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Journal of Applied Mathematics
Volume 2013, Article ID 582504, 7 pages
http://dx.doi.org/10.1155/2013/582504
Research Article

Bounds for the Arithmetic Mean in Terms of the Neuman-Sándor and Other Bivariate Means

1School of Architecture Engineering, Huzhou Vocational & Technical College, Huzhou 313000, China
2School of Mathematics and Computation Science, Hunan City University, Yiyang 413000, China
3Department of Distance Education, Huzhou Broadcast and TV University, Huzhou 313000, China

Received 10 October 2013; Revised 27 November 2013; Accepted 27 November 2013

Academic Editor: Juan Manuel Peña

Copyright © 2013 Fan Zhang 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 present the largest values , , and and the smallest values , , and such that the double inequalities , , and hold for all with , where , , , and denote the Neuman-Sándor, arithmetic, Heronian, harmonic, and harmonic root-square means of and , respectively.

1. Introduction

For with the Neuman-Sándor mean [1] is defined by where is the inverse hyperbolic sine function.

Recently, the Neuman-Sándor mean has been the object intensive research. In particular, many remarkable inequalities for the Neuman-Sándor mean can be found in the literature [110].

Let , , , , , , , , , and be the harmonic root-square, harmonic, geometric, Heronian, logarithmic, first Seiffert, arithmetic, second Seiffert, quadratic, and contraharmonic means of and , respectively. Then it is known that the inequalities hold for all with .

Neuman and Sándor [1, 2] proved that the inequalities hold for all and with and . All the results stated above are in fact particular cases of more general and stronger results for the Schwab-Borchardt means [1, 2]. Some of them are based on the sequential method of Sándor [11]. In particular, Neuman and Sándor [1] also found that the inequality holds for all and with , where , , , and .

Li et al. [3] proved that the double inequality holds for all with , where , , and is the th generalized logarithmic mean of and ; is the unique solution of the equation .

In [4], Neuman proved that the double inequalities hold for all with if and only if , , , and .

The main purpose of this paper is to find the largest values , , and and the smallest values , , and such that the double inequalities hold for all with . All numerical computations are carried out using Mathematical software.

2. Lemmas

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

Lemma 1 (see [12, Lemma 1.1]). Suppose that the power series and have the radius of convergence and for all . Let ; then the following hold.(1)If the sequence is (strictly) increasing (decreasing), then is also (strictly) increasing (decreasing) on .(2)If the sequence is (strictly) increasing (decreasing) for and (strictly) decreasing (increasing) for , then there exists such that is (strictly) increasing (decreasing) on and (strictly) decreasing (increasing) on .

Lemma 2. The function is strictly decreasing on , where and denote the hyperbolic sine and hyperbolic cosine functions, respectively.

Proof. Making use of power series and , the function can be written as
Let
Then simple computation leads to where
It follows from (11) that for all .
Equations (10) and (12) together with inequality (13) lead to the conclusion that the sequence is strictly decreasing for and strictly increasing for . Then from Lemma 1(2) and (8) together with (9) we clearly see that there exists such that is strictly decreasing on and strictly increasing on .
Let . Then simple computations lead to
It is not difficult to verify that
From the piecewise monotonicity of and inequality (15) we clearly see that , which implies that is strictly decreasing on .

Lemma 3. The inequality holds for all .

Proof. Simple computations lead to for all .

Lemma 4. The function is strictly decreasing in .

Proof. Differentiating gives Making use of the power series we get for .
Let
We divide the proof into two cases.
Case 1 (). Then from Lemma 3, (22), and (23) we have
Case 2 (). Then from Lemma 3, (19), (20), and (23) together with we get
Numerical computations show that
It follows from (29) and (30) that is strictly increasing in . Then (28) leads to the conclusion that there exists such that is strictly decreasing in and strictly increasing in .
From (27) and the piecewise monotonicity of we clearly see that there exists such that is strictly decreasing in and strictly increasing in . Therefore, for follows from (25) and (26) together with the piecewise monotonicity of .

Lemma 5. Let , , and Then and for all .

Proof. We first prove that for . From (32) one has Let Then
We divide the proof into two cases.
Case 1 (). Then we clearly see that
Case 2 (). Then we get
It follows from (32) and (33) together with Cases 1 and 2 that for . Then from (34) and (35) we know that is strictly decreasing in .
Therefore, for follows from (33) and the monotonicity of .
Next, we prove that for . From (32) we clearly see that we only have to prove that for all .
Inequality (39) can be rewritten as Let Then inequality (40) follows from Lemma 4 and (41) together with .

3. Main Results

Theorem 6. 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 Let Then can be rewritten as for all .
It follows from (45) together with Lemma 1(1) that is strictly increasing in . Note that
Therefore, Theorem 6 follows from (43) and (44) together with (46) and the monotonicity of .

Theorem 7. 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
Simple computations lead to
Therefore, Theorem 7 follows from Lemma 2, (48), and (49).

Theorem 8. 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 where is defined as in Lemma 5.
Note that
Therefore, Theorem 8 follows from (51)–(53) together with Lemma 5.

Acknowledgments

This work was supported by the Natural Science Foundation of China under Grant 61374086, the Natural Science Foundation of the Department of Education of Zhejiang Province under Grant Y201223519, and the Natural Science Foundation of Zhejiang Radio and Television University under Grant XKT-13Z04.

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