Optimal Two Parameter Bounds for the Seiffert Mean

Abstract

We obtain sharp bounds for the Seiffert mean in terms of a two parameter family of means. Our results generalize and extend the recent bounds presented in the Journal of Inequalities and Applications (2012) and Abstract and Applied Analysis (2012).

1. Introduction

For 𝑎,𝑏>0 with 𝑎≠𝑏, the Seiffert mean 𝑇(𝑎,𝑏), root mean square 𝑆(𝑎,𝑏), and contraharmonic mean 𝐶(𝑎,𝑏) are defined by 𝑇(𝑎,𝑏)=𝑎−𝑏)/(,𝑆()=2arctan[(𝑎−𝑏𝑎+𝑏)]𝑎,𝑏𝑎²+𝑏²2,𝐶()=𝑎𝑎,𝑏²+𝑏²,𝑎+𝑏 (1)(2)(3) respectively. It is well known that the inequalities 𝑇(𝑎,𝑏)<𝑆(𝑎,𝑏)<𝐶(𝑎,𝑏) (4) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Recently, 𝑇(𝑎,𝑏), 𝑆(𝑎,𝑏), and 𝐶(𝑎,𝑏) have been the subject of intensive research. In particular, many remarkable inequalities and properties for these means can be found in the literature [1–8].

For 𝛼,𝛽,𝜆,𝜇∈(1/2,1), very recently Chu et al. [9, 10] proved that the inequalities 𝑆(()()𝑎)()𝑎,𝐶(()()𝑎)()𝑎𝛼𝑎+1−𝛼𝑏,𝛼𝑏+1−𝛼<𝑇𝑎,𝑏<𝑆𝛽𝑎+1−𝛽𝑏,𝛽𝑏+1−𝛽𝜆𝑎+1−𝜆𝑏,𝜆𝑏+1−𝜆<𝑇𝑎,𝑏<𝐶𝜇𝑎+1−𝜇𝑏,𝜇𝑏+1−𝜇 (5)(6) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤(1+16/𝜋²−1)/2, 𝛽≥(3+√6)/6, 𝜆≤(1+4/𝜋−1)/2, and 𝜇≥(3+√3)/6.

Let 𝑡∈(1/2,1), 𝑝≥1/2, and 𝑄_𝑡,𝑝(𝑎,𝑏)=𝐶^𝑝(𝑡𝑎+(1−𝑡)𝑏,𝑡𝑏+(1−𝑡)𝑎)𝐴^1−𝑝(𝑎,𝑏), (7) where 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2 is the classical arithmetic mean of 𝑎 and 𝑏. Then from (2), (3), and (7) we clearly see that 𝑄_𝑡,1/2()(()()𝑎),𝑄𝑎,𝑏=𝑆𝑡𝑎+1−𝑡𝑏,𝑡𝑏+1−𝑡_𝑡,1(𝑎,𝑏)=𝐶(𝑡𝑎+(1−𝑡)𝑏,𝑡𝑏+(1−𝑡)𝑎), (8) and 𝑄_𝑡,𝑝(𝑎,𝑏) is strictly increasing with respect to 𝑡∈(1/2,1) for fixed 𝑎,𝑏>0 with 𝑎≠𝑏.

It is natural to ask what are the greatest value 𝑡₁=𝑡₁(𝑝) and the least value 𝑡₂=𝑡₂(𝑝) in (1/2,1) such that the double inequality 𝑄_𝑡1,𝑝(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝑄_𝑡2,𝑝(𝑎,𝑏) (9) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 and 𝑝≥1/2. The aim of this paper is to answer this question; our main result is the following Theorem 1.

Theorem 1. If 𝑡₁,𝑡₂∈(1/2,1) and 𝑝∈[1/2,∞), then the double inequality 𝑄_𝑡1,𝑝(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝑄_𝑡2,𝑝(𝑎,𝑏) (10) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝑡₁≤1/2+[(4/𝜋)^1/𝑝−1]/2 and 𝑡₂≥1/2+3𝑝/(6𝑝).

Remark 2. If we take 𝑝=1/2 and 𝑝=1 in Theorem 1, then inequality (10) reduces to inequalities (5) and (6), respectively.

2. Proof of Theorem 1

In order to prove Theorem 1 we need two lemmas, which we present in this section.

Lemma 3 (see [11, Theorem  1.25]). For −∞<𝑎<𝑏<∞, let 𝑓,𝑔∶[𝑎,𝑏]→ℝ be continuous on [𝑎,𝑏] and differentiable on (𝑎,𝑏); let 𝑔^′(𝑥)≠0 on (𝑎,𝑏). If 𝑓^′(𝑥)/𝑔^′(𝑥) is increasing (decreasing) on (𝑎,𝑏), then so are 𝑓(𝑥)−𝑓(𝑎)𝑔(𝑥)(𝑎),𝑓(𝑥)(𝑏)−𝑔−𝑓𝑔(𝑥)(𝑏).−𝑔 (11) If 𝑓^′(𝑥)/𝑔^′(𝑥) is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 4. Let 𝑢∈[0,1], 𝑝≥1/2 and 𝑓_𝑢,𝑝(𝑥)=𝑝log1+𝑢𝑥²−log𝑥+logarctan𝑥. (12) Then(1)𝑓_𝑢,𝑝(𝑥)>0 for 𝑥∈(0,1) if and only if 3𝑝𝑢≥1;(2)𝑓_𝑢,𝑝(𝑥)<0 for 𝑥∈(0,1) if and only if 1+𝑢≤(4/𝜋)^1/𝑝.

Proof. By (12) and simple computations one has lim_𝑥→0𝑓_𝑢,𝑝(𝑥)𝑓=0,_𝑢,𝑝(1)()𝜋=𝑝log1+𝑢+log4𝑓_′𝑢,𝑝,(𝑥)=2𝑝𝑢𝑥1+𝑢𝑥²+11+𝑥²−1arctan𝑥𝑥=𝑢2𝑝−1𝑥²1+𝑥²arctan𝑥+𝑥³−1+𝑥²×arctan𝑥−𝑥𝑥(1+𝑥²)(1+𝑢𝑥²)arctan𝑥⁻¹=2𝑝−1𝑥²1+𝑥²arctan𝑥+𝑥³𝑥1+𝑥²1+𝑢𝑥²(𝑥,arctan𝑥𝑢−𝑔) (13)(14)(15) where 𝑔(𝑥)=1+𝑥²arctan𝑥−𝑥2𝑝−1𝑥²1+𝑥²arctan𝑥+𝑥³. (16)
Let 𝑔₁(𝑥)=arctan𝑥−𝑥/(1+𝑥²) and 𝑔₂(𝑥)=(2𝑝−1)𝑥²arctan𝑥+𝑥³/(1+𝑥²). Then 𝑔₂(𝑥) is strictly increasing in (0,1): 𝑔(𝑥)=𝑔₁(𝑥)𝑔₂(𝑥),𝑔₁(0)=𝑔₂(0)𝑔=0,_′1(𝑥)𝑔^′₂(𝑥)=12𝑝−11+𝑥²²arctan𝑥/𝑥+𝑝𝑥².+𝑝+1 (17)(18)
It is not difficult to verify that the function 𝑥→[(1+𝑥²)²arctan𝑥]/𝑥 is strictly increasing from (0,1) onto (1,𝜋); hence (18) implies that 𝑔^′₁(𝑥)/𝑔^′₂(𝑥) is strictly decreasing in (0,1). Therefore, 𝑔(𝑥) is strictly decreasing in (0,1) follows from Lemma 3 and (17) together with the monotonicity of 𝑔₂(𝑥) and 𝑔^′₁(𝑥)/𝑔^′₂(𝑥). Moreover, making use of l'Hôpital's rule we get lim_𝑥→0𝑔(𝑥)=1,𝑔(1)=3𝑝𝜋−2.2𝑝−1𝜋+2 (19)(20)
We divide the proof into three cases.
Case  1 (𝑢≥1/(3𝑝)). Then from (15) and (19) together with the monotonicity of 𝑔(𝑥) we conclude that 𝑓_𝑢,𝑝(𝑥) is strictly increasing in (0,1). Therefore 𝑓_𝑢,𝑝(𝑥)>0 for 𝑥∈(0,1) follows from (13) and the monotonicity of 𝑓_𝑢,𝑝(𝑥).
Case  2 (𝑢≤(𝜋−2)/[(2𝑝−1)𝜋+2]). Then from (15) and (20) together with the monotonicity of 𝑔(𝑥) we clearly see that 𝑓_𝑢,𝑝(𝑥) is strictly decreasing in (0,1). Therefore 𝑓_𝑢,𝑝(𝑥)<0 for 𝑥∈(0,1) follows from (13) and the monotonicity of 𝑓_𝑢,𝑝(𝑥).
Case  3 ((𝜋−2)/[(2𝑝−1)𝜋+2]<𝑢<1/(3𝑝)). Then from (15), (19), and (20) together with the monotonicity of 𝑔(𝑥) we know that there exists 𝑥₀∈(0,1) such that 𝑓_𝑢,𝑝(𝑥) is strictly decreasing in (0,𝑥₀) and strictly increasing in (𝑥₀,1). Therefore, 𝑓_𝑢,𝑝(𝑥)<0 in (0,1) if and only if 𝑓_𝑢,𝑝(1)≤0 follows from (13) and the piecewise monotonicity of 𝑓_𝑢,𝑝(𝑥), which (14) gives immediately 1+𝑢≤(4/𝜋)^1/𝑝.

Proof of Theorem 1. Since both 𝑄_𝑡,𝑝(𝑎,𝑏) and 𝑇(𝑎,𝑏) are symmetric and homogeneous of degree 1, without loss of generality, we assume that 𝑎>𝑏. Let 𝑥=(𝑎−𝑏)/(𝑎+𝑏)∈(0,1). Then from (1) and (7) we get 𝑄log^𝑡,𝑝(𝑎,𝑏)𝑇()𝑄𝑎,𝑏=log^𝑡,𝑝(𝑎,𝑏)𝐴()𝑇()𝑎,𝑏−log𝑎,𝑏𝐴()()𝑎,𝑏=𝑝log1+1−2𝑡²𝑥²−log𝑥+logarctan𝑥. (21)
Therefore, Theorem 1 follows from Lemma 4 and (21).

Acknowledgments

This work was supported by the Natural Science Foundation of China (Grant nos. 11071069 and 11171307) and the Natural Science Foundation of Zhejiang Province (Grant no. LY13A010004).