Optimal Lehmer Mean Bounds for the Combinations of Identric and Logarithmic Means

Abstract

For any 𝛼∈(0,1), we answer the questions: what are the greatest values 𝑝 and 𝜆 and the least values 𝑞 and 𝜇, such that the inequalities 𝐿_𝑝(𝑎,𝑏)<𝐼^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)<𝐿_𝑞(𝑎,𝑏) and 𝐿_𝜆(𝑎,𝑏)<𝛼𝐼(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏)<𝐿_𝜇(𝑎,𝑏) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏? Here, 𝐼(𝑎,𝑏), 𝐿(𝑎,𝑏), and 𝐿_𝑝(𝑎,𝑏) denote the identric, logarithmic, and 𝑝th Lehmer means of two positive numbers 𝑎 and 𝑏, respectively.

1. Introduction

For 𝑎,𝑏>0 with 𝑎≠𝑏, the logarithmic mean 𝐿(𝑎,𝑏) and identric mean 𝐼(𝑎,𝑏) are defined by 𝐿(𝑎,𝑏)=𝑎−𝑏,𝐼()=1log𝑎−log𝑏𝑎,𝑏𝑒𝑏^𝑏𝑎^{𝑎1/(𝑏−𝑎)}, (1)(2) respectively. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for 𝐿 and 𝐼 can be found in the literature [1–19]. In [14, 17, 20], inequalities between 𝐿, 𝐼, and the classical arithmetic-geometric mean of Gauss are proved. The ratio of identric means leads to the weighted geometric mean 𝐼𝑎²,𝑏²𝐼()=𝑎𝑎,𝑏^𝑎𝑏^{𝑏1/(𝑎+𝑏)}, (3) which has been investigated in [11, 13, 21]. It might be surprising that the logarithmic mean has applications in physics, economics, and even in meteorology [22–24]. In [22], 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 [5]. In [25, 26], it is shown that 𝐿 can be expressed in terms of Gauss’ hypergeometric function ₂𝐹₁. And in [26], 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_𝑗), where 0<𝑎₁<𝑎₂<⋯<𝑎_𝑛 and 0<𝑏₁<𝑏₂<⋯<𝑏_𝑛, is positive for all 𝑛≥1.

For 𝑝∈ℝ, the power mean 𝑀_𝑝(𝑎,𝑏) and Lehmer mean 𝐿_𝑝(𝑎,𝑏) of order 𝑝 of two positive numbers 𝑎 and 𝑏 with 𝑎≠𝑏 are defined by 𝑀_𝑝()=𝑎𝑎,𝑏^𝑝+𝑏^𝑝2^1/𝑝,𝐿_𝑝()=𝑎𝑎,𝑏^𝑝+1+𝑏^𝑝+1𝑎^𝑝+𝑏^𝑝. (4)(5)

It is well known that 𝑀_𝑝(𝑎,𝑏) and 𝐿_𝑝(𝑎,𝑏) are strictly increasing with respect to 𝑝∈ℝ for fixed 𝑎,𝑏>0 with 𝑎≠𝑏. The main properties for 𝑀_𝑝 and 𝐿_𝑝 are given in [27–32].

Let 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝐺(𝑎,𝑏)=√𝑎𝑏, and 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏) be the arithmetic, geometric, and harmonic means of two positive numbers 𝑎 and 𝑏, respectively. Then it is well known that min{𝑎,𝑏}<𝐻(𝑎,𝑏)=𝐿₋₁(𝑎,𝑏)=𝑀₋₁(𝑎,𝑏)<𝐺(𝑎,𝑏)=𝐿_−1/2(𝑎,𝑏)=𝑀₀(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝐴(𝑎,𝑏)=𝐿₀(𝑎,𝑏)=𝑀₁(𝑎,𝑏)<max{𝑎,𝑏} (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 power means are proved in [2–4, 6, 8, 9, 33]: 𝑀_2/3(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝑀_log2(),𝑀𝑎,𝑏₀(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝑀_1/3(),𝑀𝑎,𝑏₀(𝑎,𝑏)<𝐼(𝑎,𝑏)𝐿(𝑎,𝑏)<𝑀_1/2(),1𝑎,𝑏2(𝐼(𝑎,𝑏)+𝐿(𝑎,𝑏))<𝑀_1/2(𝑎,𝑏) (7)(8) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Alzer and Qiu [19] proved that ()+()𝐺()()()+𝐺(),𝑀𝛼𝐴𝑎,𝑏1−𝛼𝑎,𝑏<𝐼𝑎,𝑏<𝛽𝐴𝑎,𝑏1−𝛽𝑎,𝑏_𝑝()<1𝑎,𝑏2(𝐼(𝑎,𝑏)+𝐿(𝑎,𝑏)) (9) for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤2/3, 𝛽≥2/𝑒=0.73575…, and 𝑝≤log2/(1+log2)=0.40938….

The following sharp upper and lower Lehmer mean bounds for 𝐿, 𝐼, (𝐿𝐼)^1/2, and (𝐿+𝐼)/2 are presented in [34]: 𝐿_−1/3(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝐿₀(),𝑀𝑎,𝑏_−1/6(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝐿₀(),𝐿𝑎,𝑏_−1/4(𝑎,𝑏)<𝐼^1/2(𝑎,𝑏)𝐿^1/2(𝑎,𝑏)<𝐿₀(),𝐿𝑎,𝑏_−1/4()<1𝑎,𝑏2(𝐼(𝑎,𝑏)+𝐿(𝑎,𝑏))<𝐿₀(𝑎,𝑏) (10)(11)(12) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The purpose of this paper is to present the best possible upper and lower Lehmer mean bounds of the product 𝐼^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏) and the sum 𝛼𝐼(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏) for any 𝛼∈(0,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏.

2. Lemmas

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

Lemma 1. If 𝑝≥−1/3, then the following statements are true:(1)18𝑝²+21𝑝+8>0;(2)270𝑝³+567𝑝²+384𝑝+104>0;(3)1134𝑝⁴+3429𝑝³+3879𝑝²+1950𝑝+448>0;(4)1458𝑝⁵+4941𝑝⁴+6903𝑝³+4809𝑝²+1714𝑝+320>0;(5)5346𝑝⁶+18225𝑝⁵+27297𝑝⁴+21996𝑝³+9672𝑝²+2264𝑝+320>0;(6)9𝑝³+33𝑝²+51𝑝+14>0.

Proof. (1) We clearly see that 18𝑝²+21𝑝+8>18𝑝²+−13×21+8=18𝑝²+1>0. (13)
(2) Let 𝑔₁𝑝=270𝑝³+567𝑝²+384𝑝+104. (14) Then simple computations lead to 𝑔₁−13𝑔=29>0,_′1𝑝=6135𝑝²+189𝑝+64>6135𝑝²−1+189×3+64=6135𝑝²+1>0. (15)
Therefore, Lemma 1(2) follows from (14) and (15).
(3) Let 𝑔₂𝑝=1134𝑝⁴+3429𝑝³+3879𝑝²+1950𝑝+448. (16) Then simple computations yield 𝑔₂−13𝑔=116>0,_′2𝑝=4536𝑝³+10287𝑝²𝑔+7758𝑝+1950,_′2−13𝑔=339>0,_2′′𝑝=13608𝑝²+20574𝑝+7758>13608𝑝²−1+20574×3+7758=13608𝑝²+900>0. (17)
Therefore, Lemma 1(3) follows from (16) and (17).
(4) Let 𝑔₃𝑝=1458𝑝⁵+4941𝑝⁴+6903𝑝³+4809𝑝²+1714𝑝+320. (18) Then simple computations lead to 𝑔₃−13=2473𝑔>0,_′3𝑝=7290𝑝⁴+19764𝑝³+20709𝑝²𝑔+9618𝑝+1714,_′3−13𝑔=167>0,_3′′𝑝=29160𝑝³+59292𝑝²𝑔+41418𝑝+9618,_3′′−13𝑔=1320>0,_3′′′𝑝=87480𝑝²+118584𝑝+41418>87480𝑝²−1+41418+118584×3=87480𝑝²+1890>0. (19)
Therefore, Lemma 1(4) follows from (18) and (19).
(5) Let 𝑔₄𝑝=5346𝑝⁶+18225𝑝⁵+27297𝑝⁴+21996𝑝³+9672𝑝²+2264𝑝+320. (20) Then simple computations yield 𝑔₄−13=2843𝑔>0,_′4𝑝=32076𝑝⁵+91125𝑝⁴+109188𝑝³+65988𝑝²𝑔+19344𝑝+2264,_′4−13𝑔=97>0,_4′′𝑝=160380𝑝⁴+364500𝑝³+327564𝑝²𝑔+131976𝑝+19344,_4′′−13𝑔=228>0,_4′′′𝑝=1085940𝑝³+10125𝑝²,𝑔+6066𝑝+1222_4′′′−13𝑔=11340>0,₄₍₄₎𝑝=1944990𝑝²+1125𝑝+337. (21)(22)(23)(24)(25)(26)(27)(28)
It follows from (28) and the discriminant Δ=(1125)²−4×990×337=−68895<0 of the quadratic function 𝑓(𝑝)=990𝑝²+1125𝑝+337 that 𝑔₄⁽⁴⁾𝑝>0. (29)
Therefore, Lemma 1(5) follows from (20)–(27) and (29).
(6) Let 𝑔₅𝑝=9𝑝³+33𝑝²+51𝑝+14. (30) Then we have 𝑔₅−13=13𝑔>0,_′5𝑝=27𝑝²+66𝑝+51>27𝑝²1−66×3+51=27𝑝²+29>0. (31)
Therefore, Lemma 1(6) follows from (30) and (31).

Lemma 2. Suppose that ℎ(𝑡)=(6𝑝+1)𝑡^6𝑝+6 −3𝑡^6𝑝+4−2(3𝑝+2)𝑡^6𝑝+3 +(3𝑝+1)𝑡^3𝑝+6−3𝑝𝑡^3𝑝+5 −3(𝑝+1)𝑡^3𝑝+4+3(𝑝+1)𝑡^3𝑝+2 +3𝑝𝑡^3𝑝+1−(3𝑝+1)𝑡^3𝑝 +2(3𝑝+2)𝑡³ +3𝑡²−6𝑝−1. If 𝑝∈(−1/3,−1/6), then ℎ(𝑡)>0 for 𝑡>1.

Proof. Let ℎ₁(𝑡)=ℎ^′(𝑡)/𝑡, ℎ₂(𝑡)=𝑡^4−3𝑝ℎ₁^′′(𝑡), ℎ₃(𝑡)=ℎ₂^′′′(𝑡)/𝑡, and ℎ₄(𝑡)=𝑡^4−3𝑝ℎ₃^′′′(𝑡)/[9𝑝(𝑝+1)(3𝑝+1)(3𝑝+2)²(6𝑝+1)]; then elaborated computations lead to ℎ(1)ℎ=0,₁(𝑡)=6𝑝+16𝑝+1𝑡^6𝑝+4−63𝑝+2𝑡^6𝑝+2−62𝑝+13𝑝+2𝑡^6𝑝+1+3𝑝+23𝑝+1𝑡^3𝑝+4−3𝑝3𝑝+5𝑡^3𝑝+3−3𝑝+13𝑝+4𝑡^3𝑝+2+3𝑝+13𝑝+2𝑡^3𝑝+3𝑝3𝑝+1𝑡^3𝑝−1−3𝑝3𝑝+1𝑡^3𝑝−2ℎ+63𝑝+2𝑡+6,₁(1)ℎ=0,_′1(𝑡)=12𝑝+13𝑝+26𝑝+1𝑡^6𝑝+3−123𝑝+13𝑝+2𝑡^6𝑝+1−62𝑝+13𝑝+26𝑝+1𝑡^6𝑝+3𝑝+23𝑝+13𝑝+4𝑡^3𝑝+3−9𝑝𝑝+13𝑝+5𝑡^3𝑝+2−3𝑝+13𝑝+23𝑝+4𝑡^3𝑝+1+9𝑝𝑝+13𝑝+2𝑡^3𝑝−1+3𝑝3𝑝−13𝑝+1𝑡^3𝑝−2−3𝑝3𝑝−23𝑝+1𝑡^3𝑝−3,ℎ+63𝑝+2_′1(1)ℎ=0,₂(𝑡)=36𝑝+12𝑝+13𝑝+26𝑝+1𝑡^3𝑝+6−123𝑝+13𝑝+26𝑝+1𝑡^3𝑝+4−36𝑝2𝑝+13𝑝+26𝑝+1𝑡^3𝑝+3+9𝑝+1𝑝+23𝑝+13𝑝+4𝑡⁶−9𝑝𝑝+13𝑝+23𝑝+5𝑡⁵−3𝑝+13𝑝+13𝑝+23𝑝+4𝑡⁴+9𝑝𝑝+13𝑝−13𝑝+2𝑡²𝑡,ℎ+3𝑝3𝑝−23𝑝−13𝑝+1−9𝑝𝑝−13𝑝−23𝑝+1₂(1)=123𝑝+118𝑝²ℎ+21𝑝+8,_′2(𝑡)=108𝑝+1𝑝+22𝑝+13𝑝+26𝑝+1𝑡^3𝑝+5−123𝑝+13𝑝+23𝑝+46𝑝+1𝑡^3𝑝+3−108𝑝𝑝+12𝑝+13𝑝+26𝑝+1𝑡^3𝑝+2+54𝑝+1𝑝+23𝑝+13𝑝+4𝑡⁵−45𝑝𝑝+13𝑝+23𝑝+5𝑡⁴−12𝑝+13𝑝+13𝑝+23𝑝+4𝑡³𝑡,ℎ+18𝑝𝑝+13𝑝−13𝑝+2+3𝑝3𝑝−13𝑝−23𝑝+1_′2(1)=123𝑝+19𝑝+718𝑝²,ℎ+21𝑝+8_2′′(𝑡)×𝑡=108𝑝+1𝑝+22𝑝+13𝑝+23𝑝+56𝑝+1^3𝑝+4×𝑡−36𝑝+13𝑝+13𝑝+23𝑝+46𝑝+1^3𝑝+2−108𝑝𝑝+12𝑝+13𝑝+2²6𝑝+1𝑡^3𝑝+1+270𝑝+1𝑝+23𝑝+13𝑝+4𝑡⁴−180𝑝𝑝+13𝑝+23𝑝+5𝑡³−36𝑝+13𝑝+13𝑝+23𝑝+4𝑡²,ℎ+18𝑝𝑝+13𝑝−13𝑝+2_2′′(1)=363𝑝+1𝑝+1270𝑝³+567𝑝²,ℎ+384𝑝+104₃(𝑡)×𝑡=108𝑝+1𝑝+22𝑝+13𝑝+23𝑝+43𝑝+56𝑝+1^3𝑝+2−36𝑝+13𝑝+13𝑝+2²×3𝑝+46𝑝+1𝑡^3𝑝−108𝑝𝑝+12𝑝+13𝑝+13𝑝+2²×6𝑝+1𝑡^3𝑝−1×𝑡+1080𝑝+1𝑝+23𝑝+13𝑝+4²𝑡,ℎ−540𝑝𝑝+13𝑝+23𝑝+5−72𝑝+13𝑝+13𝑝+23𝑝+4₃(1)×=363𝑝+1𝑝+11134𝑝⁴+3429𝑝³+3879𝑝²,ℎ+1950𝑝+448_′3(𝑡)=108𝑝+1𝑝+22𝑝+13𝑝+2²×3𝑝+43𝑝+56𝑝+1𝑡^3𝑝+1−108𝑝𝑝+13𝑝+13𝑝+2²×3𝑝+46𝑝+1𝑡^3𝑝−1−108𝑝𝑝+12𝑝+13𝑝−13𝑝+13𝑝+2²×6𝑝+1𝑡^3𝑝−2×𝑡,ℎ+2160𝑝+1𝑝+23𝑝+13𝑝+4−540𝑝𝑝+13𝑝+23𝑝+5_′3(1)×=108𝑝+13𝑝+11458𝑝⁵+4941𝑝⁴+6903𝑝³+4809𝑝²,ℎ+1714𝑝+320_3′′(𝑡)×=108𝑝+1𝑝+22𝑝+13𝑝+13𝑝+2²3𝑝+43𝑝+56𝑝+1𝑡^3𝑝−108𝑝𝑝+13𝑝−13𝑝+13𝑝+2²×𝑡3𝑝+46𝑝+1^3𝑝−2×−108𝑝𝑝+12𝑝+13𝑝−23𝑝−13𝑝+13𝑝+2²6𝑝+1𝑡^3𝑝−3,ℎ+2160𝑝+1𝑝+23𝑝+13𝑝+4_3′′(1)×=108𝑝+13𝑝+15346𝑝⁶+18225𝑝⁵+27297𝑝⁴+21996𝑝³+9672𝑝²,ℎ+2264𝑝+320₄(𝑡)=36𝑝+22𝑝+13𝑝+43𝑝+5𝑡³𝑡𝑡𝑡−121−3𝑝2−3𝑝3𝑝+4+361−𝑝1−3𝑝2−3𝑝2𝑝+1>36𝑝+22𝑝+13𝑝+43𝑝+5−121−3𝑝2−3𝑝3𝑝+4+361−𝑝1−3𝑝2−3𝑝2𝑝+1=243𝑝+49𝑝³+33𝑝²𝑡.+51𝑝+14+361−𝑝1−3𝑝2−3𝑝2𝑝+1 (32)(33)(34)(35)(36)(37)(38)(39)(40)(41)(42)(43)(44)(45)(46)(47)(48)(49)
It follows from Lemma 1(6) and (49) that ℎ₄(𝑡)>0 (50) for 𝑡>1.
From Lemma 1(1)–(5) and (38)–(48), we clearly see that ℎ₂(1)ℎ>0,_′2(1)ℎ>0,_2′′(1)ℎ>0,₃(1)ℎ>0,_′3(1)ℎ>0,_3′′(1)>0. (51)
Therefore, Lemma 2 follows from 9𝑝(𝑝+1)(3𝑝+1)(3𝑝+2)²(6𝑝+1)>0 and (32)–(36) together with (50) and (51).

Lemma 3. Inequality [𝑀_2/3(𝑎,𝑏)]^𝛼[𝐿_−1/3(𝑎,𝑏)]^1−𝛼>𝐿_(𝛼−2)/6(𝑎,𝑏) holds for any 𝛼∈(0,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏.

Proof. Without loss of generality, we assume that 𝑎>𝑏. Let 𝑡=(𝑎/𝑏)^1/3>1 and 𝑝=(𝛼−2)/6; then 𝑝∈(−1/3,−1/6), 𝛼=6𝑝+2, and from (1) and (4) we have 𝑀_2/3(𝑎,𝑏)^𝛼𝐿_−1/3(𝑎,𝑏)^1−𝛼−𝐿_(𝛼−2)/6()=𝑏𝑡𝑎,𝑏_3𝑝+1𝑡²+1^3𝑝+2(𝑡+1)^6𝑝+1−2^9𝑝+3𝑡^6𝑝+1𝑡^3𝑝+3×2+1_9𝑝+3𝑡^6𝑝+1𝑡^3𝑝+1⁻¹. (52) Let 𝑓(𝑡)=log𝑡^3𝑝+1+3𝑝+2log𝑡²+()−𝑡+16𝑝+1log𝑡+16𝑝+1log𝑡−log^3𝑝+3+1−33𝑝+1log2. (53) Then simple computations lead to 𝑓(1)𝑓=0,_′(𝑡)=ℎ(𝑡)𝑡(𝑡+1)𝑡²+1𝑡^3𝑝+1𝑡^3𝑝+3,+1 (54)(55) where ℎ(𝑡) is defined as in Lemma 2.
From (54) and (55) together with Lemma 2, we clearly see that 𝑓(𝑡)>0 (56) for 𝑡>1.
Therefore, Lemma 3 follows from (52) and (53) together with (56).

3. Main Results

Theorem 4. Inequality 𝐿_(𝛼−2)/6(𝑎,𝑏)<𝐼^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)<𝐿₀(𝑎,𝑏) holds for any 𝛼∈(0,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏, and 𝐿_(𝛼−2)/6(𝑎,𝑏) and 𝐿₀(𝑎,𝑏) are the best possible lower and upper Lehmer mean bounds for the product 𝐼^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏).

Proof. Inequality 𝐼^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)<𝐿₀(𝑎,𝑏) follows directly from (10) and (11).
For the other inequality, we note that 𝐼^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)−𝐿_(𝛼−2)/6(𝑎,𝑏)=𝐼^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)−𝑀_2/3(𝑎,𝑏)^𝛼𝐿_−1/3(𝑎,𝑏)^1−𝛼+𝑀_2/3(𝑎,𝑏)^𝛼𝐿_−1/3(𝑎,𝑏)^1−𝛼−𝐿_(𝛼−2)/6(𝑎,𝑏). (57)
Therefore, 𝐼^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)>𝐿_(𝛼−2)/6(𝑎,𝑏) follows from (7) and (10) together with Lemma 3 and (57).
Next, we prove that 𝐿_(𝛼−2)/6(𝑎,𝑏) and 𝐿₀(𝑎,𝑏) are the best possible lower and upper Lehmer mean bounds for the product 𝐼^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏).
For any 0<𝜀<1 and 𝑥>0, from (1) and (2) we have lim_𝑥→+∞𝐼^𝛼(1,𝑥)𝐿^1−𝛼(1,𝑥)𝐿_−𝜀(1,𝑥)=lim_𝑥→+∞𝑥^−𝜀+1𝑥^𝜀−1𝑒+1^{𝛼((log𝑥)/(𝑥−1)−1)}×11−𝑥^1−𝛼𝑥^𝜀log𝑥^1−𝛼𝐿=+∞,_{(𝛼−2)/6+𝜀}(1,1+𝑥)−𝐼^𝛼(1,1+𝑥)𝐿^1−𝛼()=()1,1+𝑥1+𝑥^{(𝛼+4)/6+𝜀}+1(1+𝑥)^{(𝛼−2)/6+𝜀}+1−𝑒^{𝛼[(1+𝑥)log(1+𝑥)/𝑥−1]}×𝑥^1−𝛼log(1+𝑥)^𝛼. (58) Letting 𝑥→0 and making use of Taylor expansion, one has (1+𝑥)^{(𝛼+4)/6+𝜀}+1(1+𝑥)^{(𝛼−2)/6+𝜀}+1−𝑒^{𝛼[(1+𝑥)log(1+𝑥)/𝑥−1]}×𝑥^1−𝛼log(1+𝑥)^𝛼1=1+2𝛼𝑥+−124+𝜀124𝑥²+𝑜𝑥²−𝛼1+21𝑥+𝛼81𝛼−6𝑥²+𝑜𝑥²×11+2−𝛼21𝑥+8𝛼²−1124𝛼−𝑥12²+𝑜𝑥²1=1+2𝛼𝑥+−124+𝜀124𝑥²−11+2𝛼𝑥+−124𝑥12²+𝑜𝑥²=𝜀4𝑥²+𝑜𝑥². (59)
Equations (58) and (59) imply that for any 0<𝜀<1, there exist 𝑋₁=𝑋₁(𝛼,𝜀)>1 and 𝛿₁=𝛿₁(𝛼,𝜀)>0, such that 𝐼^𝛼(1,𝑥)𝐿^1−𝛼(1,𝑥)>𝐿_−𝜀(1,𝑥) for 𝑥∈(𝑋₁,+∞) and 𝐿_{(𝛼−2)/6+𝜀}(1,1+𝑥) >𝐼^𝛼(1,1+𝑥)𝐿^1−𝛼(1,1+𝑥) for 𝑥∈(0,𝛿₁).

Theorem 5. Inequality 𝐿_(𝛼−2)/6(𝑎,𝑏)<𝛼𝐼(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏)<𝐿₀(𝑎,𝑏) holds for any 𝛼∈(0,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏, and 𝐿_(𝛼−2)/6(𝑎,𝑏) and 𝐿₀(𝑎,𝑏) are the best possible lower and upper Lehmer mean bounds for the sum 𝛼𝐼(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏).

Proof. Inequality 𝛼𝐼(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏)<𝐿₀(𝑎,𝑏) follows directly from (10) and (11), and inequality 𝐿_(𝛼−2)/6(𝑎,𝑏)<𝛼𝐼(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏) follows from Theorem 4.
Next, we prove that 𝐿_(𝛼−2)/6(𝑎,𝑏) and 𝐿₀(𝑎,𝑏) are the best possible lower and upper Lehmer mean bounds for the sum 𝛼𝐼(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏).
For any 0<𝜀<1 and 𝑥>0, from (1) and (2), we have lim_𝑥→+∞𝛼𝐼(1,𝑥)+(1−𝛼)𝐿(1,𝑥)𝐿_−𝜀(1,𝑥)=lim_𝑥→+∞𝑥^−𝜀+1𝑥^𝜀−1×+1𝛼𝑒^{(log𝑥)/(𝑥−1)}𝑥^𝜀+(1−𝛼)(1−1/𝑥)𝑥log𝑥^𝜀𝐿=+∞,_{(𝛼−2)/6+𝜀}()()−()𝐿()=()1,1+𝑥−𝛼𝐼1,1+𝑥1−𝛼1,1+𝑥1+𝑥^{(𝛼+4)/6+𝜀}+1(1+𝑥)^{(𝛼−2)/6+𝜀}+1−𝛼𝑒^{(1+𝑥)log(1+𝑥)/𝑥−1}−(1−𝛼)𝑥().log1+𝑥 (60)(61) Letting 𝑥→0 and making use of Taylor expansion, one has (1+𝑥)^{(𝛼+4)/6+𝜀}+1(1+𝑥)^{(𝛼−2)/6+𝜀}+1−𝛼𝑒^{(1+𝑥)log(1+𝑥)/𝑥−1}−(1−𝛼)𝑥()1log1+𝑥=1+2𝛼𝑥+−124+𝜀124𝑥²+𝑜𝑥²1−𝛼1+21𝑥−𝑥24²+𝑜𝑥²−(11−𝛼)1+21𝑥−𝑥12²+𝑜𝑥²=𝜀4𝑥²+𝑜𝑥². (62)
Equations (60)-(62) imply that for any 0<𝜀<1, there exist 𝑋₂=𝑋₂(𝛼,𝜀)>1 and 𝛿₂=𝛿₂(𝛼,𝜀)>0, such that 𝛼𝐼(1,𝑥)+(1−𝛼)𝐿(1,𝑥)>𝐿_−𝜀(1,𝑥) for 𝑥∈(𝑋₂,+∞) and 𝐿_{(𝛼−2)/6+𝜀}(1,1+𝑥)>𝛼𝐼(1,1+𝑥)+(1−𝛼)𝐿(1,1+𝑥) for 𝑥∈(0,𝛿₂).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307 and the Natural Science Foundation of Zhejiang Province under Grants LY13H070004 and LY13A010004.