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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 903982, 5 pages
http://dx.doi.org/10.1155/2013/903982
Research Article

Bounds for the Combinations of Neuman-Sándor, Arithmetic, and Second Seiffert Means in terms of Contraharmonic Mean

1Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
2School of Distance Education, Huzhou Broadcast and TV University, Huzhou 313000, China
3School of Information and Engineering, Huzhou Teachers College, Huzhou 313000, China
4School of Mathematics and Computation Science, Hunan City University, Yiyang 413000, China

Received 4 January 2013; Accepted 27 February 2013

Academic Editor: Salvatore A. Marano

Copyright © 2013 Zai-Yin He 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

We give the greatest values , and the least values , in (1/2, 1) such that the double inequalities and hold for any and all with , where , , and are the arithmetic, Neuman-Sándor, contraharmonic, and second Seiffert means of and , respectively.

1. Introduction

For with , the Neuman-Sándor mean [1], second Seiffert mean [2] are defined by respectively. Herein, is the inverse hyperbolic sine function.

Let , , , , , , , and be the harmonic, geometric, logarithmic, first Seiffert, identric, arithmetic, quadratic, and contraharmonic means of two distinct positive real numbers and , respectively. Then it is well known that the inequalities hold for all with .

Among means of two variables, the Neuman-Sándor, contraharmonic, and second Seiffert means have attracted the attention of several researchers. In particular, many remarkable inequalities and applications for these means can be found in the literature [315].

Neuman and Sándor [1, 16] proved that the inequalities hold for all with .

Let    with ,    and . Then the Ky Fan inequalities can be found in [1].

Li et al. [17] proved that the double inequality holds for all with , where , and is the th generalized logarithmic mean of and , and is the unique solution of the equation .

In [18], Neuman proved that the inequalities hold for all with if and only if , , and .

Zhao et al. [19] found the least values and the greatest values such that the double inequalities hold for all with .

In [20, 21], the authors proved that the double inequalities hold for all with if and only of , ,   , , , and .

For , Chu et al. [22, 23] proved that the inequalities hold for all with if and only if , , and .

The aim of this paper is to find the greatest values , and the least values , such that the double inequalities hold for any and all with .

2. Lemmas

In order to prove our main results, we need three lemmas, which we present in this section.

Lemma 1 (see [24, Theorem 1.25]). For , let be continuous on and differentiable on , let on . If is increasing (decreasing) on , then so are If is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2. Let and Then for all if and only if and   for all if and only if .

Proof. From (12), one has where
Let and , then
It is not difficult to verify that the function is strictly increasing on . Then (17) and (18) together with Lemma 1 lead to the conclusion that is strictly decreasing on . Moreover, making use of L'Hôpital's rule, we get
We divide the proof into four cases.
Case 1. . Then from (15) and (19) together with the monotonicity of , we clearly see that is strictly increasing on . Therefore, for all follows from (13) and the monotonicity of .
Case 2. . Then from (15) and (20) together with the monotonicity of , we clearly see that is strictly decreasing on . Therefore, for all follows from (13) and the monotonicity of .
Case 3. . Then (14) leads to
From (15), (19), and (20) together with the monotonicity of , we clearly see that there exists unique such that is strictly decreasing on and strictly increasing on . Therefore, for all follows from (13) and (21) together with the piecewise monotonicity of .
Case 4. . Then (14) leads to
It follows from (15), (19), and (20) together with the monotonicity of , there exists unique such that is strictly decreasing on and strictly increasing on . Equation (13) and inequality (22) together with the piecewise monotonicity of lead to the conclusion that there exists such that for and for .

Lemma 3. Let and Then for all if and only if and for all if and only if .

Proof. From (23) we get where
Let and , then
It is not difficult to verify that the function is strictly increasing on . Then (28) together with Lemma 1 leads to the conclusion that is strictly decreasing on . Moreover, making use of L'Hôpital's rule, we have
We divide the proof into four cases.
Case 1. . Then from (26) and (29) together with the monotonicity of , we clearly see that is strictly increasing on . Therefore, for all follows from (24) and the monotonicity of .
Case 2. . Then from (26) and (30) together with the monotonicity of , we clearly see that is strictly decreasing on . Therefore, for all follows from (24) and the monotonicity of .
Case 3. . Then (25) leads to
From (26), (29), and (30) together with the monotonicity of , we clearly see that there exists such that is strictly decreasing on and strictly increasing on . Therefore, for all follows from (24) and (31) together with the piecewise monotonicity of .
Case 4. . Then (25) leads to
It follows from (26), (29), and (30) together with the monotonicity of , there exists such that is strictly decreasing on and strictly increasing on . Equation (24) and inequality (32) together with the piecewise monotonicity of lead to the conclusion that there exists such that for and for .

3. Main Results

Theorem 4. If and , then the double inequality holds for all with if and only if and .

Proof. Since , , and are symmetric and homogeneous of degree one, without loss of generality, we assume that . Let and , then and
Therefore, Theorem 4 follows easily from Lemma 2 and (34).

Theorem 5. If and , then the double inequality holds for all with if and only if and .

Proof. Since , , and are symmetric and homogeneous of degree one, without loss of generality, we assume that . Let and , then and
Therefore, Theorem 5 follows easily from Lemma 3 and (36).

Remark 6. If , then Theorem 4 reduces to the first double inequality in (8).

Corollary 7. If , then the double inequality holds for all with if and only if and .

Proof. Corollary 7 follows easily from Theorem 5 with .

Acknowledgments

This research was supported by the Natural Science Foundation of China (Grants no.11171307, 61173123), the Natural Science Foundation of Zhejiang Province (Grants no. Z1110551, LY12F02012), and the Natural science Foundation of Huzhou City (Grant no. 2012YZ06).

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