Journal of Applied Mathematics

Volume 2013 (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

Fan Zhang,¹ Yu-Ming Chu,² and Wei-Mao Qian³

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 [1–10].

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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