Abstract and Applied Analysis

VolumeΒ 2012Β (2012), Article IDΒ 684834, 7 pages

http://dx.doi.org/10.1155/2012/684834

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

^{1}College of Mathematics and Computation Science, Hunan City University, Yiyang 413000, China^{2}Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China

Received 2 July 2012; Accepted 21 August 2012

Academic Editor: JosefΒ DiblΓk

Copyright Β© 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 [2β17]. 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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