1. Introduction

For , the generalized logarithmic mean and power mean of two positive numbers and are defined as

(1.1)
(1.2)

It is well known that and are continuous and increasing with respect to for fixed and . Let , , , , and be the arithmetic, identric, logarithmic, geometric, and harmonic means of and , respectively. Then

(1.3)

In [1], the following results are established: (1) implies that ; (2) implies that ; (3) implies that there exist such that ; (4) implies that there exist such that . Hence the question was answered: what are the least value and the greatest value such that the inequality holds for all ?

Stolarsky [2] proved that , with equality if and only if .

In [3], Pittenger proved that

(1.4)

for all , where

(1.5)

Here , are sharp and equality holds only if or or . The case reduces to Lin's results [1]. Other generalizations of Lin's results were given by Imoru [4].

Qi and Guo [5] established that

(1.6)

for all , and . The upper bound in (1.6) is the best possible.

In [6], Chu et al. established the following result:

(1.7)

for all , where the function is the logarithmic derivative of the gamma function.

Recently, some monotonicity results of the ratio between generalized logarithmic means were established in [7–9].

The purpose of this paper is to answer the following questions: what are the greatest values and , and the least value such that , , and for all ?

2. Main Results

Theorem 2.1.

for all , with inequality if and only if , and the constant cannot be improved.

Proof.

If , then from (1.1) we clearly see that Next, we assume that and , and then elementary computations yield

(2.1)

To prove that is the largest number for which the inequality holds, we take and , and we see that

(2.2)
(2.3)

where

Making use of the Taylor expansion, we have

(2.4)

Equations (2.3) and (2.4) imply that for any there exists such that for .

Theorem 2.2.

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

Proof.

Simple computations yield

(2.5)

Next we prove that is the optimal value for which the inequality holds.

For and , elementary computations yield

(2.6)
(2.7)

where

Using Taylor expansion we get

(2.8)

Equations (2.7) and (2.8) imply that for any there exists such that for .

Theorem 2.3.

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

Proof.

Form (1.1) we clearly see that if . If , then simple computations yield

(2.9)

To show that is the best possible constant for which the inequality holds, let and , and then

(2.10)

where

Using Taylor expansion we have

(2.11)

Equations (2.10) and (2.11) imply that for any there exists such that for .

Remark 2.4.

If , then

(2.12)

Therefore, we cannot get inequality for any and all

Remark 2.5.

It is easy to verify that for all