Optimal Lower Power Mean Bound for the Convex Combination of Harmonic and Logarithmic Means

Abstract

We find the least value 𝜆∈(0,1) and the greatest value 𝑝=𝑝(𝛼) such that 𝛼𝐻(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏)>𝑀_𝑝(𝑎,𝑏) for 𝛼∈[𝜆,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝐻(𝑎,𝑏), 𝐿(𝑎,𝑏), and 𝑀_𝑝(𝑎,𝑏) are the harmonic, logarithmic, and 𝑝-th power means of two positive numbers 𝑎 and 𝑏, respectively.

1. Introduction

For 𝑝∈ℝ, the 𝑝-th power mean 𝑀_𝑝(𝑎,𝑏) and logarithmic mean 𝐿(𝑎,𝑏) of two positive numbers 𝑎 and 𝑏 are defined by𝑀_𝑝(𝑎,𝑏)=⎧⎪⎨⎪⎩𝑎^𝑝+𝑏^𝑝2^1/𝑝√,𝑝≠0,𝐿(𝑎,𝑏)=⎧⎪⎨⎪⎩𝑎𝑏𝑏,−𝑎𝑝=0,(1.1)𝑎lo,g𝑏−log𝑎,𝑎𝑎≠=𝑏𝑏,,(1.2) respectively.

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. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for 𝑀_𝑝(𝑎,𝑏) and 𝐿(𝑎,𝑏) can be found in the literature [1–17]. It might be surprising that the logarithmic mean has applications in physics, economics, and even in meteorology [18–20]. In [18], the authors study a variant of Jensen's functional equation involving 𝐿, which appears in a heat conduction problem. A representation of 𝐿 as an infinite product and an iterative algorithm for computing the logarithmic mean as the common limit of two sequences of special geometric and arithmetic means are given in [8]. In [21, 22], it is shown that 𝐿 can be expressed in terms of Gauss's hypergeometric function ₂𝐹₁. And, in [21], the authors prove that the reciprocal of the logarithmic mean is strictly totally positive, that is, every 𝑛×𝑛 determinant with elements 1/𝐿(𝑎_𝑖,𝑏_𝑖), where 0<𝑎₁<𝑎₂<⋯<𝑎_𝑛 and 0<𝑏₁<𝑏₂<⋯<𝑏_𝑛, is positive for all 𝑛≥1.

Let 𝐴(𝑎,𝑏)=1/2(𝑎+𝑏),𝐼(𝑎,𝑏)=1/𝑒(𝑏^𝑏/𝑎^𝑎)^{1/(𝑏−𝑎)}(𝑏≠𝑎),𝐼(𝑎,𝑏)=𝑎 (𝑏=𝑎), 𝐺(𝑎,𝑏)=√𝑎𝑏, and 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏) be the arithmetic, identric, geometric, and harmonic means of two positive numbers 𝑎 and 𝑏, respectively, then it is well known thatmin{𝑎,𝑏}<<𝐻𝐿((𝑎𝑎,,𝑏𝑏))<=𝐼𝑀(𝑎,₋₁𝑏()𝑎<,𝑏)𝐴<(𝑎𝐺,𝑏()𝑎=,𝑏𝑀)=₁𝑀₀(𝑎,𝑏)(𝑎,𝑏)<max{𝑎,𝑏}(1.3) for all 𝑎, 𝑏>0 with 𝑎≠𝑏.

In [23], Alzer and Janous established the following best possible inequality: 𝑀_log2/log3(𝑎,𝑏)<23𝐴(𝑎,𝑏)+13𝐺(𝑎,𝑏)<𝑀^2/3(𝑎,𝑏)(1.4) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [8, 11, 24], the authors presented bounds for 𝐿 in terms of 𝐺 and 𝐴𝐺^2/3(𝑎,𝑏)𝐴^1/3(𝑎,𝑏)<𝐿(𝑎,𝑏)<23𝐺(𝑎,𝑏)+13𝐴(𝑎,𝑏)(1.5) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The following companion of (1.3) provides inequalities for the geometric and arithmetic means of 𝐿 and 𝐼. A proof can be found in [25] 𝐺^1/2(𝑎,𝑏)𝐴^1/2(𝑎,𝑏)<𝐿^1/2(𝑎,𝑏)𝐼^1/2(𝑎,𝑏)<12𝐿(𝑎,𝑏)+12𝐼(𝑎,𝑏)<12𝐺(𝑎,𝑏)+21𝐴(𝑎,𝑏)(1.6) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The following sharp bounds for 𝐿, 𝐼, (𝐿𝐼)^1/2, and (𝐿+𝐼)/2 in terms of the power means are proved in [4, 5, 7, 9, 16, 25, 26]: 𝑀𝑀₀(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝑀_1/3(𝑎,𝑏),_2/3𝑀(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝑀_log2(𝑎,𝑏),₀(𝑎,1𝑏)<𝐿^1/2(𝑎,𝑏)𝐼^1/2(𝑎,𝑏)<𝑀_1/2(𝑎,𝑏),2𝐿(𝑎,𝑏)+12𝐼(𝑎,𝑏)<𝑀^1/2(𝑎,𝑏)(1.7) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Alzer and Qiu [27] found the sharp bound of 1/2(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏)) in terms of the power mean as follows: 𝑀_𝑐(𝑎,𝑏)<12(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))(1.8) for all 𝑎,𝑏>0 with 𝑎≠𝑏, with the best possible parameter 𝑐=log2/(1+log2).

The main purpose of this paper is to find the least value 𝜆∈(0,1) and the greatest value 𝑝=𝑝(𝛼) such that 𝛼𝐻(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏)>𝑀_𝑝(𝑎,𝑏) for 𝛼∈[𝜆,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏.

2. Lemmas

In order to establish our main result we need three lemmas, which we present in this section.

Lemma 2.1. Let 𝛼∈(1/4,1), 𝑝=(1−4𝛼)/3∈(−1,0), and 𝑓(𝑡)=−4𝛼𝑝(𝑝+1)²(𝑝+2)𝑡^𝑝−1+2(1−𝛼)𝑝²(1−𝑝²)𝑡^𝑝−2+2(1−𝛼)𝑝(1−𝑝)²(2−𝑝)𝑡^𝑝−3+12(1−𝛼)(1−𝑝). Then 𝑓(𝑡)>0 for 𝑡∈[1,+∞).

Proof. Simple computations lead to 𝑓(1)=6481(1−𝛼)²56𝛼²lim_𝑡→+∞+23𝛼+11>0,(2.1)𝑓(𝑡)=𝑓12(1−𝛼)(1−𝑝)=8(1−𝛼)(1+2𝛼)>0,(2.2)_(𝑡)=−2𝑝(1−𝑝)𝑡^𝑝−4𝑓₁(𝑡),(2.3) where 𝑓₁(𝑡)=−2𝛼(𝑝+1)²(𝑝𝑓+2)𝑡²+(1−𝛼)𝑝(𝑝+1)(2−𝑝)𝑡+(1−𝛼)(1−𝑝)(2−𝑝)(3−𝑝),₁(1)=427(1−𝛼)148𝛼²lim_𝑡→+∞−11𝛼+25>0,(2.4)𝑓′₁𝑓₁(𝑡)=−∞,(2.5)(𝑡)==−−44𝛼(𝑝+1)²(𝑝+2)𝑡+(1−𝛼)𝑝(𝑝+1)(2−𝑝)27(1−𝛼)²[16𝛼(7−4𝛼)𝑡+(4𝛼−1)(4𝛼+5)]<0(2.6) for 𝑡∈[1,+∞).
Inequality (2.6) implies that 𝑓₁(𝑡) is strictly decreasing in [1,+∞), then from (2.4) and (2.5) we know that 𝜆₁>1 exists such that 𝑓₁(𝑡)>0 for 𝑡∈[1,𝜆₁) and 𝑓₁(𝑡)<0 for 𝑡∈(𝜆₁,+∞). Hence, equation (2.3) leads to the conclusion that 𝑓(𝑡) is strictly increasing in [1,𝜆₁] and strictly decreasing in [𝜆₁,+∞).

Therefore, Lemma 2.1 follows from (2.1) and (2.2) together with the piecewise monotonicity of 𝑓(𝑡).

Lemma 2.2. Let 𝛼∈(1/4,1), 𝑝=(1−4𝛼)/3∈(−1,0), and 8𝑔(𝑝𝑡₂)=−(1−𝛼)(𝑝+1)(𝑝+2)²(𝑝+3)𝑡^𝑝+(𝑝+1)(𝑝³−𝛼𝑝³−19𝛼𝑝²+3𝑝²−34𝛼𝑝+2𝑝−8𝛼)𝑡^𝑝−1+(1−𝛼)𝑝(𝑝³−−𝑝+4)𝑡^𝑝−2+(1−𝛼)(1−𝑝)(𝑝³+5𝑝²−14𝑝+4)𝑡^𝑝−3+4(1−𝛼)(7−4𝑝)−4𝑝(1−𝛼)𝑡⁻¹+4𝛼(1+𝑝)𝑡⁻², then 𝑔(𝑡)>0 for 𝑡∈[1,+∞).

Proof. Let 8𝑔₁𝑝(₂𝑡)=−(1−𝛼)(𝑝+1)(𝑝+2)²(𝑝+3)𝑡³+(𝑝+1)(𝑝³−𝛼𝑝³−19𝛼𝑝²+3𝑝²−34𝛼𝑝+2𝑝−8𝛼)𝑡²+(1−𝛼)𝑝(𝑝³−−𝑝+4)𝑡+(1−𝛼)(1−𝑝)(𝑝³+5𝑝²−14𝑝+4)+4(1−𝛼)(7−4𝑝)𝑡^3−𝑝−4𝑝(1−𝛼)𝑡^2−𝑝+4𝛼(1+𝑝)𝑡^1−𝑝. Then simple computations lead to 𝑔𝑔(𝑡)=𝑡^𝑝−3𝑔₁(𝑡),(2.7)₁(1)=1627(1−𝛼)80𝛼²+110𝛼−1>0,(2.8)8𝑔^1𝑝(₂𝑡)=−3(1−𝛼)(𝑝+1)(𝑝+2)²(𝑝+3)𝑡²+2(𝑝+1)(𝑝³−𝛼𝑝³−19𝛼𝑝²+3𝑝²−34𝛼𝑝+2𝑝−8𝛼)𝑡+(1−𝛼)𝑝(𝑝³−−𝑝+4)+4(1−𝛼)(7−4𝑝)(3−𝑝)𝑡^2−𝑝−4𝑝(1−𝛼)(2−𝑝)𝑡^1−𝑝+4𝛼(1−𝑝²)𝑡^−𝑝, g^1(1)=3227(1−𝛼)−16𝛼³+38𝛼²+176𝛼−9>0,(2.9)4𝑔𝑝^1)((𝑡3)=−𝑝−)6(2(1−−𝑝𝛼)𝑡)(𝑝_1−𝑝+1)(𝑝+2)²(𝑝+3)𝑡+2(𝑝+1)(𝑝³−𝛼𝑝³−19𝛼𝑝²+3𝑝²−34𝛼𝑝+2𝑝−8𝛼)+4(1−𝛼)(7−−4𝑝(1−𝛼)(2−𝑝)(1−𝑝)𝑡^−𝑝−4𝛼𝑝(1−𝑝²)𝑡^−𝑝−1, g^1(1)=881(1−𝛼)−128𝛼⁴>0,+896𝛼³+288𝛼²+5294𝛼−437(2.10)𝑝𝑔)𝑡^1(𝑡)_−𝑝−1=−6(1−𝛼)(𝑝+1)(𝑝+2)²(𝑝+3)+4(1−𝛼)(7−4𝑝)(3−𝑝)(2−𝑝)(1−𝑝)𝑡^−𝑝+4𝑝²(1−𝛼)(2−𝑝)(1−+4𝛼𝑝(1+𝑝)²(1−𝑝)𝑡^−𝑝−2g^1(1)=881(1−𝛼)576𝛼⁴>0,𝑔+3872𝛼³+660𝛼²+6612𝛼−785(2.11)₁₍₄₎(𝑡)=−4𝑝(1−𝑝)𝑡^−𝑝−3𝑔₂(𝑡),(2.12) where 𝑔₂(𝑡)=(1−𝛼)(7𝑔−4𝑝)(3−𝑝)(2−𝑝)𝑡²+(1−𝛼)𝑝(2−𝑝)(𝑝+1)𝑡+𝛼(𝑝+1)²(𝑝+2)₂,(1)=427(1−𝛼)96𝛼³𝑔+232𝛼²+388𝛼+175>0,(2.13)_2(𝑡)=≥2𝑔_2(1−𝛼)(7−4𝑝)(3−𝑝)(2−𝑝)𝑡+(1−𝛼)𝑝(2−𝑝)(𝑝+1)(1)=49(1−𝛼)(5+4𝛼)12𝛼²+31𝛼+23>0(2.14) for 𝑡∈[1,+∞).

From (2.13) and (2.14), we clearly see that 𝑔₂(𝑡)>0 for 𝑡∈[1,+∞), then (2.12) leads to the conclusion that 𝑔^1(𝑡) is strictly in [1,+∞).

Therefore, Lemma 2.2 follows from (2.7)–(2.11) and the monotonicity of 𝑔^1(𝑡).

Lemma 2.3. Let 𝛼∈(1/4,1), 𝑝=(1−4𝛼)/3∈(−1,0), and ℎ(𝑡)=2𝛼(1−𝑡^𝑝+1)𝑡log²𝑡+(1−𝛼)(1+𝑡^𝑝−1)(1+𝑡)²𝑡log𝑡+(1−𝛼)(1+𝑡)²(1−𝑡)(𝑡^𝑝+1), then ℎ(𝑡)>0 for 𝑡∈(1,+∞).

Proof. Let ℎ₁(𝑡)=𝑡^−𝑝ℎ^(𝑡) and ℎ₂(𝑡)=𝑡^𝑝+2ℎ^1(𝑡), then simple computations lead to ℎ(1)=0,(2.15)𝛼ℎ)^(𝑡𝑡+)=3𝛼2𝛼+[11]−lo(g𝑝+𝑡−2)(𝑡1−^𝑝+1]l𝛼o)g[(²𝑝𝑡++3[()𝑝𝑡+_𝑝+22−𝛼𝑝−6𝛼)𝑡^𝑝+1+2(1−𝛼)(𝑝+1)𝑡^𝑝+(1−𝛼)𝑝𝑡^𝑝−1+3(1−𝛼)𝑡²+4(1−+(𝑝+1)𝑡^𝑝+1−(𝑝+3)𝑡^𝑝−(𝑝+1)𝑡^𝑝−1+2𝑡²−2], ℎ^(1)=0,(2.16)6ℎ(₁1(𝑡−)=𝛼)−𝑡2𝛼_1−𝑝(𝑝+1)(𝑝+2)log²𝑡+[(𝑝²−𝛼𝑝²+3𝑝−11𝛼𝑝−14𝛼+2)+2(1−𝛼)𝑝(𝑝+1)𝑡⁻¹−(1−𝛼)𝑝(1−𝑝)𝑡⁻²+1)𝑡₋₂+4(1−𝛼)𝑡^−𝑝+4𝛼𝑡^−1−𝑝]log𝑡−(1−𝛼)(𝑝+2)(𝑝+3)𝑡+(1−𝛼)(𝑝²+5𝑝+2)𝑡⁻¹+(1−𝛼)(𝑝²+𝑝−−(1−𝛼)𝑡^1−𝑝+4(1−𝛼)𝑡^−𝑝+(1+3𝛼)𝑡^−1−𝑝−(1−𝛼)𝑝²−(1−𝛼)𝑝−5𝛼+1, ℎ₁(1)=0,(2.17)ℎ4𝛼₂((𝑡1)+=𝑝−)][4lo𝛼g(𝑝𝑡+−1()1(𝑝−+𝛼)2()𝑝𝑡+^𝑝+12+)(𝑝2(+13−)𝑡𝛼)𝑝_𝑝+2(𝑝+1)𝑡^𝑝−2(1−𝛼)𝑝(1−𝑝)𝑡^𝑝−1−6(1−𝛼)(1−𝑝)𝑡²+4(1−𝛼)𝑝𝑡+(1−𝛼)(𝑝₂+(𝑝²−𝛼𝑝²+3𝑝−11𝛼𝑝−14𝛼+2)𝑡^𝑝+1+(1−𝛼)(𝑝²−3𝑝−2)𝑡^𝑝−+3𝑝−2)𝑡^𝑝−1+(1−𝛼)(𝑝+5)𝑡²+4(1−𝛼)(1−𝑝)𝑡+𝛼−3𝛼𝑝−𝑝−1, ℎ₂(1)=0,(2.18)(ℎ1^2(−𝑡)𝛼=)(𝑝−+[4𝛼2)(₂𝑝+1)²(𝑝+2)𝑡^𝑝+2(1−𝛼)𝑝²(𝑝+1)𝑡^𝑝−1+2(1−𝛼)𝑝(1−𝑝)²𝑡^𝑝−2−12(1−𝛼)(1−𝑝)𝑡+4(1−𝛼)𝑝]log𝑡−𝑝)(𝑝₂(𝑝+3)𝑡^𝑝+1+(𝑝+1)(𝑝²−𝛼𝑝²−15𝛼𝑝+3𝑝−22𝛼+2)𝑡^𝑝+(1−𝛼)𝑝(𝑝²−5𝑝−4)𝑡^𝑝−1+(1−𝛼)(1−+5𝑝−2)𝑡^𝑝−2+4(1−𝛼)(4−𝑝)𝑡−4𝛼(𝑝+1)𝑡⁻¹+4(1−𝛼)(1−2𝑝), ℎℎ^2(1)=0,_2(𝑡)=𝑓(𝑡)log𝑡+𝑔(𝑡),(2.19) where 𝑓(𝑡) and 𝑔(𝑡) are defined as in Lemmas 2.1 and 2.2, respectively.
From (2.19) and (2.10) together with Lemmas 2.1 and 2.2, we clearly see that ℎ₂(𝑡) is strictly increasing in [1,+∞).

Therefore, Lemma 2.3 follows from (2.15)–(2.18) and the monotonicity of ℎ₂(𝑡).

3. Main Result

Theorem 3.1. Inequality 𝛼𝐻(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏)>𝑀_{(1−4𝛼)/3}(𝑎,𝑏)(3.1) holds for 𝛼∈[1/4,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏, and 𝑀_{(1−4𝛼)/3}(𝑎,𝑏) is the best possible lower power mean bound for the sum 𝛼𝐻(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏).

Proof. We divide the proof of inequality (3.1) into two cases.
Case 1 (𝛼=1/4). Without loss of generality, we assume that 𝑎>𝑏 and put 𝑡=√𝑎/𝑏>1, then from (1.1) and (1.2), we have 𝛼𝐻(=𝑎,1𝑏)+(1−𝛼)𝐿(𝑎,𝑏)−𝑀_{(1−4𝛼)/3}(𝑎,𝑏)4[𝐻(𝑎,𝑏)+3𝐿(𝑎,𝑏)]−√=3𝑡₄𝑎𝑏−42𝑡³−𝑡²+2𝑡log𝑡−38𝑡²+1log𝑡𝑏.(3.2) Let 𝐹(𝑡)=3𝑡⁴−42𝑡³−𝑡²+2𝑡log𝑡−3,(3.3) then simple computations lead to 𝐹𝐹(1)=0,_(𝑡)=43𝑡³−2𝑡²𝐹+𝑡−2−83𝑡²−𝑡+1log𝑡,_𝐹(1)=0,_(𝑡)=4𝑡𝐹¹(𝑡),(3.4) where 𝐹₁(𝑡)=9𝑡³−10𝑡²+3𝑡−2−2(6𝑡−1)𝑡log𝑡, 𝐹𝐹^(1)=𝐹₁(1)=0,_1(𝑡)=27𝑡²−𝐹32𝑡+5−2(12𝑡−1)log𝑡,_1𝐹(1)=0,_1(𝑡)=2𝑡𝐹²(𝑡),(3.5) where 𝐹₂(𝑡)=27𝑡²−12𝑡log𝑡−28𝑡+1, F𝐹^1(1)=𝐹₂(1)=0,_2(𝑡)=54𝑡−12log𝑡−40>0(3.6) for 𝑡>1.
Therefore, inequality (3.1) follows easily from (3.2)–(3.6).
Case 2 (𝛼∈(1/4,1)). Without loss of generality, we assume that 𝑎>𝑏. Let 𝑝=(1−4𝛼)/3∈(−1,0) and 𝑡=𝑎/𝑏>1, then from (1.1) and (1.2), one has 𝛼𝐻(=𝑎,𝛼𝑏𝐻)+(𝑎(1,𝑏−)+𝛼)(𝐿1−(𝑎,𝛼𝑏))𝐿−(𝑎𝑀,𝑏)−𝑀_{(1−4𝛼)/3}(_𝑝𝑎,𝑏)=𝑏2𝛼𝑡(𝑎,𝑏)𝑡+1+(1−l𝛼o)g(𝑡𝑡−1)−𝑡^𝑝+12^1/𝑝.(3.7) Let 𝐺(𝑡)=log2𝛼𝑡𝑡+1+(1−l𝛼o)g(𝑡𝑡−1)−1𝑝log𝑡^𝑝+12.(3.8) Then simple computations lead to 𝐺lim_𝑡→1𝐺(𝑡)=0,(3.9)_(𝑡)=ℎ(𝑡)𝑡(𝑡+1)(𝑡^𝑝+1)log𝑡2𝛼𝑡log𝑡+(1−𝛼)𝑡²−1,(3.10) where ℎ(𝑡) is defined as in Lemma 2.3.
From Lemma 2.3 and (3.10), we clearly see that 𝐺(𝑡) is strictly increasing in (1,+∞).

Therefore, inequality (3.1) follows from (3.7)–(3.9) and the monotonicity of 𝐺(𝑡).

Next, we prove that 𝑀_{(1−4𝛼)/3}(𝑎,𝑏) is the best possible lower power mean bound for the sum 𝛼𝐻(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏) if 𝛼∈[1/4,1).

For any 𝛼∈[1/4,1), 𝑝>(1−4𝛼)/3, and 𝑥>0, one has𝑀_𝑝(1+𝑥,1)−𝛼𝐻(1+𝑥,1)−(1−𝛼)𝐿(1+𝑥,1)=𝐽(𝑥)2^1/𝑝1+𝑥2log(1+𝑥),(3.11) where 𝐽(𝑥)=(1+𝑥/2)[1+(1+𝑥)^𝑝]^1/𝑝log(1+𝑥)−2^1/𝑝[𝛼(1+𝑥)log(1+𝑥)+(1−𝛼)𝑥(1+𝑥/2)]. Letting 𝑥→0 and making use of Taylor expansion, we have𝐽(𝑥)=2^1/𝑝8𝑝−1−34𝛼𝑥³+𝑜𝑥³.(3.12)

Equations (3.11) and (3.12) imply that for any 𝛼∈[1/4,1) and 𝑝>(1−4𝛼)/3 there exists 𝛿>0, such that 𝛼𝐻(1+𝑥,1)+(1−𝛼)𝐿(1+𝑥,1)<𝑀_𝑝(1+𝑥,1) for 𝑥∈(0,𝛿).

Remark 3.2. If 0<𝛼<1/4, then from (1.1) and (1.2), we have lim_𝑥→+∞𝑀_{(1−4𝛼)/3}(1,𝑥)𝛼𝐻(1,𝑥)+(1−𝛼)𝐿(1,𝑥)=2^{3/(4𝛼−1)}×lim_𝑥→+∞1+𝑥^{(4𝛼−1)/3}^{3/(1−4𝛼)}2𝛼/(𝑥+1)+((1−1/𝑥)(1−𝛼)/log𝑥)=+∞.(3.13)
Equation (3.13) implies that for any 0<𝛼<1/4, there exists 𝑋>1, such that 𝑀_{(1−4𝛼)/3}(1,𝑥)>𝛼𝐻(1,𝑥)+(1−𝛼)𝐿(1,𝑥) for 𝑥∈(𝑋,+∞). Therefore, 𝜆=1/4 is the least value of 𝜆 in (0,1) such that inequality (3.1) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Acknowledgments

This research is supported by the N. S. Foundation of China under Grant 11071069, N. S. Foundation of Zhejiang province under Grants nos. Y7080106 and Y6100170, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant no. T200924.