Abstract and Applied Analysis
Volume 2013 (2013), Article ID 348326, 12 pages
http://dx.doi.org/10.1155/2013/348326
Research Article

## Best Possible Bounds for Neuman-Sándor Mean by the Identric, Quadratic and Contraharmonic Means

1Department of Mathematics, Hangzhou Normal University, Hangzhou 310036, China
2School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China
3School of Information & Engineering, Huzhou Teachers College, Huzhou 313000, China
4School of Automation, Southeast University, Nanjing 210096, China

Received 19 January 2013; Accepted 1 February 2013

Copyright © 2013 Tie-Hong Zhao 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 prove that the double inequalities hold for all with if and only if , , , and , where , , , and are the identric, Neuman-Sándor, quadratic, and contraharmonic means of and , respectively.

#### 1. Introduction

For and with , the identric mean , Neuman-Sándor mean [1], quadratic mean , contraharmonic mean , and th power mean are defined by respectively, where is the inverse hyperbolic sine function.

Recently, the identric, Neuman-Sándor, quadratic, and contraharmonic means have attracted the interest of numerous eminent mathematicians. In particular, many remarkable inequalities for these means can be found in the literature [118].

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

Neuman and Sándor [1, 8] established that for all with .

Let with , , and . Then the Ky Fan inequalities were presented in [1].

Li et al. [19] found the best possible bounds for the Neuman-Sándor mean in terms of the generalized logarithmic mean . Neuman [20] and Zhao et al. [21] proved that the inequalities hold for all with if and only if , , , , , , , and .

In [22], Chu and Long gave the best possible constants , and such that the double inequalities and hold for all with .

The ratio of identric means leads to the weighted geometric mean which has been investigated in [2325]. Alzer [26] proved that the inequalities hold for all with .

The following sharp bounds for , , and in terms of the power mean and the convex combination of arithmetic and geometric means are given in [27] as for all with .

Chu et al. [28] presented the optimal constants , and such that the double inequalities hold for all with .

The aim of this paper is to find the best possible constants and such that the double inequalities hold for all with . All numerical computations are carried out using MATHEMATICA software.

#### 2. Lemmas

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

Lemma 1. The double inequality holds for .

Proof. To prove Lemma 1, it suffices to prove that for .
From the expressions of and , we get where for .
Therefore, inequality (12) follows from (14)–(16), and inequality (13) follows from (14)–(17).

Lemma 2. Let Then for , and for .

Proof. To prove inequalities (19) and (20), it suffices to show that for , and for .
From (21) and (22), one has for , and for .
Therefore, inequality (21) follows from (23) and (24), and inequality (22) follows from (23) and (25).

Lemma 3. Let
Then the double inequality holds for .

Proof. To prove inequality (27), it suffices to show that for .
First, we prove inequality (28). From the expression of , we have where
Note that for , and for .
It follows from (32) and (34)–(36) together with Lemma 1 that for , and for .
From (33), (37), and (38), we clearly see that there exists such that for and for . Then (31) leads to the conclusion that is strictly increasing on and strictly decreasing on .
Therefore, inequality (28) follows from (30) and the piecewise monotonicity of .
Next, we prove inequality (29). From the expression of , we get where
It follows from Lemma 1 and (40) that for .
Therefore, inequality (29) follows from (39) together with (41).

Lemma 4. Let Then the double inequality holds for .

Proof. To prove Lemma 4, it suffices to prove that for .
We first prove inequality (44). From the expression of , we obtain where
Note that for , and for .
It follows from Lemma 1, (48), and (51)–(53) that for , and for .
From (50) and (55), we know that is strictly decreasing on , and this in conjunction with (49) and (54) leads to the conclusion that there exists such that for and for . Then (47) implies that is strictly increasing on and strictly decreasing on . Therefore, inequality (44) follows from (46) and the piecewise monotonicity of .
Next, we prove inequality (45). From the expression of one has where
It follows from Lemma 1 and (52) that for .
Therefore, inequality (45) follows from (56) together with (58).

Lemma 5. Let be defined as in Lemma 2 and
Then the double inequality holds for .

Proof. From Lemma 2, one has for .
Therefore, Lemma 5 follows easily from (61).

Lemma 6. Let be defined as in Lemma 2 and
Then the double inequality holds for .

Proof. It follows from Lemma 2 that for .
Therefore, Lemma 6 follows from (64).

Lemma 7. The inequality holds for .

Proof. Let Then where
It follows from Lemma 1 and (68) that for .
Therefore, Lemma 7 follows from (67) together with (69).

Lemma 8. Let
Then for .

Proof. Let Then
Lemma 7 and give and for . This in turn implies that for .
On the other hand, from the expression of , we get where for .
From (75)–(76), we clearly see that and for . This in turn implies that for .
Equation (72) together with inequalities (74) and (77) lead to the conclusion that for .

Lemma 9. Let Then for .

Proof. Let then
From (74), we clearly see that for .
On the other hand, from the expression of together with Lemma 1, we get for .
From (83), we clearly see that and for . This in turn implies that for .
Equation (81) together with inequalities (82) and (84) lead to the conclusion that for .

Lemma 10. Let be defined as in Lemma 2 and Then for .

Proof. Differentiating yields
It follows from (19) and (87) that for .
Therefore, for follows from (88).

Lemma 11. Let be defined as in Lemma 2 and Then for .

Proof. Differentiating yields
It follows from (19) and (90) together with the monotonicity of the function on that for .
Equation (91) leads to the conclusion that for .

Lemma 12. Let and be defined, respectively, as in Lemmas 3 and 5, and . Then is strictly decreasing on if .

Proof. Differentiating with respect to and making use of Lemmas 8 and 10, we get for and . This in turn implies that is strictly decreasing on if .

Lemma 13. Let and be defined, respectively, as in Lemmas 4 and 6, and . Then is strictly decreasing on if .

Proof. Differentiating with respect to and making use of Lemmas 9 and 11, we have for and . This in turn implies that is strictly decreasing on if .

#### 3. Main Results

Theorem 14. The double inequality holds for all with if and only if and .

Proof. Since , , and are symmetric and homogeneous of degree one, then without loss of generality, we assume that . Let , , and . Then , and
The difference between the convex combination of and is as follows:
Equation (99) leads to where , and are defined as in Lemmas 2, 3, 5, and 12, respectively.
It follows from (101) together with Lemmas 3 and 5 that for . Moreover, we see clearly, from Lemma 12, that is strictly decreasing on and so for . This in conjunction with (100) and (102) implies that for .
On the other hand, (101) and Lemmas 3 and 5 together with the monotonicity of the function on lead to for .
It follows from Lemma 12 that is strictly decreasing on . Note that From (104) and (105) together with the monotonicity of on , we clearly see that there exists such that is strictly increasing on and strictly decreasing on . This in conjunction with (100) implies that for .
Equation (99) together with inequalities (103) and (106) gives rise to
Therefore, Theorem 14 follows from (107) together with the following statements.(i)If , then (96) and (97) imply that there exists such that for all with .(ii)If , then (96) and (98) imply that there exists such that for all with .

Theorem 15. The double inequality holds for all with if and only if and .

Proof. We will follow the same idea in the proof of Theorem 14. Since , , and are symmetric and homogeneous of degree one. Without loss of generality, we assume that . Let ,