Abstract and Applied Analysis

Abstract and Applied Analysis / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 684834 | 7 pages | https://doi.org/10.1155/2012/684834

A Sharp Double Inequality between Seiffert, Arithmetic, and Geometric Means

Academic Editor: Josef Diblík
Received02 Jul 2012
Accepted21 Aug 2012
Published02 Sep 2012

Abstract

For fixed 𝑠≥1 and any 𝑡1,𝑡2∈(0,1/2) we prove that the double inequality 𝐺𝑠(𝑡1ğ‘Ž+(1−𝑡1)𝑏,𝑡1𝑏+(1−𝑡1)ğ‘Ž)𝐴1−𝑠(ğ‘Ž,𝑏)<𝑃(ğ‘Ž,𝑏)<𝐺𝑠(𝑡2ğ‘Ž+(1−𝑡2)𝑏,𝑡2𝑏+(1−𝑡2)ğ‘Ž)𝐴1−𝑠(ğ‘Ž,𝑏) holds for all ğ‘Ž,𝑏>0 with ğ‘Žâ‰ ğ‘ if and only if 𝑡1ò(1−1−(2/𝜋)2/𝑠)/2 and 𝑡2ó(1−1/3𝑠)/2. Here, 𝑃(ğ‘Ž,𝑏), 𝐴(ğ‘Ž,𝑏) and 𝐺(ğ‘Ž,𝑏) denote the Seiffert, arithmetic, and geometric means of two positive numbers ğ‘Ž and 𝑏, respectively.

1. Introduction

The Seiffert mean 𝑃(ğ‘Ž,𝑏) [1] of two distinct positive numbers ğ‘Ž and 𝑏 is defined by 𝑃(ğ‘Ž,𝑏)=ğ‘Žâˆ’ğ‘î‚€âˆš4arctan.ğ‘Ž/𝑏−𝜋(1.1)

Recently, the Seiffert mean 𝑃(ğ‘Ž,𝑏) has been the subject of intensive research. In particular, many remarkable inequalities for 𝑃(ğ‘Ž,𝑏) can be found in the literature [2–17]. The Seiffert mean 𝑃(ğ‘Ž,𝑏) can be rewritten as (see [6, (2.4)]) 𝑃(ğ‘Ž,𝑏)=ğ‘Žâˆ’ğ‘.2arcsin((ğ‘Žâˆ’ğ‘)/(ğ‘Ž+𝑏))(1.2)

Let 𝐴(ğ‘Ž,𝑏)=(ğ‘Ž+𝑏)/2, √𝐺(ğ‘Ž,𝑏)=ğ‘Žğ‘ and 𝐻(ğ‘Ž,𝑏)=2ğ‘Žğ‘/(ğ‘Ž+𝑏) be the classical arithmetic, geometric, and harmonic means of two positive numbers ğ‘Ž and 𝑏, respectively. Then it is well known that inequalities 𝐻(ğ‘Ž,𝑏)<𝐺(ğ‘Ž,𝑏)<𝑃(ğ‘Ž,𝑏)<𝐴(ğ‘Ž,𝑏) hold for all ğ‘Ž,𝑏>0 with ğ‘Žâ‰ ğ‘.

For 𝛼,𝛽,𝜆,𝜇∈(0,1/2), Chu et al. [18, 19] proved that the double inequalities 𝐻𝐺(ğ›¼ğ‘Ž+(1−𝛼)𝑏,𝛼𝑏+(1−𝛼)ğ‘Ž)<𝑃(ğ‘Ž,𝑏)<𝐺(ğ›½ğ‘Ž+(1−𝛽)𝑏,𝛽𝑏+(1−𝛽)ğ‘Ž),(ğœ†ğ‘Ž+(1−𝜆)𝑏,𝜆𝑏+(1−𝜆)ğ‘Ž)<𝑃(ğ‘Ž,𝑏)<𝐻(ğœ‡ğ‘Ž+(1−𝜇)𝑏,𝜇𝑏+(1−𝜇)ğ‘Ž)(1.3) hold for all ğ‘Ž,𝑏>0 with ğ‘Žâ‰ ğ‘ if and only if √𝛼≤(1−1−4/𝜋2)/2, √𝛽≥(3−3)/6, √𝜆≤(1−1−2/𝜋)/2 and √𝜇≥(6−6)/12.

Let 𝑡∈(0,1/2), 𝑠≥1 and 𝑄𝑡,𝑠(ğ‘Ž,𝑏)=𝐺𝑠(ğ‘¡ğ‘Ž+(1−𝑡)𝑏,𝑡𝑏+(1−𝑡)ğ‘Ž)𝐴1−𝑠(ğ‘Ž,𝑏),(1.4) then it is not difficult to verify that 𝑄𝑡,1𝑄(ğ‘Ž,𝑏)=𝐺(ğ‘¡ğ‘Ž+(1−𝑡)𝑏,𝑡𝑏+(1−𝑡)ğ‘Ž),𝑡,2(ğ‘Ž,𝑏)=𝐻(ğ‘¡ğ‘Ž+(1−𝑡)𝑏,𝑡𝑏+(1−𝑡)ğ‘Ž)(1.5) and 𝑄𝑡,𝑠(ğ‘Ž,𝑏) is strictly increasing with respect to 𝑡∈(0,1/2) for fixed ğ‘Ž,𝑏>0 with ğ‘Žâ‰ ğ‘.

It is natural to ask what are the largest value 𝑡1=𝑡1(𝑠) and the least value 𝑡2=𝑡2(𝑠) in (0,1/2) such that the double inequality 𝑄𝑡1,𝑠(ğ‘Ž,𝑏)<𝑃(ğ‘Ž,𝑏)<𝑄𝑡2,𝑠(ğ‘Ž,𝑏) holds for all ğ‘Ž,𝑏>0 with ğ‘Žâ‰ ğ‘ and 𝑠≥1. The main purpose of this paper is to answer this question.

2. Main Result

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

Lemma 2.1. If 𝑠≥1, then 1/(3𝑠)+(2/𝜋)2/𝑠<1.

Proof. Consider the following: 1𝑓(𝑠)=+23𝑠𝜋2/𝑠.(2.1)
Then simple computations lead to lim𝑠→+âˆžğ‘“ğ‘“(𝑠)=1,(2.2)2(𝑠)=𝑠2𝜋log22𝜋2/𝑠−1≥26log(𝜋/2)𝑠2𝜋log22𝜋2−1=6log(𝜋/2)24log(𝜋/2)−𝜋23𝜋2𝑠2(2.3) for 𝑠≥1.
Computational and numerical experiments show that 𝜋24log2−𝜋2=0.968⋯>0.(2.4)
Inequalities (2.3) and (2.4) imply that 𝑓(𝑠) is strictly increasing in [1,+∞). Therefore, Lemma 2.1 follows from (2.1) and (2.2) together with the monotonicity of 𝑓(𝑠).

Lemma 2.2. Let 0≤𝑢≤1, 𝑠≥1 and 𝑓𝑢,𝑠𝑠(𝑥)=2log1−𝑢𝑥2−log𝑥+log(arcsin𝑥).(2.5)
Then inequality 𝑓𝑢,𝑠(𝑥)>0 holds for all 𝑥∈(0,1) if and only if 3𝑠𝑢≤1, and inequality 𝑓𝑢,𝑠(𝑥)<0 holds for all 𝑥∈(0,1) if and only if 𝑢+(2/𝜋)2/𝑠≥1.

Proof. If 𝑢=0, then we clearly see that 3𝑠𝑢≤1, 𝑢+(2/𝜋)2/𝑠<1 and 𝑓0,𝑠(𝑥)=log[(arcsin𝑥)/𝑥]>0 for all 𝑠≥1 and 𝑥∈(0,1). In the following discussion, we assume that 0<𝑢≤1.
From (2.5) and simple computations we have lim𝑥→0+𝑓𝑢,𝑠𝑓(𝑥)=0,(2.6)î…žğ‘¢,𝑠1(𝑥)=√1−𝑥2−arcsin𝑥1+𝑢(𝑠−1)𝑥2𝑥1−𝑢𝑥2=1+𝑢(𝑠−1)𝑥2𝑥1−𝑢𝑥2𝑔arcsin𝑥𝑢,𝑠(𝑥),(2.7) where 𝑔𝑢,𝑠𝑥(𝑥)=1−𝑢𝑥2√1−𝑥21+𝑢(𝑠−1)𝑥2𝑔−arcsin𝑥,(2.8)𝑢,𝑠𝑔(0)=0,(2.9)î…žğ‘¢,𝑠𝑥(𝑥)=21−𝑥23/21+𝑢(𝑠−1)𝑥22â„Žğ‘¢,𝑠(𝑥),(2.10) where â„Žğ‘¢,𝑠(𝑥)=𝑢2(𝑠−1)2𝑥4+𝑢−𝑠2𝑥𝑢+𝑢𝑠+4𝑠−22ℎ+1−3𝑠𝑢,(2.11)𝑢,ğ‘ â„Ž(0)=1−3𝑠𝑢,(2.12)𝑢,𝑠(1)=𝑢𝑠(1−𝑢)+(1−𝑢)2.(2.13)
We divide the proof into four cases.
Case 1 (3𝑠𝑢≤1). Then from (2.11) and (2.12) together with the fact that −𝑢𝑠2+𝑢𝑠+4𝑠−2=2(𝑠−1)+𝑠(𝑢+2𝑠𝑢+1)+𝑠(1−3𝑠𝑢)>0,(2.14) we clearly see that â„Žğ‘¢,𝑠(0)≥0,(2.15) and â„Žğ‘¢,𝑠(𝑥) is strictly increasing in [0,1].
Equation (2.12) and the monotonicity of â„Žğ‘¢,𝑠(𝑥) imply that â„Žğ‘¢,𝑠(𝑥)>0(2.16) for 𝑥∈(0,1].
Equation (2.10) and inequality (2.16) lead to the conclusion that 𝑔𝑢,𝑠(𝑥) is strictly increasing in [0,1). Then from (2.9) we know that 𝑔𝑢,𝑠(𝑥)>0(2.17) for 𝑥∈(0,1).
It follows from (2.7) and inequality (2.17) that 𝑓𝑢,𝑠(𝑥) is strictly increasing in (0,1].
Therefore, 𝑓𝑢,𝑠(𝑥)>0 for all 𝑥∈(0,1) follows from (2.6) and the monotonicity of 𝑓𝑢,𝑠(𝑥).
Case 2 (3𝑠𝑢>1). Then (2.12) and the continuity of â„Žğ‘¢,𝑠(𝑥) imply that there exists 0<𝜆<1 such that â„Žğ‘¢,𝑠(𝑥)<0(2.18) for 𝑥∈[0,𝜆).
Therefore, 𝑓𝑢,𝑠(𝑥)<0 for 𝑥∈(0,𝜆) follows easily from (2.6), (2.7), (2.9) and (2.10) together with inequality (2.18).
Case 3 (𝑢+(2/𝜋)2/𝑠≥1). Then Lemma 2.1 and (2.12) lead to â„Žğ‘¢,𝑠2(0)=1−3𝑠𝑢≤1−3𝑠1−𝜋2/𝑠<0.(2.19)
We divide the proof into two subcases.
Subcase 3.1 (𝑢=1). Then (2.13) becomes â„Žğ‘¢,𝑠(1)=0.(2.20)
Let 𝑡=𝑥2, then from (2.11) we clearly see that the function â„Žğ‘¢,𝑠 is a quadratic function of variable 𝑡. It follows from inequality (2.19) and (2.20) that â„Žğ‘¢,𝑠(𝑥)<0(2.21) for all 𝑥∈[0,1).
Therefore, 𝑓𝑢,𝑠(𝑥)<0 for 𝑥∈(0,1) follows easily from (2.6), (2.7), (2.9) and (2.10) together with inequality (2.21).
Subcase 3.2 (0<𝑢<1). Then from (2.5), (2.8), and (2.13) we have 𝑓𝑢,𝑠𝜋(1)=log2(1−𝑢)𝑠/2≤0,(2.22)lim𝑥→1−𝑔𝑢,𝑠(â„Žğ‘¥)=+∞,(2.23)𝑢,𝑠(1)>0.(2.24)
From (2.11), (2.19), and (2.24) we clearly see that there exists 0<𝜆1<1 such that â„Žğ‘¢,𝑠(𝑥)<0 for 𝑥∈[0,𝜆1) and â„Žğ‘¢,𝑠(𝑥)>0 for 𝑥∈(𝜆1,1]. Then (2.10) implies that 𝑔𝑢,𝑠(𝑥) is strictly decreasing in [0,𝜆1] and strictly increasing in [𝜆1,1).
From (2.9) and (2.23) together with the piecewise monotonicity of 𝑔𝑢,𝑠(𝑥) we clearly see that there exists 0<𝜆2<1 such that 𝑔𝑢,𝑠(𝑥)<0, for 𝑥∈(0,𝜆2) and 𝑔𝑢,𝑠(𝑥)>0 for 𝑥∈(𝜆2,1). Then (2.7) implies that 𝑓𝑢,𝑠(𝑥) is strictly decreasing in (0,𝜆2] and strictly increasing in [𝜆2,1].
Therefore, 𝑓𝑢,𝑠(𝑥)<0 for 𝑥∈(0,1) follows from (2.6) and (2.22) together with the piecewise monotonicity of 𝑓𝑢,𝑠(𝑥).
Case 4 (𝑢+(2/𝜋)2/𝑠<1). Then (2.5) leads to 𝑓𝑢,𝑠𝜋(1)=log2(1−𝑢)𝑠/2>0.(2.25)
From inequality (2.25) and the continuity of 𝑓𝑢,𝑠(𝑥) we know that there exists 0<𝜇<1 such that 𝑓𝑢,𝑠(𝑥)>0 for 𝑥∈(𝜇,1].

Theorem 2.3. If 𝑡1,𝑡2∈(0,1/2) and 𝑠≥1, then the double inequality 𝑄𝑡1,𝑠(ğ‘Ž,𝑏)<𝑃(ğ‘Ž,𝑏)<𝑄𝑡2,𝑠(ğ‘Ž,𝑏)(2.26) holds for all ğ‘Ž,𝑏>0 with ğ‘Žâ‰ ğ‘ if and only if 𝑡1ò(1−1−(2/𝜋)2/𝑠)/2 and 𝑡2ó(1−1/3𝑠)/2.

Proof. Since both 𝑄𝑡,𝑠(ğ‘Ž,𝑏) and 𝑃(ğ‘Ž,𝑏) are symmetric and homogeneous of degree 1. Without loss of generality, we assume that ğ‘Ž>𝑏. Let 𝑥=(ğ‘Žâˆ’ğ‘)/(ğ‘Ž+𝑏)∈(0,1). Then from (1.2) and (1.4) we have 𝑄log𝑡,𝑠(ğ‘Ž,𝑏)𝑄𝑃(ğ‘Ž,𝑏)=log𝑡,𝑠(ğ‘Ž,𝑏)𝐴(ğ‘Ž,𝑏)−log𝑃(ğ‘Ž,𝑏)=𝑠𝐴(ğ‘Ž,𝑏)2log1−(1−2𝑡)2𝑥2−log𝑥+log(arcsin𝑥).(2.27)
Therefore, Theorem 2.3 follows easily from Lemma 2.2 and (2.27).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grant 11071069 and the Natural Science Foundation of Hunan Province under Grant 09JJ6003.

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Copyright © 2012 Wei-Ming Gong et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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