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

Tie-Hong Zhao,¹ Yu-Ming Chu,² Yun-Liang Jiang,³ and Yong-Min Li⁴

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

Academic Editor: Khalil Ezzinbi

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 [1–18].

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 [23–25]. 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

Equation (32) leads to

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

Equation (48) leads to

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 , , and . Then .

Making use of (95) together with gives

The difference between the convex combination of and is as follows:

Equation (112) leads to

where , and are defined as in Lemmas 2, 4, 6, and 13, respectively.

It follows from Lemmas 4, 6, and 13 together with (114) that

for and is strictly decreasing on . Thus, we have for . This in conjunction with (113) and (115) implies that

for .

On the other hand, Lemmas 4, 6, and 13 together with (114) lead to

for and is strictly decreasing on . Note that

From (117) and (118) 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 (113) implies that

for .

Equation (112) together with inequalities (116) and (119) lead to the conclusion that

Therefore, Theorem 15 follows from (120) together with the following statements.

(i)If , then (109) and (110) imply that there exists such that for all with .(ii)If , then (109) and (111) imply that there exists such that for all with .

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 11171307, 61174076, and 61173123 and the Natural Science Foundation of Zhejiang Province under Grants Z1110551 and LY12F02012.

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