Abstract
We find the least values , , and in (0, 1/2) such that the inequalities , , , and hold for all with , respectively. Here , and denote the arithmetic-geometric, harmonic, geometric, and logarithmic means of two positive numbers and , respectively.
1. Introduction
The classical arithmetic-geometric mean of two positive real numbers and is defined as the common limit of sequences and , which are given by
Let , , , , and and be the harmonic, geometric, logarithmic, identric, arithmetic, and -th power means of two positive numbers and with , respectively. Then it is well known that for all with .
Recently, the inequalities for means have been the subject of intensive research. In particular, many remarkable inequalities for arithmetic-geometric mean can be found in the literature [1–9].
Carlson and Vuorinen [2], and Bracken [9] proved that for all with .
In [3], Vamanamurthy and Vuorinen established the following inequalities: for all with .
We recall the Gauss identity [6, 7] for and . As usual, and denote the complete elliptic integrals [8] given by
where for , and .
For fixed with and , let
Then it is not difficult to verify that , and are continuous and strictly increasing in , respectively. Note that , = and .
Therefore, it is natural to ask what are the least values , , and in such that the inequalities , + , and hold for all with , respectively. The main purpose of this paper is to answer these questions. Our main results are Theorems 1.1–1.3.
Theorem 1.1. If , then inequality holds for all with if and only if .
Theorem 1.2. If , then inequality holds for all with if and only if .
Theorem 1.3. If , then inequality holds for all with if and only if .
2. Lemmas
In order to establish our main results we need several formulas and lemmas, which we present in this section.
For , the following derivative formulas were presented in [6, Appendix , pp. 474-475]:
The following Lemma 2.1 can be found in [6, Theorem and Exercise ].
Lemma 2.1. is strictly increasing from onto ;
is strictly increasing from onto .
Lemma 2.2. Inequality holds for all .
Proof. Let
Then simple computations lead to
It follows from Lemma 2.1 (1) and (2.6) that for , which implies that is strictly increasing in .
Therefore, inequality (2.3) follows from (2.4) and (2.5) together with the monotonicity of .
Lemma 2.3. Inequality holds for all .
Proof. Let
Then simple computations lead to
Clearly the function is strictly decreasing from onto . Then (2.10) and Lemma 2.1 (2) lead to the conclusion that for . Thus, is strictly increasing in .
Therefore, inequality (2.7) follows from (2.8) and (2.9) together with the monotonicity of .
3. Proof of Theorems 1.1–1.3
Proof of Theorem 1.1. Let , then from the monotonicity of the function in we know that to prove inequality (1.10) we only need to prove that
for all with .
From (1.1) and (1.7) we clearly see that both and are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that . Let and , then from (1.5) we have
Let
Then making use of (2.2) we get
where . Note that
Therefore, inequality (3.1) follows from (3.2)–(3.5).
Next, we prove that the parameter is the best possible parameter in such that inequality (1.10) holds for all with .
Since for and small ,
It follows from (3.6) and (3.7) that inequality holds for small only .
Remark 3.1. For and , one has
Equation (3.8) implies that there does not exist such that for all .
Proof of Theorem 1.2. Let , then from the monotonicity of the function in we know that to prove inequality (1.11) we only need to prove that
for all with .
From (1.1) and (1.8) we clearly see that both and are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that . Let and , then from (1.5) we have
Let
Then making use of (2.2) we have
Therefore, inequality (3.9) follows from (3.10)–(3.12) together with Lemma 2.2.
Next, we prove that the parameter is the best possible parameter in such that inequality (1.11) holds for all with .
Since for and small ,
It follows from (3.6) and (3.13) that inequality holds for small only .
Remark 3.2. For and , one has
Equation (3.14) implies that there does not exist such that for all .
Proof of Theorem 1.3. Let , then from the monotonicity of in we know that to prove inequality (1.12) we only need to prove that
for all with .
From (1.1) and (1.9) we clearly see that both and are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that . Let and , then from (1.5) one has
Let
Then from (2.2) we get
where is defined as in Lemma 2.3.
Therefore, inequality (3.15) follows from (3.16)–(3.18) together with Lemma 2.3.
Next, we prove that the parameter is the best possible parameter in such that inequality (1.12) holds for all with .
Since for and small ,
It follows from (3.6) and (3.19) that inequality holds for small only .
Remark 3.3. For and , one has
Equation (3.20) implies that there exist no values such that for all .
Acknowledgments
The authors wish to thank the anonymous referees for their careful reading of the paper and their fruitful comments and suggestions. This research was supported by the Natural Science Foundation of China under Grant 11071069, and Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.