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Research Article | Open Access

Volume 2011 |Article ID 657935 | 7 pages | https://doi.org/10.1155/2011/657935

# A Sharp Double Inequality between Harmonic and Identric Means

Accepted06 Aug 2011
Published12 Oct 2011

#### Abstract

We find the greatest value and the least value in such that the double inequality holds for all with . Here, , and denote the harmonic and identric means of two positive numbers and , respectively.

#### 1. Introduction

The classical harmonic mean and identric mean of two positive numbers and are defined by respectively. Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for and can be found in the literature [1β17].

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

Long and Chu [18] answered the question: what are the greatest value and the least value such that for all with and with .

In [19], the authors proved that the double inequality holds for all with if and only if and .

The following sharp bounds for , and in terms of power means are presented in [20]:

for all with .

Alzer and Qiu [21] proved that the inequalities hold for all positive real numbers and with if and only if and , and so forth.

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 the 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 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 . Then from the monotonicity of the function in we know that to prove inequality (1.8) we only need to prove that inequalities hold for all with .
Without loss of generality, we assume that . Let and , then from (1.1) and (1.2) one has
Let Then simple computations lead to where where
We divide the proof into two cases. Case 1 (). Then (2.19), (2.22), (2.25), and (2.28) lead to
From (2.27) we clearly see that is strictly increasing in , then inequality (2.32) leads to the conclusion that for , hence is strictly increasing in .
It follows from inequality (2.31) and the monotonicity of that is strictly increasing in . Then (2.30) implies that for , so is strictly increasing in .
From (2.29) and the monotonicity of we clearly see that is strictly increasing in .
From (2.5), (2.7), (2.9), (2.11), (2.13), (2.16), and the monotonicity of we conclude that for .
Therefore, inequality (2.1) follows from (2.3) and (2.4) together with inequality (2.33).

Case 2 (). Then (2.19), (2.22), (2.25), and (2.28) lead to From (2.27) and (2.37) we know that is strictly increasing in . Then (2.26) and (2.36) lead to the conclusion that there exists such that for and for , hence is strictly decreasing in and strictly increasing in .
It follows from (2.23) and (2.35) together with the piecewise monotonicity of that there exists such that is strictly decreasing in and strictly increasing in . Then (2.20) and (2.34) lead to the conclusion that there exists such that is strictly decreasing in and strictly increasing in .
From (2.16) and (2.17) together with the piecewise monotonicity of we clearly see that there exists such that for and for . Therefore, is strictly decreasing in and strictly increasing in . Then (2.11)β(2.14) 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 increasingin .
Note that (2.6) becomes for .
From (2.5) and (2.38) together with the piecewise monotonicity of we clearly see that for .
Therefore, inequality (2.2) follows from (2.3) and (2.4) together with inequality (2.39).
Next, we prove that the parameter is the best possible parameter in such that inequality (2.1) holds for all with . In fact, if , then (2.19) leads to . From the continuity of we know that there exists such that for .
It follows from (2.3)β(2.5), (2.7), (2.9), (2.11), (2.13), and (2.16) that for .
Finally, we prove that the parameter is the best possible parameter in such that inequality (2.2) holds for all with . In fact, if , then (2.6) leads to . Hence, there exists such that for .
Therefore, for , follows from (2.3) and (2.4) together with inequality (2.41).

#### Acknowledgments

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

#### References

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