that . If , then inequality (3.5) holds. So we just need to prove the following.

Let be is a control ordered set. If with and , then
(3.6)

We will verify inequality (3.6) by induction.

For , we find from the inequality (3.4) that
(3.7)
Since the control ordered set is nonempty and finite set by using Definition 3.1, we can suppose that
(3.8)
From Hardy's inequality (see [17, page 74]), we have that
(3.9)
By means of , inequality (3.7), and Chebyshev's inequality (3.3), it is easy to obtain that
(3.10)

which implies that the inequality (3.6) holds for .

Assume that the inequality (3.6) is true for , that is,
(3.11)
For , from and , we have Thus,
(3.12)

The inequality (3.6) is proved by induction. The proof of Theorem 3.2 is hence completed.

As an application of the inequality (3.6), we have the following.

Theorem 3.3.

Let , let be a control ordered set, that is,
(3.13)
If then
(3.14)

Proof.

The right-hand inequality of (3.14) is proved in [4]. Now, we will give the demonstration of the left-hand inequality in (3.14).

We can suppose that . By means of , , and , we find from the inequality (3.6) that
(3.15)
From
(3.16)
and Hardy's inequality (see [17, page 74]), we obtain that
(3.17)
Therefore, we deduce that
(3.18)

The proof of Theorem 3.3 is thus completed.

3.2. Remarks

Remark 3.4.

If , then Theorems 3.2 and 3.3 are also true.

Remark 3.5.

If and , then is a control ordered set.

In fact,
(3.19)

Remark 3.6.

By using the proof of Theorem 3.2 and Remark 3.5, we know the following: if with and
(3.20)

then the inequality (3.6) holds.

Remark 3.7.

The inequality (3.6) is also a Chebyshev type inequality involving homogeneous symmetric polynomials.

3.3. An Open Problem

According to Theorem 3.3, we pose the following open problem.

Conjecture 3.8.

Under the hypotheses of Theorem 3.3, one has
(3.21)

Authors’ Affiliations

(1)
College of Mathematics and Information Science, Chengdu University
(2)
Department of Mathematics, Shili Senior High School in Zixing

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Copyright

© J.-J.Wen and Z.-H. Zhang. 2010

This article is published under license to BioMed Central Ltd. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.