# An Optimal Double Inequality for Means

- Wei-Mao Qian
^{1}Email author and - Ning-Guo Zheng
^{1}

**2010**:578310

https://doi.org/10.1155/2010/578310

© W.-M. Qian and N.-G. Zheng. 2010

**Received: **3 September 2010

**Accepted: **27 September 2010

**Published: **30 November 2010

## 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

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 [1–9]. It might be surprising that the generalized logarithmic mean, has applications in physics, economics, and even in meteorology [10–13].

In [17–19] the authors present bounds for and in terms of and .

Theorem A.

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

Theorem B.

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

Theorem C.

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

Theorem D.

Theorem E.

with the best possible parameter .

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.

Proof.

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

Lemma 2.2.

Proof.

## 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 .

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

Case 1.

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

Case 2.

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

Case 3.

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

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.

Case 2.

Case 3.

Case 4.

Case 5.

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 .

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

Next, we prove that the constant cannot be improved.

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

## Declarations

### Acknowledgment

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

## Authors’ Affiliations

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