Journal of Applied Mathematics

Volume 2012 (2012), Article ID 379785, 12 pages

doi:10.1155/2012/379785

Research Article

A Best Possible Double Inequality for Power Mean

Yong-Min Li,¹ Bo-Yong Long,² and Yu-Ming Chu¹

¹Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China

²School of Mathematics Science, Anhui University, Hefei 230039, China

Received 29 February 2012; Accepted 9 September 2012

Academic Editor: Huijun Gao

Copyright © 2012 Yong-Min Li 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 answer the question: for any with and , what are the greatest value and the least value , such that the double inequality holds for all with ? Where is the th power mean of two positive numbers and .

1. Introduction

For , the th power mean of two positive numbers and is defined by

It is well known that is continuous and strictly increasing with respect to for fixed , with . Many classical means are special case of the power mean, for example, , , and are the harmonic, geometric, and arithmetic means of and , respectively. Recently, the power mean has been the subject of intensive research. In particular, many remarkable inequalities and properties for the power mean can be found in literature [1–15].

Let and be the logarithmic and identric means of two positive numbers and with , respectively. Then it is well known that

for all with .

In [16–22], the authors presented the sharp power mean bounds for , , , and as follows:

for all with .

Alzer and Qiu [12] proved that the inequality

holds for all with if and only if .

The following sharp bounds for the sum , and the products and in terms of power means were proved in [5, 8] as follows:

for any and all with .

In [2, 7], the authors answered the question: for any , what are the greatest values ,  ,   , and , and the least values , ,  , and , such that the inequalities

hold for all with ?

In [4], the authors presented the greatest value and the least value such that the double inequality

holds for all with and with .

It is the aim of this paper to answer the question: for any , with and , what are the greatest value and the least value , such that the double inequality

holds for all with ?

2. Main Result

In order to establish our main result, we need a lemma which we present in this section.

Lemma 2.1. Let , ,   and . Then

for , and

for .

Proof. From (1.1), we have

Let

then simple computations lead to

Equation (2.6) implies that

for , and

for .

Therefore, inequality (2.1) follows from (2.3)–(2.5) and inequality (2.7), and inequality (2.2) follows from (2.3)–(2.5) and inequality (2.8).

Let

Then we clearly see that , and it is not difficult to verify that the identity holds for all , if . Let

Then we have Theorem 2.2 as follows.

Theorem 2.2. The double inequality

holds for all , with , and and are the best possible lower and upper power mean bounds for the geometric mean of and .

Proof. From (1.1), we clearly see that is symmetric and homogenous of degree 1. Without loss of generality, we assume that , and . We divide the proof of inequality (2.11) into three cases.

Case 1. . Then from Lemma 2.1, we clearly see that

for , and

for .

From (1.1), we get

Let

then simple computations lead to

where

where

where

where

If , then (2.15), (2.18), (2.21), (2.22), (2.24), (2.25), (2.27), and (2.28) lead to

We divide the discussion into two subcases.

Subcase 1.1. . Then (2.26) and (2.29) together with inequalities (2.36) and (2.37) imply that is strictly decreasing in . In fact, if , then (2.29) and inequality (2.37) imply that for . If , then (2.29) and inequality (2.36) lead to the conclusion that for .

From inequalities (2.34) and (2.35) together with the monotonicity of , we know that there exists such that for and for . Then (2.23) leads to the conclusion that is strictly increasing in and strictly decreasing in .

It follows from (2.32) and (2.33) together with the piecewise monotonicity of that there exists such that for and for . Then (2.20) leads to the conclusion that is strictly increasing in and strictly decreasing in .

From (2.17), (2.19) and (2.31) together with the piecewise monotonicity of , we clearly see that there exists such that is strictly increasing in and strictly decreasing in .

Therefore, follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of .

Subcase 1.2.   . Then (2.30) and (2.35) again hold, and (2.18), (2.21), (2.22), and (2.28) lead to

It follows from (2.29) and inequalities and (2.41) that for . Then (2.26) and inequality (2.35) lead to the conclusion that for . Therefore, is strictly increasing in follows from (2.23).

It follows from (2.20) and (2.39) together with inequality (2.40) and the monotonicity of that there exists such that is strictly decreasing in and strictly increasing in .

From (2.17), (2.19) and (2.38) together with the piecewise monotonicity of , we clearly see that there exists such that is strictly decreasing in and strictly increasing in .

Therefore, follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of .

Case 2. . Clearly, we have for and for . Therefore, we need only to prove that

for , and

for .

From (1.1), one has

Let

then simple computations lead to

If (or , resp.), then (2.47) leads to the conclusion that is strictly decreasing (or increasing, resp.) in . Therefore, inequalities (2.42) and (2.43) follow from (2.44)–(2.46) and the monotonicity of .

Case 3.. Then from Lemma 2.1, we clearly see that for and for . Therefore, we need only to prove that

for , and

for .

From (1.1), we get

Let

then simple computations lead to

If (or , resp.), then (2.53) implies that is strictly increasing (or decreasing, resp.) in . Therefore, inequalities (2.48) and (2.49) follow from (2.50)–(2.52) and the monotonicity of .

Next, we prove that and are the best possible lower and upper power mean bounds for the geometric mean of and . We divide the proof into six cases.

Case . . Then for any and , from (1.1), one has

Letting and making use of Taylor expansion, we get

Equations (2.54) and (2.56) together with inequality (2.55) imply that for any , there exist and such that for and for .

Case . . Then for and , making use of (1.1) and Taylor expansion, we have

Equation (2.57) and inequality (2.58) imply that for any , there exist and such that for and for .

Case . . Then for and , making use of (1.1) and Taylor expansion, we have

Equation (2.59) leads to the conclusion that for any , there exist and such that for and for .

Case . . Then for and , making use of (1.1) and Taylor expansion, we have

Equation (2.60) implies that for any , there exist and such that for and for .

Case . . Then for any and , making use of (1.1) and Taylor expansion, one has

Equation (2.61) leads to the conclusion that for any , there exist and such that for and for .

Case . . Then for any and , making use of (1.1) and Taylor expansion, one has

Equation (2.62) shows that for any , there exist and such that for and for .

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

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