Abstract and Applied Analysis

Volume 2012 (2012), Article ID 684834, 7 pages

doi:10.1155/2012/684834

Research Article

A Sharp Double Inequality between Seiffert, Arithmetic, and Geometric Means

Wei-Ming Gong,¹ Ying-Qing Song,¹ Miao-Kun Wang,² and Yu-Ming Chu²

¹College of Mathematics and Computation Science, Hunan City University, Yiyang 413000, China

²Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China

Received 2 July 2012; Accepted 21 August 2012

Academic Editor: Josef Diblík

Copyright © 2012 Wei-Ming Gong et al. 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.

Abstract.

For fixed and any we prove that the double inequality holds for all with if and only if and . Here, , and denote the Seiffert, arithmetic, and geometric means of two positive numbers and , respectively.

1. Introduction

The Seiffert mean [1] of two distinct positive numbers and is defined by

Recently, the Seiffert mean has been the subject of intensive research. In particular, many remarkable inequalities for can be found in the literature [2–17]. The Seiffert mean can be rewritten as (see [6, (2.4)])

Let , and be the classical arithmetic, geometric, and harmonic means of two positive numbers and , respectively. Then it is well known that inequalities hold for all with .

For , Chu et al. [18, 19] proved that the double inequalities

hold for all with if and only if , , and .

Let , and

then it is not difficult to verify that

and is strictly increasing with respect to for fixed with .

It is natural to ask what are the largest value and the least value in such that the double inequality holds for all with and . The main purpose of this paper is to answer this question.

2. Main Result

In order to establish our main result we need two lemmas, which we present in the following.

Lemma 2.1. If , then .

Proof. Consider the following:

Then simple computations lead to

for .

Computational and numerical experiments show that

Inequalities (2.3) and (2.4) imply that is strictly increasing in . Therefore, Lemma 2.1 follows from (2.1) and (2.2) together with the monotonicity of .

Lemma 2.2. Let , and

Then inequality holds for all if and only if , and inequality holds for all if and only if .

Proof. If , then we clearly see that , and for all and . In the following discussion, we assume that .

From (2.5) and simple computations we have

where

where

We divide the proof into four cases.

Case 1 (). Then from (2.11) and (2.12) together with the fact that

we clearly see that

and is strictly increasing in .

Equation (2.12) and the monotonicity of imply that

for .

Equation (2.10) and inequality (2.16) lead to the conclusion that is strictly increasing in . Then from (2.9) we know that

for .

It follows from (2.7) and inequality (2.17) that is strictly increasing in .

Therefore, for all follows from (2.6) and the monotonicity of .

Case 2 (). Then (2.12) and the continuity of imply that there exists such that

for .

Therefore, for follows easily from (2.6), (2.7), (2.9) and (2.10) together with inequality (2.18).

Case 3 (). Then Lemma 2.1 and (2.12) lead to

We divide the proof into two subcases.

Subcase 3.1 (). Then (2.13) becomes

Let , then from (2.11) we clearly see that the function is a quadratic function of variable . It follows from inequality (2.19) and (2.20) that

for all .

Therefore, for follows easily from (2.6), (2.7), (2.9) and (2.10) together with inequality (2.21).

Subcase 3.2 (). Then from (2.5), (2.8), and (2.13) we have

From (2.11), (2.19), and (2.24) we clearly see that there exists such that for and for . Then (2.10) implies that is strictly decreasing in and strictly increasing in .

From (2.9) and (2.23) together with the piecewise monotonicity of we clearly see that there exists such that , for and for . Then (2.7) implies that is strictly decreasing in and strictly increasing in .

Therefore, for follows from (2.6) and (2.22) together with the piecewise monotonicity of .

Case 4 (). Then (2.5) leads to

From inequality (2.25) and the continuity of we know that there exists such that for .

Theorem 2.3. If and , then the double inequality

holds for all with if and only if and .

Proof. Since both and are symmetric and homogeneous of degree 1. Without loss of generality, we assume that . Let . Then from (1.2) and (1.4) we have

Therefore, Theorem 2.3 follows easily from Lemma 2.2 and (2.27).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grant 11071069 and the Natural Science Foundation of Hunan Province under Grant 09JJ6003.

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