Abstract

For fixed 𝑠1 and any 𝑡1,𝑡2(0,1/2) we prove that the double inequality 𝐺𝑠(𝑡1𝑎+(1𝑡1)𝑏,𝑡1𝑏+(1𝑡1)𝑎)𝐴1𝑠(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝐺𝑠(𝑡2𝑎+(1𝑡2)𝑏,𝑡2𝑏+(1𝑡2)𝑎)𝐴1𝑠(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎𝑏 if and only if 𝑡1(11(2/𝜋)2/𝑠)/2 and 𝑡2(11/3𝑠)/2. Here, 𝑃(𝑎,𝑏), 𝐴(𝑎,𝑏) and 𝐺(𝑎,𝑏) denote the Seiffert, arithmetic, and geometric means of two positive numbers 𝑎 and 𝑏, respectively.

1. Introduction

The Seiffert mean 𝑃(𝑎,𝑏) [1] of two distinct positive numbers 𝑎 and 𝑏 is defined by 𝑃(𝑎,𝑏)=𝑎𝑏4arctan.𝑎/𝑏𝜋(1.1)

Recently, the Seiffert mean 𝑃(𝑎,𝑏) has been the subject of intensive research. In particular, many remarkable inequalities for 𝑃(𝑎,𝑏) can be found in the literature [217]. The Seiffert mean 𝑃(𝑎,𝑏) can be rewritten as (see [6, (2.4)]) 𝑃(𝑎,𝑏)=𝑎𝑏.2arcsin((𝑎𝑏)/(𝑎+𝑏))(1.2)

Let 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝐺(𝑎,𝑏)=𝑎𝑏 and 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏) be the classical arithmetic, geometric, and harmonic means of two positive numbers 𝑎 and 𝑏, respectively. Then it is well known that inequalities 𝐻(𝑎,𝑏)<𝐺(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝐴(𝑎,𝑏) hold for all 𝑎,𝑏>0 with 𝑎𝑏.

For 𝛼,𝛽,𝜆,𝜇(0,1/2), Chu et al. [18, 19] proved that the double inequalities 𝐻𝐺(𝛼𝑎+(1𝛼)𝑏,𝛼𝑏+(1𝛼)𝑎)<𝑃(𝑎,𝑏)<𝐺(𝛽𝑎+(1𝛽)𝑏,𝛽𝑏+(1𝛽)𝑎),(𝜆𝑎+(1𝜆)𝑏,𝜆𝑏+(1𝜆)𝑎)<𝑃(𝑎,𝑏)<𝐻(𝜇𝑎+(1𝜇)𝑏,𝜇𝑏+(1𝜇)𝑎)(1.3) hold for all 𝑎,𝑏>0 with 𝑎𝑏 if and only if 𝛼(114/𝜋2)/2, 𝛽(33)/6, 𝜆(112/𝜋)/2 and 𝜇(66)/12.

Let 𝑡(0,1/2), 𝑠1 and 𝑄𝑡,𝑠(𝑎,𝑏)=𝐺𝑠(𝑡𝑎+(1𝑡)𝑏,𝑡𝑏+(1𝑡)𝑎)𝐴1𝑠(𝑎,𝑏),(1.4) then it is not difficult to verify that 𝑄𝑡,1𝑄(𝑎,𝑏)=𝐺(𝑡𝑎+(1𝑡)𝑏,𝑡𝑏+(1𝑡)𝑎),𝑡,2(𝑎,𝑏)=𝐻(𝑡𝑎+(1𝑡)𝑏,𝑡𝑏+(1𝑡)𝑎)(1.5) and 𝑄𝑡,𝑠(𝑎,𝑏) is strictly increasing with respect to 𝑡(0,1/2) for fixed 𝑎,𝑏>0 with 𝑎𝑏.

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(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝑄𝑡2,𝑠(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎𝑏 and 𝑠1. 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 𝑠1, then 1/(3𝑠)+(2/𝜋)2/𝑠<1.

Proof. Consider the following: 1𝑓(𝑠)=+23𝑠𝜋2/𝑠.(2.1)
Then simple computations lead to lim𝑠+𝑓𝑓(𝑠)=1,(2.2)2(𝑠)=𝑠2𝜋log22𝜋2/𝑠126log(𝜋/2)𝑠2𝜋log22𝜋21=6log(𝜋/2)24log(𝜋/2)𝜋23𝜋2𝑠2(2.3) for 𝑠1.
Computational and numerical experiments show that 𝜋24log2𝜋2=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 0𝑢1, 𝑠1 and 𝑓𝑢,𝑠𝑠(𝑥)=2log1𝑢𝑥2log𝑥+log(arcsin𝑥).(2.5)
Then inequality 𝑓𝑢,𝑠(𝑥)>0 holds for all 𝑥(0,1) if and only if 3𝑠𝑢1, and inequality 𝑓𝑢,𝑠(𝑥)<0 holds for all 𝑥(0,1) if and only if 𝑢+(2/𝜋)2/𝑠1.

Proof. If 𝑢=0, then we clearly see that 3𝑠𝑢1, 𝑢+(2/𝜋)2/𝑠<1 and 𝑓0,𝑠(𝑥)=log[(arcsin𝑥)/𝑥]>0 for all 𝑠1 and 𝑥(0,1). In the following discussion, we assume that 0<𝑢1.
From (2.5) and simple computations we have lim𝑥0+𝑓𝑢,𝑠𝑓(𝑥)=0,(2.6)𝑢,𝑠1(𝑥)=1𝑥2arcsin𝑥1+𝑢(𝑠1)𝑥2𝑥1𝑢𝑥2=1+𝑢(𝑠1)𝑥2𝑥1𝑢𝑥2𝑔arcsin𝑥𝑢,𝑠(𝑥),(2.7) where 𝑔𝑢,𝑠𝑥(𝑥)=1𝑢𝑥21𝑥21+𝑢(𝑠1)𝑥2𝑔arcsin𝑥,(2.8)𝑢,𝑠𝑔(0)=0,(2.9)𝑢,𝑠𝑥(𝑥)=21𝑥23/21+𝑢(𝑠1)𝑥22𝑢,𝑠(𝑥),(2.10) where 𝑢,𝑠(𝑥)=𝑢2(𝑠1)2𝑥4+𝑢𝑠2𝑥𝑢+𝑢𝑠+4𝑠22+13𝑠𝑢,(2.11)𝑢,𝑠(0)=13𝑠𝑢,(2.12)𝑢,𝑠(1)=𝑢𝑠(1𝑢)+(1𝑢)2.(2.13)
We divide the proof into four cases.
Case 1 (3𝑠𝑢1). Then from (2.11) and (2.12) together with the fact that 𝑢𝑠2+𝑢𝑠+4𝑠2=2(𝑠1)+𝑠(𝑢+2𝑠𝑢+1)+𝑠(13𝑠𝑢)>0,(2.14) we clearly see that 𝑢,𝑠(0)0,(2.15) and 𝑢,𝑠(𝑥) is strictly increasing in [0,1].
Equation (2.12) and the monotonicity of 𝑢,𝑠(𝑥) imply that 𝑢,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 𝑢,𝑠(𝑥) imply that there exists 0<𝜆<1 such that 𝑢,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/𝑠1). Then Lemma 2.1 and (2.12) lead to 𝑢,𝑠2(0)=13𝑠𝑢13𝑠1𝜋2/𝑠<0.(2.19)
We divide the proof into two subcases.
Subcase 3.1 (𝑢=1). Then (2.13) becomes 𝑢,𝑠(1)=0.(2.20)
Let 𝑡=𝑥2, then from (2.11) we clearly see that the function 𝑢,𝑠 is a quadratic function of variable 𝑡. It follows from inequality (2.19) and (2.20) that 𝑢,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,𝑠𝜋(1)=log2(1𝑢)𝑠/20,(2.22)lim𝑥1𝑔𝑢,𝑠(𝑥)=+,(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 𝑢,𝑠(𝑥)<0 for 𝑥[0,𝜆1) and 𝑢,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,𝑠𝜋(1)=log2(1𝑢)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 𝑠1, then the double inequality 𝑄𝑡1,𝑠(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝑄𝑡2,𝑠(𝑎,𝑏)(2.26) holds for all 𝑎,𝑏>0 with 𝑎𝑏 if and only if 𝑡1(11(2/𝜋)2/𝑠)/2 and 𝑡2(11/3𝑠)/2.

Proof. Since both 𝑄𝑡,𝑠(𝑎,𝑏) and 𝑃(𝑎,𝑏) are symmetric and homogeneous of degree 1. Without loss of generality, we assume that 𝑎>𝑏. Let 𝑥=(𝑎𝑏)/(𝑎+𝑏)(0,1). Then from (1.2) and (1.4) we have 𝑄log𝑡,𝑠(𝑎,𝑏)𝑄𝑃(𝑎,𝑏)=log𝑡,𝑠(𝑎,𝑏)𝐴(𝑎,𝑏)log𝑃(𝑎,𝑏)=𝑠𝐴(𝑎,𝑏)2log1(12𝑡)2𝑥2log𝑥+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.