• Research Article
• Open Access

# Proof of One Optimal Inequality for Generalized Logarithmic, Arithmetic, and Geometric Means

Journal of Inequalities and Applications20102010:902432

https://doi.org/10.1155/2010/902432

• Received: 11 July 2010
• Accepted: 31 October 2010
• Published:

## Abstract

Two open problems were posed in the work of Long and Chu (2010). In this paper, we give the solutions of these problems.

## Keywords

• Real Number
• Unique Solution
• Open Problem
• Simple Computation
• Elementary Calculation

## 1. Introduction

The arithmetic and geometric means of two positive numbers and are defined by , , respectively. If is a real number, then the generalized logarithmic mean with parameter of two positive numbers , is defined by

In the paper , Long and Chu propose the two following open problems:

Open Problem 1.

What is the least value such that the inequality

holds for and all with Open Problem 2.

What is the greatest value such that the inequality

holds for and all with For information on the history, background, properties, and applications of inequalities for generalized logarithmic, arithmetic, and geometric means, please refer to  and related references there in.

## 2. Main Result

Theorem 2.1.

Let , , , . Let be a solution of
and is the best constant,

and is the best constant.

## 3. Proof of Theorem 2.1

Because is increasing with respect to for fixed and , it suffices to prove that for any (resp., ) there exists such that (resp., ), and is the best constant. Without loss of generality, we assume that . Let , . Equations (2.2), (2.3) are equivalent to
On putting , we obtain (3.1) is equivalent to
Introduce the function by
Simple computations yield for Let and the unique solution to
To see that is optimal in both cases (2.2), (2.3), note that . Thus, if the constant is decreased (resp., increased), then the desired bound for would not hold for small . This follows from the fact that for a fixed , the function

is nondecreasing.

From now on, let for . To show the estimates for this , we start from observing that . Furthermore, one easily checks that
Thus, it suffices to verify that has exactly one zero inside the interval . It follows from the mean value theorem. After some computations, this is equivalent to saying that the function given by

has exactly one root in . Here, the expression under the logarithm may be nonpositive, so we define on a maximal interval, contained in . It is easy to see that this interval must be of the form , for some . This follows from the fact that is strictly positive on and is strictly increasing on this interval.

Since and , we will be done if we show that has exactly one root in . After some computations, we obtain that the equation is equivalent to
Because is a quadratic polynomial in the variable , all that remains is to show that
or, in virtue of the definition of ,

This can be easily established by some elementary calculations. It completes the proof.

## Declarations

### Acknowledgments

The author is indebted to the anonymous referee for many valuable comments, for a correction of one part of the proof, and for his improving of the organization of the paper. This work was supported by Vega no. 1/0157/08 and Kega no. 3/7414/09.

## Authors’ Affiliations

(1)
Faculty of Industrial Technologies in Púchov, Alexander Dubček University in Trenčín, I. Krasku 491/30, 02001 Púchov, Slovakia

## References 