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An Optimal Double Inequality for Means

Abstract

For , the generalized logarithmic mean , arithmetic mean and geometric mean of two positive numbers and are defined by , ; , , , ; , , ; , , ; and , respectively. In this paper, we give an answer to the open problem: for , what are the greatest value and the least value , such that the double inequality holds for all ?

1. Introduction

For , the generalized logarithmic mean of two positive numbers and is defined by

(1.1)

It is wellknown that is continuous and increasing with respect to for fixed and . In the recent past, the generalized logarithmic mean has been the subject of intensive research. Many remarkable inequalities and monotonicity results can be found in the literature [19]. It might be surprising that the generalized logarithmic mean, has applications in physics, economics, and even in meteorology [1013].

If we denote by ,, , and the arithmetic mean, identric mean, logarithmic mean, geometric mean and harmonic mean of two positive numbers and , respectively, then

(1.2)

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

(1.3)

In [14], Alzer and Janous established the following sharp double inequality (see also [15], Page 350):

(1.4)

for all .

For , Janous [16] found the greatest value and the least value such that

(1.5)

for all .

In [1719] the authors present bounds for and in terms of and .

Theorem A.

For all positive real numbers and with , one has

(1.6)

The proof of the following Theorem B can be found in [20].

Theorem B.

For all positive real numbers and with , one has

(1.7)

The following Theorems C–E were established by Alzer and Qiu in [21].

Theorem C.

The inequalities

(1.8)

hold for all positive real numbers and with if and only if and .

Theorem D.

Let and be real numbers with . If , then

(1.9)

And if , then

(1.10)

Theorem E.

For all real numbers and with , one has

(1.11)

with the best possible parameter .

However, the following problem is still open: for , what are the greatest value and the least value , such that the double inequality

(1.12)

holds for all ? The purpose of this paper is to give the solution to this open problem.

2. Lemmas

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

Lemma 2.1.

If , then

(2.1)

Proof.

Let , then simple computation yields

(2.2)
(2.3)

where

(2.4)

where

(2.5)
(2.6)
(2.7)

If , then from (2.7) we clearly see that

(2.8)

Therefore, Lemma 2.1 follows from (2.3)–(2.6) and (2.8).

Lemma 2.2.

If , then

(2.9)

Proof.

Let , then simple computation leads to

(2.10)

where

(2.11)

where

(2.12)

where

(2.13)
(2.14)

If , then from (2.14) we clearly see that

(2.15)

From (2.10)–(2.13) and (2.15) we know that for .

3. Main Results

Theorem 3.1.

If , then for all , with equality if and only if  , and the constant in , cannot be improved.

Proof.

If , then we clearly see that .

If , without loss of generality, we assume that . Let and

(3.1)

Firstly, we prove . The proof is divided into three cases.

Case 1.

. We note that (1.1) leads to the following identity:

(3.2)

From (3.2) and Lemma 2.1 we clearly see that for and .

Case 2.

. Equation (1.1) leads to the following identity:

(3.3)

From (3.3) and Lemma 2.2 we clearly see that for and .

Case 3.

. From (1.1) we have the following identity:

(3.4)

Equation (3.4) and elementary computation yields

(3.5)
(3.6)

where

(3.7)
(3.8)

If , then (3.8) implies

(3.9)

for . Therefore, follows from (3.5)–(3.7) and (3.9).

If , then (3.8) leads to

(3.10)

for . Therefore, follows from (3.5)–(3.7) and (3.10).

Next, we prove that the constant in the inequality cannot be improved. The proof is divided into five cases.

Case 1.

. For any , let , then (1.1) leads to

(3.11)

where .

Making use of Taylor expansion we get

(3.12)

Case 2.

. For any , let , then

(3.13)

where .

Using Taylor expansion we have

(3.14)

Case 3.

. For any , let , then

(3.15)

where .

Making use of Taylor expansion and elaborated calculation we have

(3.16)

Case 4.

. For any , let , then

(3.17)

where .

Using Taylor expansion and elaborated calculation we have

(3.18)

Case 5.

. For any , let , then

(3.19)

where .

Using Taylor expansion and elaborated calculation we get

(3.20)

Cases 1–5 show that for any , there exists , for any there exists such that for .

Theorem 3.2.

If , then for all , with equality if and only if , and the constant in cannot be improved.

Proof.

If , then we clearly see that .

If , without loss of generality, we assume that . Let and

(3.21)

Firstly, we prove for . Simple computation leads to

(3.22)

where

(3.23)

for and .

From (3.23) we clearly see that

(3.24)

for .

Since , we have for . Therefore, follows from (3.22) and (3.24).

Next, we prove that the constant cannot be improved.

For any , we have

(3.25)

Equation (3.25) imply that for any there exists , such that for .

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Acknowledgment

This work was supported by the Natural Science Foundation of Zhejiang Broadcast and TV University (Grant no. XKT-09G21).

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Correspondence to Wei-Mao Qian.

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Qian, WM., Zheng, NG. An Optimal Double Inequality for Means. J Inequal Appl 2010, 578310 (2010). https://doi.org/10.1155/2010/578310

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