Abstract

We prove that 𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)>𝑀(14𝛼)/3(𝑎,𝑏) for 𝛼(0,1) and all 𝑎,𝑏>0 with 𝑎𝑏 if and only if 𝛼[1/4,1) and 𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)<𝑀(14𝛼)/3(𝑎,𝑏) if and only if 𝛼(0,3345/8011/16), and the parameter (14𝛼)/3 is the best possible in either case. Here, 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐿(𝑎,𝑏)=(𝑎𝑏)/(log𝑎log𝑏), and 𝑀𝑝(𝑎,𝑏)=((𝑎𝑝+𝑏𝑝)/2)1/𝑝(𝑝0) and 𝑀0(𝑎,𝑏)=𝑎𝑏 are the harmonic, logarithmic, and pth power means of a and b, respectively.

1. Introduction

The classical logarithmic mean 𝐿(𝑎,𝑏) of two positive real numbers 𝑎 and 𝑏 with 𝑎𝑏 is defined by 𝐿(𝑎,𝑏)=𝑎𝑏.log𝑎log𝑏(1.1)

In the recent past, the bivariate means have been the subject of intensive research. In particular, many remarkable inequalities for 𝐿(𝑎,𝑏) can be found in the literature [121]. It might be surprising that the logarithmic mean has applications in physics, economics, and even in meteorology [2224]. In [22] the authors study a variant of Jensen's functional equation involving the logarithmic mean, which appears in a heat conduction problem. A representation of 𝐿(𝑎,𝑏) as an infinite product and an iterative algorithm for computing it as the common limit of two sequences of special geometric and arithmetic means are given in [4]. In [25, 26] it is shown that 𝐿(𝑎,𝑏) 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(𝑎𝑖,𝑏𝑖), where 0<𝑎1<𝑎2<<𝑎𝑛 and 0<𝑏1<𝑏2<<𝑏𝑛, is positive for all 𝑛1.

Let 𝐺(𝑎,𝑏)=𝑎𝑏, 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐼(𝑎,𝑏)=1/𝑒(𝑎𝑎/𝑏𝑏)1/(𝑎𝑏), 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝑀𝑝(𝑎,𝑏)=((𝑎𝑝+𝑏𝑝)/2)1/𝑝(𝑝0) and 𝑀0(𝑎,𝑏)=𝑎𝑏, and 𝐿𝑝(𝑎,𝑏)=(𝑎𝑝+1+𝑏𝑝+1)/(𝑎𝑝+𝑏𝑝) be the geometric, harmonic, identric, arithmetic, 𝑝th power, and 𝑝th Lehmer means of two positive numbers 𝑎 and 𝑏, respectively. Then it is well known that both 𝑀𝑝(𝑎,𝑏) and 𝐿𝑝(𝑎,𝑏) are continuous and strictly increasing with respect to 𝑝 for fixed 𝑎,𝑏>0 with 𝑎𝑏, and the inequalitiesmin{𝑎,𝑏}<𝐻(𝑎,𝑏)=𝑀1(𝑎,𝑏)=𝐿1(𝑎,𝑏)<𝐺(𝑎,𝑏)=𝑀0(𝑎,𝑏)=𝐿1/2(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝐴(𝑎,𝑏)=𝑀1(𝑎,𝑏)=𝐿0(𝑎,𝑏)<max{𝑎,𝑏}(1.2) hold for all 𝑎,𝑏>0 with 𝑎𝑏.

In [4], Carlson proves that the double inequality 𝐺(𝑎,𝑏)(𝐴(𝑎,𝑏)+𝐺(𝑎,𝑏))21<𝐿(𝑎,𝑏)<2(𝐴(𝑎,𝑏)+𝐺(𝑎,𝑏))(1.3) holds for all 𝑎,𝑏>0 with 𝑎𝑏.

In [5], Lin finds the best possible upper and lower power bounds for the logarithmic mean as follows: 𝑀0(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝑀1/3(𝑎,𝑏)(1.4) for all 𝑎,𝑏>0 with 𝑎𝑏.

In [9], Sándor establishes that 𝐺(𝑎,𝑏)𝐼(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝐴(𝑎,𝑏)+𝐺(𝑎,𝑏)𝐼(𝑎,𝑏)(1.5) for all 𝑎,𝑏>0 with 𝑎𝑏.

In [27], Alzer gives the optimal Lehmer mean bounds for 𝐿, (𝐿𝐼)1/2, and (𝐿+𝐼)/2 as follows: 𝐿1/3(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝐿0𝐿(𝑎,𝑏),1/4(𝑎,𝑏)<𝐿(𝑎,𝑏)𝐼(𝑎,𝑏)<𝐿0𝐿(𝑎,𝑏),1/41(a,𝑏)<2(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))<𝐿0(𝑎,𝑏)(1.6) for all 𝑎,𝑏>0 with 𝑎𝑏.

The following sharp bounds for (𝐿𝐼)1/2 and (𝐿+𝐼)/2 in terms of power mean are presented in [28]: 𝑀0(𝑎,𝑏)<𝐿(𝑎,𝑏)𝐼(𝑎,𝑏)<𝑀1/2𝑀(𝑎,𝑏),log2/(1+log2)1(𝑎,𝑏)<2(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))<𝑀1/2(𝑎,𝑏)(1.7) for all 𝑎,𝑏>0 with 𝑎𝑏.

In [29, 30], the authors obtain the sharp bounds for the products 𝐴𝛼(𝑎,𝑏)𝐿1𝛼(𝑎,𝑏) and 𝐺𝛼(𝑎,𝑏)𝐿1𝛼(𝑎,𝑏) and the sum 𝛼𝐴(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏) in terms of power mean as follows: 𝑀0(𝑎,𝑏)<𝐴𝛼(𝑎,𝑏)𝐿1𝛼(𝑎,𝑏)<𝑀(1+2𝛼)/3𝑀(𝑎,𝑏),0(𝑎,𝑏)<𝐺𝛼(𝑎,𝑏)𝐿1𝛼(𝑎,𝑏)<𝑀(1𝛼)/3𝑀(𝑎,𝑏),log2/(log2log𝛼)(𝑎,𝑏)<𝛼𝐴(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)<𝑀(1+2𝛼)/3(𝑎,𝑏)(1.8) for any 𝛼(0,1) and all 𝑎,𝑏>0 with 𝑎𝑏.

In [31], Zhu presents some bounds for 𝐼(𝑎,𝑏) in terms of 𝐴(𝑎,𝑏) and 𝐿(𝑎,𝑏) and 𝐿(𝑎,𝑏) in terms of 𝐺(𝑎,𝑏) and 𝐼(𝑎,𝑏).

In [32], Chu et al. prove that the double inequality 𝛼𝐴(𝑎,𝑏)+(1𝛼)𝐻(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝛽𝐴(𝑎,𝑏)+(1𝛽)𝐻(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎𝑏 if and only if 𝛼2/𝜋 and 𝛽5/6.

It is the aim of this paper to give the optimal power mean bounds for the convex combination of harmonic and logarithmic means. Our main result is the following theorem.

Theorem 1.1. For 𝛼(0,1) and all 𝑎,𝑏>0 with 𝑎𝑏, one has(1)𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)>𝑀(14𝛼)/3(𝑎,𝑏) if and only if 𝛼[1/4,1);(2)𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)<𝑀(14𝛼)/3(𝑎,𝑏) if and only if 𝛼(0,3345/8011/16).
In particular, the parameter (14𝛼)/3 is the best possible in either case.

2. Lemmas

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

Lemma 2.1. Let 𝛼(1/4,1),𝑝=(14𝛼)/3(1,0), and 𝑔(𝑡)=4𝛼𝑝(𝑝+1)2(𝑝+2)𝑡𝑝1+2(1𝛼)𝑝2(1𝑝2)𝑡𝑝2+2(1𝛼)𝑝(1𝑝)2(2𝑝)𝑡𝑝3+12(1𝛼)(1𝑝). Then 𝑔(𝑡)>0 for 𝑡[1,+).

Proof. Simple computations lead to 𝑔(1)=6481(1𝛼)256𝛼2+23𝛼+11>0,(2.1)lim𝑡+𝑔𝑔(𝑡)=12(1𝛼)(1𝑝)=8(1𝛼)(1+2𝛼)>0,(2.2)(𝑡)=2𝑝(1𝑝)𝑡𝑝4𝑔1(𝑡),(2.3) where 𝑔1(𝑡)=2𝛼(𝑝+1)2(𝑝+2)𝑡2𝑔+(1𝛼)𝑝(𝑝+1)(2𝑝)𝑡+(1𝛼)(1𝑝)(2𝑝)(3𝑝),14(1)=27(1𝛼)148𝛼211𝛼+25>0,(2.4)lim𝑡+𝑔1𝑔(𝑡)=,(2.5)1(𝑡)=4𝛼(𝑝+1)24(𝑝+2)𝑡+(1𝛼)𝑝(𝑝+1)(2𝑝)=27(1𝛼)2[]16𝛼(74𝛼)𝑡+(4𝛼1)(4𝛼+5)<0(2.6) for 𝑡[1,+).
Inequality (2.6) implies that 𝑔1(𝑡) is strictly decreasing in [1,+). Then (2.4) and (2.5) lead to the conclusion that there exists 𝜆1>1 such that 𝑔1(𝑡)>0 for 𝑡[1,𝜆1) and 𝑔1(𝑡)<0 for 𝑡(𝜆1,+). It follows from (2.3) that 𝑔(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 𝑔(𝑡).

Lemma 2.2. Let 𝛼(1/4,1),𝑝=(14𝛼)/3(1,0), and (𝑡)=(1𝛼)(𝑝+1)(𝑝+2)2(𝑝+3)𝑡𝑝+(𝑝+1)(𝑝3𝛼𝑝319𝛼𝑝2+3𝑝234𝛼𝑝+2𝑝8𝛼)𝑡𝑝1+(1𝛼)𝑝(𝑝38𝑝2𝑝+4)𝑡𝑝2+(1𝛼)(1𝑝)(𝑝3+5𝑝214𝑝+4)𝑡𝑝3+4(1𝛼)(74𝑝)4𝑝(1𝛼)𝑡1+4𝛼(1+𝑝)𝑡2. Then (𝑡)>0 for 𝑡[1,+).

Proof. Let 1(𝑡)=𝑡3𝑝(𝑡).(2.7) Then simple computations lead to 1(1)=1627(1𝛼)80𝛼2+110𝛼1>0,(2.8)1(𝑡)=3(1𝛼)(𝑝+1)(𝑝+2)2(𝑝+3)𝑡2×𝑝+2(𝑝+1)3𝛼𝑝319𝛼𝑝2+3𝑝2𝑝34𝛼𝑝+2𝑝8𝛼𝑡+(1𝛼)𝑝38𝑝2𝑝+4+4(1𝛼)(74𝑝)(3𝑝)𝑡2𝑝4𝑝(1𝛼)(2𝑝)𝑡1𝑝+4𝛼1𝑝2𝑡𝑝,1(1)=32(271𝛼)16𝛼3+38𝛼2+176𝛼9>0,(2.9)1(𝑡)=6(1𝛼)(𝑝+1)(𝑝+2)2×𝑝(𝑝+3)𝑡+2(𝑝+1)3𝛼𝑝319𝛼𝑝2+3𝑝234𝛼𝑝+2𝑝8𝛼+4(1𝛼)(74𝑝)(3𝑝)(2𝑝)𝑡1𝑝4𝑝(1𝛼)(2𝑝)(1𝑝)𝑡𝑝4𝛼𝑝1𝑝2𝑡𝑝1,18(1)=81(1𝛼)128𝛼4+896𝛼3+288𝛼2+5294𝛼437>0,(2.10)1(𝑡)=6(1𝛼)(𝑝+1)(𝑝+2)2(𝑝+3)+4(1𝛼)(74𝑝)(3𝑝)(2𝑝)(1𝑝)𝑡𝑝+4𝑝2(1𝛼)(2𝑝)(1𝑝)𝑡𝑝1+4𝛼𝑝(1+𝑝)2(1𝑝)𝑡𝑝2,18(1)=81(1𝛼)576𝛼4+3872𝛼3+660𝛼2+6612𝛼785>0,(2.11)1(4)(𝑡)=4𝑝(1𝑝)𝑡𝑝32(𝑡),(2.12) where 2(𝑡)=(1𝛼)(74𝑝)(3𝑝)(2𝑝)𝑡2+(1𝛼)𝑝(2𝑝)(𝑝+1)𝑡+𝛼(𝑝+1)2(𝑝+2),24(1)=27(1𝛼)96𝛼3+232𝛼2+388𝛼+175>0,2(𝑡)=2(1𝛼)(74𝑝)(3𝑝)(2𝑝)𝑡+(1𝛼)𝑝(2𝑝)(𝑝+1)24(1)=9(1𝛼)(5+4𝛼)12𝛼2+31𝛼+23>0(2.13) for 𝑡[1,+).
From inequalities (2.13) we clearly see that 2(𝑡)>0 for 𝑡[1,+). Then (2.12) leads to the conclusion that 1(𝑡) is strictly increasing in [1,+).
Therefore, Lemma 2.2 follows from (2.7)–(2.11) and the monotonicity of 1(𝑡).

Lemma 2.3. Let 𝛼(0,1),𝑝=(14𝛼)/3,𝜆0=3345/8011/16=0.00903, and 𝑓(𝑡)=2𝛼(1𝑡𝑝+1)𝑡log2𝑡+(1𝛼)(1+𝑡𝑝1)(1+𝑡)2𝑡log𝑡+(1𝛼)(1+𝑡)2(1𝑡)(𝑡𝑝+1). Then the following two statements are true:(1)if 𝛼(1/4,1), then 𝑓(𝑡)>0 for 𝑡(1,+);(2)if 𝛼(0,𝜆0], then 𝑓(𝑡)<0 for 𝑡(1,+).

Proof. Let 𝑓1(𝑡)=𝑡𝑝𝑓(𝑡), 𝑓2(𝑡)=𝑡𝑝+2𝑓1(𝑡), 𝑓3(𝑡)=𝑡4𝑝𝑓2(𝑡), 𝑓4(𝑡)=𝑡𝑝+2𝑓3(𝑡) and 𝑓5(𝑡)=𝑡4𝑝𝑓4(𝑡). Then simple computations lead to 𝑓𝑓(1)=0,(2.14)(𝑡)=2𝛼1(𝑝+2)𝑡𝑝+1log2𝑡+(𝑝+2𝛼𝑝6𝛼)𝑡𝑝+1+2(1𝛼)(𝑝+1)𝑡𝑝+(1𝛼)𝑝𝑡𝑝1+3(1𝛼)𝑡2+4(1𝛼)𝑡+3𝛼+1log𝑡(1𝛼)(𝑝+3)𝑡𝑝+2+(𝑝+1)𝑡𝑝+1(𝑝+3)𝑡𝑝(𝑝+1)𝑡𝑝1+2𝑡2,𝑓2𝑓(1)=0,(2.15)1(𝑡)=2𝛼(𝑝+1)(𝑝+2)log2𝑡+𝑝2𝛼𝑝2+3𝑝11𝛼𝑝14𝛼+2+2(1𝛼)𝑝(𝑝+1)𝑡1(1𝛼)𝑝(1𝑝)𝑡2+6(1𝛼)𝑡1𝑝+4(1𝛼)𝑡𝑝+4𝛼𝑝𝑡1𝑝𝑝log𝑡(1𝛼)(𝑝+2)(𝑝+3)𝑡+(1𝛼)2𝑡+5𝑝+21𝑝+(1𝛼)2𝑡+𝑝12(1𝛼)𝑡1𝑝+4(1𝛼)𝑡𝑝+(1+3𝛼)𝑡1𝑝(1𝛼)𝑝2𝑓(1𝛼)𝑝5𝛼+1,1𝑓(1)=0,(2.16)2(𝑡)=4𝛼(𝑝+1)(𝑝+2)𝑡𝑝+1+2(1𝛼)𝑝(𝑝+1)𝑡𝑝2(1𝛼)𝑝(1𝑝)𝑡𝑝16(1𝛼)(1𝑝)𝑡2+4(1𝛼)𝑝𝑡+4𝛼(1+𝑝)log𝑡(1𝛼)(𝑝+2)(𝑝+3)𝑡𝑝+2+𝑝2𝛼𝑝2𝑡+3𝑝11𝛼𝑝14𝛼+2𝑝+1+𝑝(1𝛼)2𝑡3𝑝2𝑝𝑝(1𝛼)2𝑡+3𝑝2𝑝1+(1𝛼)(𝑝+5)𝑡2𝑓+4(1𝛼)(1𝑝)𝑡+𝛼3𝛼𝑝𝑝1,2(𝑓1)=0,(2.17)2(𝑡)=4𝛼(𝑝+1)2(𝑝+2)𝑡𝑝+2(1𝛼)𝑝2(𝑝+1)𝑡𝑝1+2(1𝛼)𝑝(1𝑝)2𝑡𝑝2]12(1𝛼)(1𝑝)𝑡+4(1𝛼)𝑝log𝑡(1𝛼)(𝑝+2)2(𝑝+3)𝑡𝑝+1𝑝+(𝑝+1)2𝛼𝑝2𝑡15𝛼𝑝+3𝑝22𝛼+2𝑝𝑝+(1𝛼)𝑝2𝑡5𝑝4𝑝1𝑝+(1𝛼)(1𝑝)2𝑡+5𝑝2𝑝2+4(1𝛼)(4𝑝)𝑡4𝛼(1+𝑝)𝑡1𝑓+4(1𝛼)(12𝑝),2(𝑓1)=0,(2.18)2(𝑡)=4𝛼𝑝(𝑝+1)2(𝑝+2)𝑡𝑝12(1𝛼)𝑝21𝑝2𝑡𝑝22(1𝛼)𝑝(1𝑝)2×(2𝑝)𝑡𝑝312(1𝛼)(1𝑝)log𝑡(1𝛼)(𝑝+1)(𝑝+2)2×(𝑝+3)𝑡𝑝𝑝+(𝑝+1)3𝛼𝑝319𝛼𝑝2+3𝑝2𝑡34𝛼𝑝+2𝑝8𝛼𝑝1+𝑝(1𝛼)𝑝38𝑝2𝑡𝑝+4𝑝2+𝑝(1𝛼)(1𝑝)3+5𝑝2𝑡14𝑝+4𝑝34(1𝛼)𝑝𝑡1+4𝛼(1+𝑝)𝑡2+4(1𝛼)(74𝑝).(2.19)
(1) If 𝛼(1/4,1), then from (2.19) we note that 𝑓2(𝑡)=𝑔(𝑡)log𝑡+(𝑡),(2.20) where 𝑔(𝑡) and (𝑡) are defined as in Lemmas 2.1 and 2.2, respectively.
Lemmas 2.1 and 2.2 together with (2.20) imply that 𝑓2(𝑡) is strictly increasing in [1,+). Therefore, 𝑓(𝑡)>0 for 𝑡(1,+) follows from (2.14)–(2.18) and the monotonicity of 𝑓2(𝑡).
(2) If 𝛼(0,𝜆0], then from (2.19) we have 𝑓2(1)=1627(1𝛼)80𝛼2=+110𝛼1128027(1𝛼)𝛼𝜆03𝛼+345+8011𝑓160,(2.21)3(𝑡)=4𝛼𝑝(1𝑝)(𝑝+1)2(𝑝+2)𝑡22(1𝛼)𝑝21𝑝2(2𝑝)𝑡2(1𝛼)𝑝(1𝑝)2(2𝑝)(3𝑝)log𝑡(1𝛼)𝑝(𝑝+1)(𝑝+2)2(𝑝+3)𝑡3𝑝+(𝑝+1)4𝛼𝑝422𝛼𝑝3+2𝑝327𝛼𝑝2𝑝2𝑡2𝑝+18𝛼𝑝+8𝛼2𝑝+(1𝛼)𝑝412𝑝3+15𝑝2𝑝+8𝑝8𝑡+(1𝛼)(1𝑝)4+4𝑝335𝑝2+50𝑝12+12(1𝛼)(1𝑝)𝑡3𝑝+4𝑝(1𝛼)𝑡2𝑝8𝛼(1+𝑝)𝑡1𝑝,𝑓3(1)=3281(1𝛼)(14𝛼)80𝛼2=+110𝛼110240181(1𝛼)4𝛼𝛼𝜆03𝛼+345+8011𝑓160,(2.22)3(𝑡)=8𝛼𝑝(1𝑝)(𝑝+1)2(𝑝+2)𝑡2(1𝛼)𝑝21𝑝2(2𝑝)log𝑡3(1𝛼)𝑝(𝑝+1)(𝑝+2)2×(𝑝+3)𝑡2(𝑝+1)3𝛼𝑝4𝑝4+26𝛼𝑝32𝑝3+25𝛼𝑝2+𝑝2𝑡𝑝22𝛼𝑝+2𝑝8𝛼(1𝛼)𝑝4+8𝑝317𝑝24𝑝+82(1𝛼)𝑝(1𝑝)2(2𝑝)(3𝑝)𝑡1+12(1𝛼)(1𝑝)(3𝑝)𝑡2𝑝+4(1𝛼)𝑝(2𝑝)𝑡1𝑝8𝛼1𝑝2𝑡𝑝,𝑓38(1)=243(1𝛼)3328𝛼4+128𝛼37248𝛼2<8+7118𝛼167243(1𝛼)3328𝜆40+128𝜆30+7118𝜆0<8167243(1𝛼)3328×(0.01)4+128×(0.01)3𝑓+7118×0.01167<0,(2.23)3(𝑡)=8𝛼𝑝(1𝑝)(𝑝+1)2(𝑝+2)log𝑡6(1𝛼)𝑝(𝑝+1)(𝑝+2)2(𝑝+3)𝑡2(1𝛼)𝑝21𝑝2(2𝑝)𝑡1+2(1𝛼)𝑝(1𝑝)2(2𝑝)(3𝑝)𝑡2+12(1𝛼)(1𝑝)(2𝑝)(3𝑝)𝑡1𝑝+4(1𝛼)𝑝(1𝑝)(2𝑝)𝑡𝑝+8𝛼𝑝1𝑝2𝑡1𝑝×2(𝑝+1)7𝛼𝑝4𝑝4+34𝛼𝑝32𝑝3+21𝛼𝑝2+𝑝2,𝑓30𝛼𝑝+2𝑝8𝛼38(1)=243(1𝛼)(74𝛼)256𝛼464𝛼31152𝛼2<8+2066𝛼53243(1𝛼)(74𝛼)256𝜆40+2066𝜆0<853[243(1𝛼)74𝛼)256×(0.01)4𝑓+2066×0.0153<0,(2.24)4(𝑡)=6(1𝛼)𝑝(𝑝+1)(𝑝+2)2(𝑝+3)𝑡𝑝+2+8𝛼𝑝(1𝑝)(𝑝+1)2(𝑝+2)𝑡𝑝+1+2(1𝛼)𝑝21𝑝2(2𝑝)𝑡𝑝4(1𝛼)𝑝(1𝑝)2(2𝑝)(3𝑝)𝑡𝑝1+12(1𝛼)(1𝑝)2(2𝑝)(3𝑝)𝑡24(1𝛼)𝑝2(1𝑝)(2𝑝)𝑡8𝛼𝑝(1𝑝)(1+𝑝)2,𝑓48(1)=(2431𝛼)1024𝛼4+21952𝛼310968𝛼2<8+13474𝛼835243(1𝛼)21952𝜆30+13474𝜆0<8835243(1𝛼)21952×(0.01)3𝑓+13474×0.01835<0,(2.25)4(𝑡)=6(1𝛼)𝑝(𝑝+1)(𝑝+2)3(𝑝+3)𝑡𝑝+1+8𝛼𝑝(1𝑝)(𝑝+1)3(𝑝+2)𝑡𝑝+2(1𝛼)𝑝31𝑝2(2𝑝)𝑡𝑝1+4(1𝛼)𝑝(1𝑝)3(2𝑝)(3𝑝)𝑡𝑝2+24(1𝛼)(1𝑝)2(2𝑝)(3𝑝)𝑡4(1𝛼)𝑝2(𝑓1𝑝)(2𝑝),48(1)=729(1𝛼)(74𝛼)1024𝛼4+21952𝛼310968𝛼2<8+13474𝛼835(7291𝛼)(74𝛼)21952𝜆30+13474𝜆0<8835729(1𝛼)(74𝛼)21952×(0.01)3𝑓+13474×0.01835<0,(2.26)4(𝑡)=6(1𝛼)𝑝(𝑝+1)2(𝑝+2)3(𝑝+3)𝑡𝑝+8𝛼𝑝2(1𝑝)(𝑝+1)3(𝑝+2)𝑡𝑝12(1𝛼)𝑝3(1+𝑝)(1𝑝)2(2𝑝)𝑡𝑝24(1𝛼)𝑝(1𝑝)3(2𝑝)2(3𝑝)𝑡𝑝3+24(1𝛼)(1𝑝)2(𝑓2𝑝)(3𝑝),4(1)=322187(1𝛼)4096𝛼6+136320𝛼5241440𝛼4+383672𝛼3209850𝛼2<+100113𝛼7255322187(1𝛼)136320𝜆50+383672𝜆30+100113𝜆0<7255322187(1𝛼)136320×(0.01)5+383672×(0.01)3𝑓+100113×0.017255<0,(2.27)5(𝑡)=6(1𝛼)𝑝2(𝑝+1)2(𝑝+2)3(𝑝+3)𝑡38𝛼𝑝2(1𝑝)2(𝑝+1)3(𝑝+2)𝑡2+2(1𝛼)𝑝3(1+𝑝)(1𝑝)2(2𝑝)2𝑡+4(1𝛼)𝑝(1𝑝)3(2𝑝)2(3𝑝)2,𝑓5(1)=326561(1𝛼)(14𝛼)4096𝛼6+173568𝛼5190368𝛼4+439136𝛼3191370𝛼2<+96723𝛼8665326561(1𝛼)(14𝛼)173568𝜆50+439136𝜆30+96723𝜆0<8665326561(1𝛼)(14𝛼)173568×(0.01)5+439136×(0.01)3𝑓+96723×0.018665<0,(2.28)5(𝑡)=18(1𝛼)𝑝2(𝑝+1)2(𝑝+2)3(𝑝+3)𝑡216𝛼𝑝2(1𝑝)2(𝑝+1)3(𝑝+2)𝑡+2(1𝛼)𝑝3(1+𝑝)(1𝑝)2(2𝑝)2,𝑓5(1)=326561(1𝛼)2(14𝛼)217408𝛼4+69920𝛼3119136𝛼2<+95282𝛼30845326561(1𝛼)(14𝛼)269920𝜆30+95282𝜆0<30845326561(1𝛼)(14𝛼)269920×(0.01)3𝑓+95282×0.0130845<0,(2.29)5(𝑡)=36(1𝛼)𝑝2(𝑝+1)2(𝑝+2)3(𝑝+3)𝑡16𝛼𝑝2(1𝑝)2(𝑝+1)3𝑓(𝑝+2),5(1)=1286561(1𝛼)3(14𝛼)2(74𝛼)160𝛼31856𝛼2<+3370𝛼22051286561(1𝛼)3(14𝛼)2(74𝛼)160𝜆30+3370𝜆0<22051286561(1𝛼)3(14𝛼)2(74𝛼)160×(0.01)3𝑓+3370×0.012205<0,5(𝑡)=36(1𝛼)𝑝2(𝑝+1)2(𝑝+2)3(𝑝+3)=128(72952𝛼)(14𝛼)2(1𝛼)3(74𝛼)3<0.(2.30)
Inequalities (2.30) imply that 𝑓5(𝑡) is strictly decreasing in [1,+). Then (2.29) leads to the conclusion that 𝑓5(𝑡) is strictly decreasing in [1,+).
It follows from (2.28) and the monotonicity of 𝑓5(𝑡) that 𝑓4(𝑡) is strictly decreasing in [1,+). Then inequalities (2.25)–(2.27) lead to the conclusion that 𝑓4(𝑡)<0 for 𝑡[1,+). Thus, 𝑓3(𝑡) is strictly decreasing in [1,+).
From inequalities (2.22)–(2.24) and the monotonicity of 𝑓3(𝑡) we clearly see that 𝑓3(𝑡)<0 for 𝑡(1,+). Thus, 𝑓2(𝑡) is strictly decreasing in [1,+).
It follows from (2.17) and (2.18) and inequality (2.21) together with the monotonicity of 𝑓2(𝑡) that 𝑓2(𝑡)<0 for 𝑡(1,+), which implies that 𝑓1(𝑡) is strictly decreasing in [1,+).
Therefore, 𝑓(𝑡)<0 for 𝑡(1,+) follows from (2.14)–(2.16) and the monotonicity of 𝑓1(𝑡).

Lemma 2.4. 3𝑡44𝑡(2𝑡2𝑡+2)log𝑡3>0 for 𝑡>1.

Proof. Let 𝐽(𝑡)=3𝑡44𝑡2𝑡2𝑡+2log𝑡3.(2.31) Then simple computations lead to 𝐽𝐽(1)=0,(𝑡)=43𝑡32𝑡2+𝑡283𝑡2𝐽𝑡+1log𝑡,𝐽(1)=0,4(𝑡)=𝑡𝐽1(𝑡),(2.32) where 𝐽1(𝑡)=9𝑡310𝑡2+3𝑡22(6𝑡1)𝑡log𝑡, 𝐽1𝐽(1)=0,(2.33)1(𝑡)=27𝑡2𝐽32𝑡+52(12𝑡1)log𝑡,1(𝐽1)=0,(2.34)12(𝑡)=𝑡𝐽2(𝑡),(2.35) where 𝐽2(𝑡)=27𝑡212𝑡log𝑡28𝑡+1, 𝐽2𝐽(1)=0,2(𝑡)=54𝑡12log𝑡40>0(2.36) for 𝑡>1.

Therefore, Lemma 2.4 follows from (2.31)–(2.36).

3. Proof of Theorem 1.1

Proof of Theorem 1.1. For all 𝑎,𝑏>0 with 𝑎𝑏, we first prove that 𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)>𝑀(14𝛼)/3(𝑎,𝑏)(3.1) for 𝛼[1/4,1), 𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)<𝑀(14𝛼)/3(𝑎,𝑏)(3.2) for 𝛼(0,3345/8011/16).
Without loss of generality, we assume that 𝑎>𝑏,𝑡=𝑎/𝑏>1 and 𝑝=(14𝛼)/3. We divide the proof into two cases.
Case 1 (𝛼=1/4). Let 𝑥=𝑡=𝑎/𝑏>1. Then we clearly see that 𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)𝑀(14𝛼)/31(𝑎,𝑏)=4[]𝐻(𝑎,𝑏)+3𝐿(𝑎,𝑏)=𝑏𝑎𝑏3𝑥44𝑥2𝑥2𝑥+2log𝑥38𝑥2.+1log𝑥(3.3) Therefore, inequality (3.1) follows from (3.3) and Lemma 2.4.
Case 2 (𝛼(0,3345/8011/16)(1/4,1)). Then we have 𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)𝑀(14𝛼)/3(𝑎,𝑏)=𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)𝑀𝑝(𝑎,𝑏)=𝑏2𝛼𝑡+𝑡+1(1𝛼)(𝑡1)𝑡log𝑡𝑝+121/𝑝.(3.4)
Let 𝐹(𝑡)=log2𝛼𝑡+𝑡+1(1𝛼)(𝑡1)1log𝑡𝑝𝑡log𝑝+12.(3.5) Then simple computations lead to lim𝑡1𝐹𝐹(𝑡)=0,(𝑡)=𝑓(𝑡)𝑡(𝑡+1)(𝑡𝑝𝑡+1)2𝛼𝑡log𝑡+(1𝛼)2,1log𝑡(3.6) where 𝑓(𝑡) is defined as in Lemma 2.3.
If 𝛼(1/4,1), then inequality (3.1) follows from (3.4)–(3.6) and Lemma 2.3(1). If 𝛼(0,3345/8011/16), then inequality (3.2) follows from (3.4)–(3.6) and Lemma 2.3(2).
Next, we prove that the parameter (14𝛼)/3 in inequalities (3.1) and (3.2) is the best possible.
For any 𝛼(0,3345/8011/16)(1/4,1), 𝑝0, and 𝑥>0, one has 𝛼𝐻(1,1+𝑥)+(1𝛼)𝐿(1,1+𝑥)𝑀𝑝(1,1+𝑥)=𝑄(𝑥)21/𝑝,(1+𝑥/2)log(1+𝑥)(3.7) where 𝑄(𝑥)=21/𝑝𝛼(1+𝑥)log(1+𝑥)+21/𝑝(1𝛼)𝑥(1+𝑥/2)(1+𝑥/2)log(1+𝑥)[1+(1+𝑥)𝑝]1/𝑝.
Letting 𝑥0 and making use of Taylor expansion, we get 2𝑄(𝑥)=1/𝑝814𝛼3𝑥𝑝3𝑥+𝑜3.(3.8)
If 𝛼[1/4,1) and 𝑝>(14𝛼)/3, then (3.7) and (3.8) imply that there exists 𝛿1=𝛿1(𝛼,𝑝)>0 such that 𝛼𝐻(1,1+𝑥)+(1𝛼)𝐿(1,1+𝑥)<𝑀𝑝(1,1+𝑥) for 𝑥(0,𝛿1). If 𝛼(0,3345/8011/16) and 𝑝<(14𝛼)/3, then (3.7) and (3.8) imply that there exists 𝛿2=𝛿2(𝛼,𝑝)>0 such that 𝛼𝐻(1,1+𝑥)+(1𝛼)𝐿(1,1+𝑥)>𝑀𝑝(1,1+𝑥) for 𝑥(0,𝛿2).
Finally, we prove that there exist 𝑎1,𝑏1,𝑎2,𝑏2>0 with 𝑎1𝑏1 and 𝑎2𝑏2 such that 𝛼𝐻(𝑎1,𝑏1)+(1𝛼)𝐿(𝑎1,𝑏1)<𝑀(14𝛼)/3(𝑎1,𝑏1) and 𝛼𝐻(𝑎2,𝑏2)+(1𝛼)𝐿(𝑎2,𝑏2)>𝑀(14𝛼)/3(𝑎2,𝑏2) for any 3345/8011/16<𝛼<1/4.
If 3345/8011/16<𝛼<1/4, then from the expression of 𝑓2(1) in (2.21) we clearly see that 𝑓2(1)>0, which leads to the conclusion that there exists 𝜆>1 such that 𝑓2(𝑡)>0(3.9) for 𝑡[1,𝜆).
From (2.14)–(2.18) and inequality (3.9) we know that 𝑓(𝑡)>0(3.10) for 𝑡(1,𝜆). Equations (3.4)–(3.6) and inequality (3.10) lead to the conclusion that 𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)>𝑀(14𝛼)/3(𝑎,𝑏) for all 𝑎/𝑏(1,𝜆).
On the other hand, simple computations lead to lim𝑥+𝑀(14𝛼)/3(1,𝑥)𝛼𝐻(1,𝑥)+(1𝛼)𝐿(1,𝑥)=23/(4𝛼1)lim𝑥+1+𝑥(4𝛼1)/33/(14𝛼)2𝛼/(𝑥+1)+(1𝛼)(11/𝑥)/log𝑥=+.(3.11)
Equation (3.11) implies that there exists 𝑋=𝑋(𝛼)>1 such that 𝛼𝐻(𝑎,𝑏)+(1𝛼)𝐿(𝑎,𝑏)<𝑀(14𝛼)/3(𝑎,𝑏) for all 𝑎/𝑏(𝑋,+).

Acknowledgment

This work was supported by the Natural Science Foundation of Zhejiang Broadcast and TV University (Grant no. XKT-09G21).