Sharp Bounds for Seiffert Mean in Terms of Contraharmonic Mean

Abstract

We find the greatest value 𝛼 and the least value 𝛽 in (1/2,1) such that the double inequality 𝐶(𝛼𝑎+(1−𝛼)𝑏,𝛼𝑏+(1−𝛼)𝑎)<𝑇(𝑎,𝑏)<𝐶(𝛽𝑎+(1−𝛽)𝑏,𝛽𝑏+(1−𝛽)𝑎) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏. Here, 𝑇(𝑎,𝑏)=(𝑎−𝑏)/[2 arctan((𝑎−𝑏)/(𝑎+𝑏))] and 𝐶(𝑎,𝑏)=(𝑎²+𝑏²)/(𝑎+𝑏) are the Seiffert and contraharmonic means of 𝑎 and 𝑏, respectively.

1. Introduction

For 𝑎,𝑏>0 with 𝑎≠𝑏, the Seiffert mean 𝑇(𝑎,𝑏) and contraharmonic mean 𝐶(𝑎,𝑏) are defined by 𝑇(𝑎,𝑏)=𝑎−𝑏)/(,𝐶()=𝑎2arctan((𝑎−𝑏𝑎+𝑏))𝑎,𝑏²+𝑏²,𝑎+𝑏 (1.1)(1.2)

respectively. Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities and properties for these means can be found in the literature [1–12].

Let 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝐺(𝑎,𝑏)=√𝑎𝑏, 𝑆(𝑎,𝑏)=(𝑎²+𝑏²)/2, and let 𝑀_𝑝(𝑎,𝑏)=((𝑎^𝑝+𝑏^𝑝)/2)^1/𝑝(𝑝≠0) and 𝑀₀(𝑎,𝑏)=√𝑎𝑏 be the arithmetic, geometric, square root, and 𝑝th power means of two positive numbers 𝑎 and 𝑏, respectively. Then it is well known that 𝑀_𝑝(𝑎,𝑏) is continuous and strictly increasing with respect to 𝑝∈ℝ for fixed 𝑎,𝑏>0 with 𝑎≠𝑏, and the inequalities 𝐺(𝑎,𝑏)=𝑀₀(𝑎,𝑏)<𝐴(𝑎,𝑏)=𝑀₁(𝑎,𝑏)<𝑆(𝑎,𝑏)=𝑀₂(𝑎,𝑏)<𝐶(𝑎,𝑏) (1.3) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Seiffert [12] proved that the double inequality 𝐴(𝑎,𝑏)=𝑀₁(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝑀₂(𝑎,𝑏)=𝑆(𝑎,𝑏) (1.4) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Hästö [13] proved that the function 𝑇(1,𝑥)/𝑀_𝑝(1,𝑥) is increasing in (0,∞) if 𝑝≤1.

In [14], the authors found the greatest value 𝑝 and the least value 𝑞 such that the double inequality 𝐻_𝑝(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝐻_𝑞(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏. Here, 𝐻_𝑘(𝑎,𝑏)=((𝑎^𝑘+(𝑎𝑏)^𝑘/2+𝑏^𝑘)/3)^1/𝑘(𝑘≠0), and 𝐻₀(𝑎,𝑏)=√𝑎𝑏 is the 𝑘th power-type Heron mean of 𝑎 and 𝑏.

Wang et al. [15] answered the question: what are the best possible parameters 𝜆 and 𝜇 such that the double inequality 𝐿_𝜆(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝐿_𝜇(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝐿_𝑟(𝑎,𝑏)=(𝑎^𝑟+1+𝑏^𝑟+1)/(𝑎^𝑟+𝑏^𝑟) is the 𝑟th Lehmer mean of 𝑎 and 𝑏.

In [16, 17], the authors proved that the inequalities 𝛼₁𝑇(𝑎,𝑏)+1−𝛼₁𝐺(𝑎,𝑏)<𝐴(𝑎,𝑏)<𝛽₁𝑇(𝑎,𝑏)+1−𝛽₁𝐺(),𝛼𝑎,𝑏₂𝑆(𝑎,𝑏)+1−𝛼₂𝐴(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝛽₂𝑆(𝑎,𝑏)+1−𝛽₂𝐴(),𝑆𝑎,𝑏_𝛼3(𝑎,𝑏)𝐴^1−𝛼3(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝑆^𝛽3(𝑎,𝑏)𝐴^1−𝛽3(𝑎,𝑏) (1.5) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₁≤3/5, 𝛽₁≥𝜋/4, 𝛼₂≤(4−𝜋)/[(√2−1)𝜋], 𝛽₂≥2/3, 𝛼₃≤2/3 and 𝛽₃≥4−2log𝜋/log2.

For fixed 𝑎,𝑏>0 with 𝑎≠𝑏, let 𝑥∈[1/2,1] and 𝐽(𝑥)=𝐶(𝑥𝑎+(1−𝑥)𝑏,𝑥𝑏+(1−𝑥)𝑎). (1.6)

Then it is not difficult to verify that 𝐽(𝑥) is continuous and strictly increasing in [1/2,1]. Note that 𝐽(1/2)=𝐴(𝑎,𝑏)<𝑇(𝑎,𝑏) and 𝐽(1)=𝐶(𝑎,𝑏)>𝑇(𝑎,𝑏). Therefore, it is natural to ask what are the greatest value 𝛼 and the least value 𝛽 in (1/2,1) such that the double inequality 𝐶(𝛼𝑎+(1−𝛼)𝑏,𝛼𝑏+(1−𝛼)𝑎)<𝑇(𝑎,𝑏)<𝐶𝛽𝑎+1−𝛽𝑏,𝛽𝑏+1−𝛽𝑎 (1.7) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏. The main purpose of this paper is to answer this question. Our main result is the following Theorem 1.1.

Theorem 1.1. If 𝛼,𝛽∈(1/2,1), then the double inequality 𝐶(𝛼𝑎+(1−𝛼)𝑏,𝛼𝑏+(1−𝛼)𝑎)<𝑇(𝑎,𝑏)<𝐶𝛽𝑎+1−𝛽𝑏,𝛽𝑏+1−𝛽𝑎 (1.8) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤(1+4/𝜋−1)/2 and 𝛽≥(3+√3)/6.

2. Proof of Theorem 1.1

Proof of Theorem 1.1. Let 𝜆=(1+4/𝜋−1)/2 and 𝜇=(3+√3)/6. We first proof that the inequalities 𝑇()(()()𝑎),𝑇()𝑎𝑎,𝑏>𝐶𝜆𝑎+1−𝜆𝑏,𝜆𝑏+1−𝜆𝑎,𝑏<𝐶𝜇𝑎+1−𝜇𝑏,𝜇𝑏+1−𝜇 (2.1)(2.2) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.
From (1.1) and (1.2) we clearly see that both 𝑇(𝑎,𝑏) and 𝐶(𝑎,𝑏) are symmetric and homogenous of degree 1. Without loss of generality, we assume that 𝑎>𝑏. Let 𝑡=𝑎/𝑏>1 and 𝑝∈(1/2,1), then from (1.1) and (1.2) one has 𝐶𝑝𝑎+1−𝑝𝑏,𝑝𝑏+1−𝑝𝑎−𝑇(𝑎,𝑏)=𝑏𝑝𝑡+1−𝑝²+1−𝑝𝑡+𝑝²2())/(×𝑡+1arctan((𝑡−1𝑡+1))2arctan𝑡−1−𝑡𝑡+1²−1𝑝𝑡+1−𝑝²+1−𝑝𝑡+𝑝². (2.3) Let 𝑓(𝑡)=2arctan𝑡−1−𝑡𝑡+1²−1𝑝𝑡+1−𝑝²+1−𝑝𝑡+𝑝². (2.4) Then simple computations lead to 𝑓(1)=0,lim_𝑡→+∞𝑓(𝑡)=𝜋2−1𝑝²+1−𝑝²𝑓_′,(𝑡)=2𝑓₁(𝑡)𝑝𝑡+1−𝑝²+1−𝑝𝑡+𝑝²²𝑡²,+1 (2.5)(2.6)(2.7) where 𝑓₁(𝑡)=4𝑝⁴−8𝑝³+10𝑝²−6𝑝+1𝑡⁴−22𝑝−1²2𝑝²−2𝑝+1𝑡³+212𝑝⁴−24𝑝³+18𝑝²−6𝑝+1𝑡²−22𝑝−1²2𝑝²−2𝑝+1𝑡+4𝑝⁴−8𝑝³+10𝑝²𝑓−6𝑝+1,₁(1)=0. (2.8)(2.9)
Let 𝑓₂(𝑡)=𝑓^′₁(𝑡)/2, 𝑓₃(𝑡)=𝑓^′₂(𝑡)/2, 𝑓₄(𝑡)=𝑓^′₃(𝑡)/3. Then from (2.8) we get 𝑓₂(𝑡)=24𝑝⁴−8𝑝³+10𝑝²−6𝑝+1𝑡³−32𝑝−1²2𝑝²−2𝑝+1𝑡²+212𝑝⁴−24𝑝³+18𝑝²−6𝑝+1𝑡−2𝑝−1²2𝑝²−2𝑝+1, (2.10)𝑓₂(1)=0, (2.11)𝑓₃(𝑡)=34𝑝⁴−8𝑝³+10𝑝²−6𝑝+1𝑡²−32𝑝−1²2𝑝²−2𝑝+1𝑡+12𝑝⁴−24𝑝³+18𝑝²−6𝑝+1, (2.12)𝑓₃(1)=6𝑝²−6𝑝+1, (2.13)𝑓₄(𝑡)=24𝑝⁴−8𝑝³+10𝑝²−6𝑝+1𝑡−2𝑝−1²2𝑝²−2𝑝+1, (2.14)𝑓₄(1)=6𝑝²−6𝑝+1. (2.15)
We divide the proof into two cases.
Case 1 (𝑝=𝜆=(1+4/𝜋−1)/2). Then (2.6), (2.13), and (2.15) lead to lim_𝑡→+∞𝑓(𝑡)𝑓=0,₃(1)2()=−𝜋−3𝜋𝑓<0,₄(1)2()=−𝜋−3𝜋<0. (2.16)(2.17)(2.18)
Note that 4𝑝⁴−8𝑝³+10𝑝²−6𝑝+1=4+2𝜋−𝜋²𝜋²>0. (2.19)
It follows from (2.8), (2.10), (2.12), (2.14), and (2.19) that lim_𝑡→+∞𝑓₁(𝑡)=+∞,lim_𝑡→+∞𝑓₂(𝑡)=+∞,lim_𝑡→+∞𝑓₃(𝑡)=+∞,lim_𝑡→+∞𝑓₄(𝑡)=+∞. (2.20)(2.21)(2.22)(2.23)
From (2.14) and inequality (2.19), we clearly see that 𝑓₄(𝑡) is strictly increasing in [1,+∞). Then (2.18) and (2.23) lead to the conclusion that there exists 𝑡₀>1 such that 𝑓₄(𝑡)<0 for 𝑡∈[1,𝑡₀) and 𝑓₄(𝑡)>0 for 𝑡∈(𝑡₀,+∞). Hence, 𝑓₃(𝑡) is strictly decreasing in [1,𝑡₀] and strictly increasing in [𝑡₀,+∞).
It follows from (2.17) and (2.22) together with the piecewise monotonicity of 𝑓₃(𝑡) that there exists 𝑡₁>𝑡₀>1 such that 𝑓₂(𝑡) is strictly decreasing in [1,𝑡₁] and strictly increasing in [𝑡₁,+∞).
From (2.11) and (2.21) together with the piecewise monotonicity of 𝑓₂(𝑡), we conclude that there exists 𝑡₂>𝑡₁>1 such that 𝑓₁(𝑡) is strictly decreasing in [1,𝑡₂] and strictly increasing in [𝑡₂,+∞).
Equations (2.7), (2.9), and (2.20) together with the piecewise monotonicity of 𝑓₁(𝑡) imply that there exists 𝑡₃>𝑡₂>1 such that 𝑓(𝑡) is strictly decreasing in [1,𝑡₃] and strictly increasing in [𝑡₃,+∞).
Therefore, inequality (2.1) follows from (2.3)–(2.5) and (2.16) together with the piecewise monotonicity of 𝑓(𝑡).Case 2 (𝑝=𝜇=(3+√3)/6). Then (2.8) leads to 𝑓₁(𝑡)=(𝑡−1)⁴9>0 (2.24) for 𝑡>1.
Inequality (2.24) and (2.7) imply that 𝑓(𝑡) is strictly increasing in [1,+∞). Therefore, inequality (2.2) follows from (2.3)–(2.5) together with the monotonicity of 𝑓(𝑡).
From inequalities (2.1) and (2.2) together with the monotonicity of 𝐽(𝑥)=𝐶(𝑥𝑎+(1−𝑥)𝑏,𝑥𝑏+(1−𝑥)𝑎) in [1/2,1], we know that inequality (1.8) holds for all 𝛼≤(1+4/𝜋−1)/2, 𝛽≥(3+√3)/6, and all 𝑎,𝑏>0 with 𝑎≠𝑏.
Next, we prove that 𝜆=(1+4/𝜋−1)/2 is the best possible parameter in [1/2,1] such that inequality (2.1) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.
For any 1>𝑝>𝜆=(1+4/𝜋−1)/2, from (2.6) one has lim_𝑡→+∞𝑓(𝑡)=𝜋2−1𝑝²+1−𝑝²>0. (2.25)
Equations (2.3) and (2.4) together with inequality (2.25) imply that for any 1>𝑝>𝜆=(1+4/𝜋−1)/2 there exists 𝑇₀=𝑇₀(𝑝)>1 such that 𝐶𝑝𝑎+1−𝑝𝑏,𝑝𝑏+1−𝑝𝑎>𝑇(𝑎,𝑏) (2.26) for 𝑎/𝑏∈(𝑇₀,+∞).
Finally, we prove that 𝜇=(3+√3)/6 is the best possible parameter such that inequality (2.2) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.
For any 1/2<𝑝<𝜇=(3+√3)/6, from (2.13) one has 𝑓₃(1)=6𝑝²−6𝑝+1<0. (2.27)
From inequality (2.27) and the continuity of 𝑓₃(𝑡), we know that there exists 𝛿=𝛿(𝑝)>0 such that 𝑓₃(𝑡)<0 (2.28) for 𝑡∈(1,1+𝛿).
Equations (2.3)–(2.5), (2.7), (2.9), and (2.11) together with inequality (2.28) imply that for any 1/2<𝑝<𝜇=(3+√3)/6 there exists 𝛿=𝛿(𝑝)>0 such that 𝑇(𝑎,𝑏)>𝐶𝑝𝑎+1−𝑝𝑏,𝑝𝑏+1−𝑝𝑎 (2.29) for 𝑎/𝑏∈(1,1+𝛿).

Acknowledgments

This research was supported by the National Natural Science Foundation of China under Grants 11071069 and 11171307, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.