An Optimal Double Inequality between Seiffert and Geometric Means

Abstract

For 𝛼,𝛽∈(0,1/2) we prove that the double inequality 𝐺(𝛼𝑎+(1−𝛼)𝑏,𝛼𝑏+(1−𝛼)𝑎)<𝑃(𝑎,𝑏)<𝐺(𝛽𝑎+(1−𝛽)𝑏,𝛽𝑏+(1−𝛽)𝑎) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤(1−√1−4/𝜋²)/2 and 𝛽≥(3−√3)/6. Here, 𝐺(𝑎,𝑏) and 𝑃(𝑎,𝑏) denote the geometric and Seiffert means of two positive numbers 𝑎 and 𝑏, respectively.

1. Introduction

For 𝑎,𝑏>0 with 𝑎≠𝑏 the Seiffert mean 𝑃(𝑎,𝑏) was introduced by Seiffert [1] as follows:𝑃(𝑎,𝑏)=𝑎−𝑏4arctan√𝑎/𝑏−𝜋.(1.1)

Recently, the bivariate mean values have been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert mean can be found in the literature [1–9].

Let 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐺(𝑎,𝑏)=√𝑎𝑏, 𝐿(𝑎,𝑏)=(𝑎−𝑏)/(log𝑎−log𝑏), 𝐼(𝑎,𝑏)=1/𝑒(𝑏^𝑏/𝑎^𝑎)^{1/(𝑏−𝑎)}, 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝐶(𝑎,𝑏)=(𝑎²+𝑏²)/(𝑎+𝑏), and 𝑀_𝑝(𝑎,𝑏)=[(𝑎^𝑝+𝑏^𝑝)/2]^1/𝑝(𝑝≠0) and 𝑀₀(𝑎,𝑏)=√𝑎𝑏 be the harmonic, geometric, logarithmic, identric, arithmetic, contraharmonic, and 𝑝th power means of two different positive numbers 𝑎 and 𝑏, respectively. Then it is well known that min{𝑎,𝑏}<𝐻(𝑎,𝑏)=<𝑀𝐼(𝑎₋₁,𝑏(𝑎),<𝑏)𝐴<(𝑎𝐺,𝑏(𝑎),=𝑏)𝑀=₁𝑀₀(𝑎,𝑏)<𝐿(𝑎,𝑏)(𝑎,𝑏)<𝐶(𝑎,𝑏)<max{𝑎,𝑏}(1.2) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

For all 𝑎,𝑏>0 with 𝑎≠𝑏, Seiffert [1] established that 𝐿(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝐼(𝑎,𝑏); Jagers [4] proved that 𝑀_1/2(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝑀_2/3(𝑎,𝑏) and 𝑀_2/3(𝑎,𝑏) is the best possible upper power mean bound for the Seiffert mean 𝑃(𝑎,𝑏); Seiffert [7] established that 𝑃(𝑎,𝑏)>𝐴(𝑎,𝑏)𝐺(𝑎,𝑏)/𝐿(𝑎,𝑏) and 𝑃(𝑎,𝑏)>2𝐴(𝑎,𝑏)/𝜋; Sándor [6] presented that (𝐴(𝑎,𝑏)+𝐺(𝑎,𝑏))/2<𝑃(𝑎,𝑏)<√𝐴(𝑎,𝑏)(𝐴(𝑎,𝑏)+𝐺(𝑎,𝑏))/2 and √³𝐴²(𝑎,𝑏)𝐺(𝑎,𝑏)<𝑃(𝑎,𝑏)<(𝐺(𝑎,𝑏)+2𝐴(𝑎,𝑏))/3; Hästö [3] proved that 𝑃(𝑎,𝑏)>𝑀_log2/log𝜋(𝑎,𝑏) and 𝑀_log2/log𝜋(𝑎,𝑏) is the best possible lower power mean bound for the Seiffert mean 𝑃(𝑎,𝑏).

Very recently, Wang and Chu [8] found the greatest value 𝛼 and the least value 𝛽 such that the double inequality 𝐴^𝛼(𝑎,𝑏)𝐻^1−𝛼(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝐴^𝛽(𝑎,𝑏)𝐻^1−𝛽(𝑎,𝑏) holds for 𝑎,𝑏>0 with 𝑎≠𝑏; For any 𝛼∈(0,1), Chu et al. [10] presented the best possible bounds for 𝑃^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏) in terms of the power mean; In [2] the authors proved that the double inequality 𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝐻(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝛽𝐴(𝑎,𝑏)+(1−𝛽)𝐻(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤2/𝜋 and 𝛽≥5/6; Liu and Meng [5] proved that the inequalities 𝛼𝛼₁𝐶(𝑎,𝑏)+1−𝛼₁𝐺(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝛽₁𝐶(𝑎,𝑏)+1−𝛽₁𝐺(𝑎,𝑏),₂𝐶(𝑎,𝑏)+1−𝛼₂𝐻(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝛽₂𝐶(𝑎,𝑏)+1−𝛽₂𝐻(𝑎,𝑏)(1.3) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₁≤2/9, 𝛽₁≥1/𝜋, 𝛼₂≤1/𝜋 and 𝛽₂≥5/12.

For fixed 𝑎,𝑏>0 with 𝑎≠𝑏 and 𝑥∈[0,1/2], let 𝑔(𝑥)=𝐺(𝑥𝑎+(1−𝑥)𝑏,𝑥𝑏+(1−𝑥)𝑎).(1.4)

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

Theorem 1.1. If 𝛼,𝛽∈(0,1/2), then the double inequality 𝐺(𝛼𝑎+(1−𝛼)𝑏,𝛼𝑏+(1−𝛼)𝑎)<𝑃(𝑎,𝑏)<𝐺(𝛽𝑎+(1−𝛽)𝑏,𝛽𝑏+(1−𝛽)𝑎)(1.5) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤(1−√1−4/𝜋²)/2 and 𝛽≥(3−√3)/6.

2. Proof of Theorem 1.1

Proof of Theorem 1.1. Let 𝜆=(1−√1−4/𝜋²)/2 and 𝜇=(3−√3)/6. We first prove that inequalities 𝑃𝑃((𝑎𝑎,,𝑏𝑏))<>𝐺𝐺((𝜇𝜆𝑎𝑎++((11−−𝜆𝜇))𝑏𝑏,,𝜆𝜇𝑏𝑏++((11−−𝜆𝜇))𝑎𝑎)),((22..12)) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.
Without loss of generality, we assume that 𝑎>𝑏. Let 𝑡=√𝑎/𝑏>1 and 𝑝∈(0,1/2), then from (1.1) one has log𝐺=(𝑝1𝑎+(1−𝑝)𝑏,𝑝𝑏+(1−𝑝)𝑎)−log𝑃(𝑎,𝑏)2log𝑝𝑡²+(1−𝑝)(1−𝑝)𝑡²+𝑝−log𝑡²−14arctan𝑡−𝜋.(2.3) Let 𝑓(𝑡)=12log𝑝𝑡²+(1−𝑝)(1−𝑝)𝑡²+𝑝−log𝑡²−14arctan𝑡−𝜋,(2.4) then simple computations lead to lim_𝑡→+∞𝑓(1)=0,(2.5)𝑓(𝑡)=1𝑓2log[𝑝(1−𝑝)]+log𝜋,(2.6)_(𝑡)=𝑡𝑡²+1𝑡²−1(4arctan𝑡−𝜋)𝑝𝑡²+(1−𝑝)(1−𝑝)𝑡²+𝑝𝑓¹(𝑡),(2.7) where 𝑓₁(𝑡)=4𝑡²−1𝑝𝑡²+1−𝑝(1−𝑝)𝑡²+𝑝𝑡𝑡²+1²𝑓−4arctan𝑡+𝜋.(2.8)₁lim_𝑡→+∞(1)=0,(2.9)𝑓𝑓₁(𝑡)=+∞,(2.10)_1(𝑡)=4𝑓²𝑡²𝑡²𝑡²+1⁴,(2.11) where 𝑓₂(𝑡)=𝑝(1−𝑝)𝑡⁵−(3𝑝−2)(3𝑝−1)𝑡⁴+2(5𝑝²−5𝑝+1)𝑡³+2(5𝑝²−5𝑝+1)𝑡²−(3𝑝−2)(3𝑝−1)𝑡+𝑝(1−𝑝).
Note that lim_𝑡→+∞𝑓₂(1)=0,(2.12)𝑓𝑓₂(𝑡)=+∞,(2.13)_2(𝑡)=5𝑝(1−𝑝)𝑡⁴−4(3𝑝−2)(3𝑝−1)𝑡³+65𝑝²−5𝑝+1𝑡²+45𝑝²−5𝑝+1𝑡𝑓−(3𝑝−2)(3𝑝−1),(2.14)_2lim_𝑡→+∞(1)=0,(2.15)𝑓𝑓^2(𝑡)=+∞,(2.16)_2(𝑡)=20𝑝(1−𝑝)𝑡³−12(3𝑝−𝑓2)(3𝑝−1)𝑡²+125𝑝²−5𝑝+1𝑡+45𝑝²−5𝑝+1,(2.17)_2(𝑡)li=m_𝑡→+∞−86𝑝²−6𝑝+1,(2.18)𝑓𝑓^2(𝑡)=+∞,(2.19)_3(𝑡)=60𝑝(1−𝑝𝑓)𝑡²−24(3𝑝−2)(3𝑝−1)𝑡+125𝑝²−5𝑝+1,(2.20)_2(1)li=m_𝑡→+∞−366𝑝²−6𝑝+1,(2.21)𝑓𝑓^2(𝑡)=+∞,(2.22)₂₍₄₎(𝑡)=𝑓120𝑝(1−𝑝)𝑡−24(3𝑝−2)(3𝑝−1),(2.23)₂₍₄₎(1)li=m_𝑡→+∞−487𝑝²−7𝑝+1,(2.24)𝑓²⁽⁴⁾(𝑡)=+∞.(2.25)
We divide the proof into two cases.
Case 1 (𝑝=𝜆=(1−√1−4/𝜋²)/2). Then (2.6), (2.18), (2.21), and (2.24) become 𝑓lim_𝑡→+∞𝑓(𝑡)=0,(2.26)_2(1)=−8𝜋²−6𝜋²𝑓<0,(2.27)_2(1)=−36𝜋²−6𝜋²𝑓<0,(2.28)₂₍₄₎(1)=−48𝜋²−7𝜋²<0.(2.29)
From (2.23) we clearly see that 𝑓²⁽⁴⁾(𝑡) is strictly increasing in [1,+∞), then (2.25) and inequality (2.29) lead to the conclusion that there exists 𝜆₁>1 such that 𝑓²⁽⁴⁾(𝑡)<0 for 𝑡∈[1,𝜆₁) and 𝑓²⁽⁴⁾(𝑡)>0 for 𝑡∈(𝜆₁,+∞). Thus, 𝑓^2(𝑡) is strictly decreasing in [1,𝜆₁] and strictly increasing in [𝜆₁,+∞).
It follows from (2.22) and inequality (2.28) together with the piecewise monotonicity of 𝑓^2(𝑡) that there exists 𝜆₂>𝜆₁>1 such that 𝑓^2(𝑡) is strictly decreasing in [1,𝜆₂] and strictly increasing in [𝜆₂,+∞). Then (2.19) and inequality (2.27) lead to the conclusion that there exists 𝜆₃>𝜆₂>1 such that 𝑓^2(𝑡) is strictly decreasing in [1,𝜆₃] and strictly increasing in [𝜆₃,+∞).
From (2.15) and (2.16) together with the piecewise monotonicity of 𝑓^2(𝑡) we know that there exists 𝜆₄>𝜆₃>1 such that 𝑓₂(𝑡) is strictly decreasing in [1,𝜆₄] and strictly increasing in [𝜆₄,+∞). Then (2.11)–(2.13) lead to the conclusion that there exists 𝜆₅>𝜆₄>1 such that 𝑓₁(𝑡) is strictly decreasing in [1,√𝜆₅] and strictly increasing in [√𝜆₅,+∞).
It follows from (2.7)–(2.10) and the piecewise monotonicity of 𝑓₁(𝑡) that there exists 𝜆₆>√𝜆₅>1 such that 𝑓(𝑡) is strictly decreasing in [1,𝜆₆] and strictly increasing in [𝜆₆,+∞).
Therefore, inequality (2.1) follows from (2.3)–(2.5) and the piecewise monotonicity of 𝑓(𝑡).
Case 2 (𝑝=𝜇=(3−√3)/6). Then (2.18), (2.21) and (2.24) become 𝑓𝑓^2(1)=0,(2.30)_2𝑓(1)=0,(2.31)₂₍₄₎(1)=8>0.(2.32)
From (2.23) we clearly see that 𝑓²⁽⁴⁾(𝑡) is strictly increasing in [1,+∞), then inequality (2.32) leads to the conclusion that 𝑓^2(𝑡) is strictly increasing in [1,+∞).
Therefore, inequality (2.2) follows from (2.3)–(2.5), (2.7)–(2.9), (2.11), (2.12), (2.15), and inequalities (2.30) and (2.31) together with the monotonicity of 𝑓^2(𝑡).
Next, we prove that 𝜆=(1−√1−4/𝜋²)/2 is the best possible parameter such that inequality (2.1) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏. In fact, if (1−√1−4/𝜋²)/2=𝜆<𝑝<1/2, then (2.6) leads to lim_𝑡→+∞𝑓(𝑡)=12log[𝑝(1−𝑝)]+log𝜋>0.(2.33)
Inequality (2.33) implies that there exists 𝑇=𝑇(𝑝)>1 such that 𝑓(𝑡)>0(2.34) for 𝑡∈(𝑇,+∞).
It follows from (2.3) and (2.4) together with inequality (2.34) that 𝑃(𝑎,𝑏)<𝐺(𝑝𝑎+(1−𝑝)𝑏,𝑝𝑏+(1−𝑝)𝑎) for 𝑎/𝑏∈(𝑇²,+∞).
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. In fact, if 0<𝑝<𝜇=(3−√3)/6, then from (2.18) we get 𝑓^2(1)<0, which implies that there exists 𝛿>0 such that 𝑓^2(𝑡)<0(2.35) for 𝑡∈[1,1+𝛿).
Therefore, 𝑃(𝑎,𝑏)>𝐺(𝑝𝑎+(1−𝑝)𝑏,𝑝𝑏+(1−𝑝)𝑎) for 𝑎/𝑏∈(1,(1+𝛿)²) follows from (2.3)–(2.5), (2.7)–(2.9), (2.11), (2.12), and (2.15) together with inequality (2.35).

Acknowledgments

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