`Journal of Applied MathematicsVolumeΒ 2011, Article IDΒ 261237, 6 pageshttp://dx.doi.org/10.1155/2011/261237`
Research Article

## An Optimal Double Inequality between Seiffert and Geometric Means

1Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
2Department of Mathematics, Hangzhou Normal University, Hangzhou 310012, China

Received 30 June 2011; Accepted 14 October 2011

Copyright Β© 2011 Yu-Ming Chu 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 we prove that the double inequality holds for all with if and only if and . Here, and denote the geometric and Seiffert means of two positive numbers and , respectively.

#### 1. Introduction

For with the Seiffert mean was introduced by Seiffert [1] as follows:

Recently, the bivariate mean values have been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert mean can be found in the literature [1β9].

Let , , , , , , and and be the harmonic, geometric, logarithmic, identric, arithmetic, contraharmonic, and th power means of two different positive numbers and , respectively. Then it is well known that for all with .

For all with , Seiffert [1] established that ; Jagers [4] proved that and is the best possible upper power mean bound for the Seiffert mean ; Seiffert [7] established that and ; SΓ‘ndor [6] presented that and ; HΓ€stΓΆ [3] proved that and is the best possible lower power mean bound for the Seiffert mean .

Very recently, Wang and Chu [8] found the greatest value and the least value such that the double inequality holds for with ; For any , Chu et al. [10] presented the best possible bounds for in terms of the power mean; In [2] the authors proved that the double inequality holds for all with if and only if and ; Liu and Meng [5] proved that the inequalities hold for all with if and only if , , and .

For fixed with and , let

Then it is not difficult to verify that is continuous and strictly increasing in . Note that and . Therefore, it is natural to ask what are the greatest value and least value in such that the double inequality holds for all with . The main purpose of this paper is to answer these questions. Our main result is the following Theorem 1.1.

Theorem 1.1. If , then the double inequality holds for all with if and only if and .

#### 2. Proof of Theorem 1.1

Proof of Theorem 1.1. Let and . We first prove that inequalities hold for all with .
Without loss of generality, we assume that . Let and , then from (1.1) one has Let then simple computations lead to where where .
Note that
We divide the proof into two cases.
Case 1 (). Then (2.6), (2.18), (2.21), and (2.24) become
From (2.23) we clearly see that is strictly increasing in , then (2.25) and inequality (2.29) lead to the conclusion that there exists such that for and for . Thus, is strictly decreasing in and strictly increasing in .
It follows from (2.22) and inequality (2.28) together with the piecewise monotonicity of that there exists such that is strictly decreasing in and strictly increasing in . Then (2.19) and inequality (2.27) lead to the conclusion that there exists such that is strictly decreasing in and strictly increasing in .
From (2.15) and (2.16) together with the piecewise monotonicity of we know that there exists such that is strictly decreasing in and strictly increasing in . Then (2.11)β(2.13) lead to the conclusion that there exists such that is strictly decreasing in and strictly increasing in .
It follows from (2.7)β(2.10) and the piecewise monotonicity of that there exists such that is strictly decreasing in and strictly increasing in .
Therefore, inequality (2.1) follows from (2.3)β(2.5) and the piecewise monotonicity of .

Case 2 (). Then (2.18), (2.21) and (2.24) become
From (2.23) we clearly see that is strictly increasing in , then inequality (2.32) leads to the conclusion that is strictly increasing in .
Therefore, inequality (2.2) follows from (2.3)β(2.5), (2.7)β(2.9), (2.11), (2.12), (2.15), and inequalities (2.30) and (2.31) together with the monotonicity of .
Next, we prove that is the best possible parameter such that inequality (2.1) holds for all with . In fact, if , then (2.6) leads to
Inequality (2.33) implies that there exists such that for .
It follows from (2.3) and (2.4) together with inequality (2.34) that for .
Finally, we prove that is the best possible parameter such that inequality (2.2) holds for all with . In fact, if , then from (2.18) we get , which implies that there exists such that for .
Therefore, for follows from (2.3)β(2.5), (2.7)β(2.9), (2.11), (2.12), and (2.15) together with inequality (2.35).

#### Acknowledgments

This research was supported by the Natural Science Foundation of China under Grant 11071069 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

#### References

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