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Journal of Applied Mathematics
VolumeΒ 2012Β (2012), Article IDΒ 471096, 14 pages
http://dx.doi.org/10.1155/2012/471096
Research Article

Inequalities between Power Means and Convex Combinations of the Harmonic and Logarithmic Means

1School of Distance Education, Huzhou Broadcast and TV University, Huzhou 313000, China
2Department of Mathematics, Hangzhou Normal University, Hangzhou 310036, China

Received 10 October 2011; Accepted 1 December 2011

Academic Editor: Md. SazzadΒ Chowdhury

Copyright Β© 2012 Wei-Mao Qian and Zhong-Hua Shen. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We prove that 𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)>𝑀(1βˆ’4𝛼)/3(π‘Ž,𝑏) for π›Όβˆˆ(0,1) and all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if π›Όβˆˆ[1/4,1) and 𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)<𝑀(1βˆ’4𝛼)/3(π‘Ž,𝑏) if and only if βˆšπ›Όβˆˆ(0,3345/80βˆ’11/16), and the parameter (1βˆ’4𝛼)/3 is the best possible in either case. Here, 𝐻(π‘Ž,𝑏)=2π‘Žπ‘/(π‘Ž+𝑏), 𝐿(π‘Ž,𝑏)=(π‘Žβˆ’π‘)/(logπ‘Žβˆ’log𝑏), and 𝑀𝑝(π‘Ž,𝑏)=((π‘Žπ‘+𝑏𝑝)/2)1/𝑝(𝑝≠0) and 𝑀0√(π‘Ž,𝑏)=π‘Žπ‘ are the harmonic, logarithmic, and pth power means of a and b, respectively.

1. Introduction

The classical logarithmic mean 𝐿(π‘Ž,𝑏) of two positive real numbers π‘Ž and 𝑏 with π‘Žβ‰ π‘ is defined by 𝐿(π‘Ž,𝑏)=π‘Žβˆ’π‘.logπ‘Žβˆ’log𝑏(1.1)

In the recent past, the bivariate means have been the subject of intensive research. In particular, many remarkable inequalities for 𝐿(π‘Ž,𝑏) can be found in the literature [1–21]. 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 the logarithmic mean, which appears in a heat conduction problem. A representation of 𝐿(π‘Ž,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(π‘Ž,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(π‘Žπ‘–,𝑏𝑖), where 0<π‘Ž1<π‘Ž2<β‹―<π‘Žπ‘› and 0<𝑏1<𝑏2<β‹―<𝑏𝑛, is positive for all 𝑛β‰₯1.

Let √𝐺(π‘Ž,𝑏)=π‘Žπ‘, 𝐻(π‘Ž,𝑏)=2π‘Žπ‘/(π‘Ž+𝑏), 𝐼(π‘Ž,𝑏)=1/𝑒(π‘Žπ‘Ž/𝑏𝑏)1/(π‘Žβˆ’π‘), 𝐴(π‘Ž,𝑏)=(π‘Ž+𝑏)/2, 𝑀𝑝(π‘Ž,𝑏)=((π‘Žπ‘+𝑏𝑝)/2)1/𝑝(𝑝≠0) and 𝑀0√(π‘Ž,𝑏)=π‘Žπ‘, and 𝐿𝑝(π‘Ž,𝑏)=(π‘Žπ‘+1+𝑏𝑝+1)/(π‘Žπ‘+𝑏𝑝) be the geometric, harmonic, identric, arithmetic, 𝑝th power, and 𝑝th Lehmer means of two positive numbers π‘Ž and 𝑏, respectively. Then it is well known that both 𝑀𝑝(π‘Ž,𝑏) and 𝐿𝑝(π‘Ž,𝑏) are continuous and strictly increasing with respect to π‘βˆˆβ„ for fixed π‘Ž,𝑏>0 with π‘Žβ‰ π‘, and the inequalitiesmin{π‘Ž,𝑏}<𝐻(π‘Ž,𝑏)=π‘€βˆ’1(π‘Ž,𝑏)=πΏβˆ’1(π‘Ž,𝑏)<𝐺(π‘Ž,𝑏)=𝑀0(π‘Ž,𝑏)=πΏβˆ’1/2(π‘Ž,𝑏)<𝐿(π‘Ž,𝑏)<𝐼(π‘Ž,𝑏)<𝐴(π‘Ž,𝑏)=𝑀1(π‘Ž,𝑏)=𝐿0(π‘Ž,𝑏)<max{π‘Ž,𝑏}(1.2) hold for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

In [4], Carlson proves that the double inequality 𝐺(π‘Ž,𝑏)(𝐴(π‘Ž,𝑏)+𝐺(π‘Ž,𝑏))21<𝐿(π‘Ž,𝑏)<2(𝐴(π‘Ž,𝑏)+𝐺(π‘Ž,𝑏))(1.3) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

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

In [9], SΓ‘ndor establishes that √𝐺(π‘Ž,𝑏)𝐼(π‘Ž,𝑏)<𝐿(π‘Ž,𝑏)<𝐴(π‘Ž,𝑏)+𝐺(π‘Ž,𝑏)βˆ’πΌ(π‘Ž,𝑏)(1.5) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

In [27], Alzer gives the optimal Lehmer mean bounds for 𝐿, (𝐿𝐼)1/2, and (𝐿+𝐼)/2 as follows: πΏβˆ’1/3(π‘Ž,𝑏)<𝐿(π‘Ž,𝑏)<𝐿0𝐿(π‘Ž,𝑏),βˆ’1/4√(π‘Ž,𝑏)<𝐿(π‘Ž,𝑏)𝐼(π‘Ž,𝑏)<𝐿0𝐿(π‘Ž,𝑏),βˆ’1/41(a,𝑏)<2(𝐿(π‘Ž,𝑏)+𝐼(π‘Ž,𝑏))<𝐿0(π‘Ž,𝑏)(1.6) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

The following sharp bounds for (𝐿𝐼)1/2 and (𝐿+𝐼)/2 in terms of power mean are presented in [28]: 𝑀0√(π‘Ž,𝑏)<𝐿(π‘Ž,𝑏)𝐼(π‘Ž,𝑏)<𝑀1/2𝑀(π‘Ž,𝑏),log2/(1+log2)1(π‘Ž,𝑏)<2(𝐿(π‘Ž,𝑏)+𝐼(π‘Ž,𝑏))<𝑀1/2(π‘Ž,𝑏)(1.7) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

In [29, 30], the authors obtain the sharp bounds for the products 𝐴𝛼(π‘Ž,𝑏)𝐿1βˆ’π›Ό(π‘Ž,𝑏) and 𝐺𝛼(π‘Ž,𝑏)𝐿1βˆ’π›Ό(π‘Ž,𝑏) and the sum 𝛼𝐴(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏) in terms of power mean as follows: 𝑀0(π‘Ž,𝑏)<𝐴𝛼(π‘Ž,𝑏)𝐿1βˆ’π›Ό(π‘Ž,𝑏)<𝑀(1+2𝛼)/3𝑀(π‘Ž,𝑏),0(π‘Ž,𝑏)<𝐺𝛼(π‘Ž,𝑏)𝐿1βˆ’π›Ό(π‘Ž,𝑏)<𝑀(1βˆ’π›Ό)/3𝑀(π‘Ž,𝑏),log2/(log2βˆ’log𝛼)(π‘Ž,𝑏)<𝛼𝐴(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)<𝑀(1+2𝛼)/3(π‘Ž,𝑏)(1.8) for any π›Όβˆˆ(0,1) and all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

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

In [32], Chu et al. prove that the double inequality 𝛼𝐴(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐻(π‘Ž,𝑏)<𝑃(π‘Ž,𝑏)<𝛽𝐴(π‘Ž,𝑏)+(1βˆ’π›½)𝐻(π‘Ž,𝑏) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if 𝛼≀2/πœ‹ and 𝛽β‰₯5/6.

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 π›Όβˆˆ(0,1) and all π‘Ž,𝑏>0 with π‘Žβ‰ π‘, one has(1)𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)>𝑀(1βˆ’4𝛼)/3(π‘Ž,𝑏) if and only if π›Όβˆˆ[1/4,1);(2)𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)<𝑀(1βˆ’4𝛼)/3(π‘Ž,𝑏) if and only if βˆšπ›Όβˆˆ(0,3345/80βˆ’11/16).
In particular, the parameter (1βˆ’4𝛼)/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 π›Όβˆˆ(1/4,1),𝑝=(1βˆ’4𝛼)/3∈(βˆ’1,0), and 𝑔(𝑑)=βˆ’4𝛼𝑝(𝑝+1)2(𝑝+2)π‘‘π‘βˆ’1+2(1βˆ’π›Ό)𝑝2(1βˆ’π‘2)π‘‘π‘βˆ’2+2(1βˆ’π›Ό)𝑝(1βˆ’π‘)2(2βˆ’π‘)π‘‘π‘βˆ’3+12(1βˆ’π›Ό)(1βˆ’π‘). Then 𝑔(𝑑)>0 for π‘‘βˆˆ[1,+∞).

Proof. Simple computations lead to 𝑔(1)=6481(1βˆ’π›Ό)2ξ€·56𝛼2ξ€Έ+23𝛼+11>0,(2.1)lim𝑑→+βˆžπ‘”π‘”(𝑑)=12(1βˆ’π›Ό)(1βˆ’π‘)=8(1βˆ’π›Ό)(1+2𝛼)>0,(2.2)β€²(𝑑)=βˆ’2𝑝(1βˆ’π‘)π‘‘π‘βˆ’4𝑔1(𝑑),(2.3) where 𝑔1(𝑑)=βˆ’2𝛼(𝑝+1)2(𝑝+2)𝑑2𝑔+(1βˆ’π›Ό)𝑝(𝑝+1)(2βˆ’π‘)𝑑+(1βˆ’π›Ό)(1βˆ’π‘)(2βˆ’π‘)(3βˆ’π‘),14(1)=ξ€·27(1βˆ’π›Ό)148𝛼2ξ€Έβˆ’11𝛼+25>0,(2.4)lim𝑑→+βˆžπ‘”1𝑔(𝑑)=βˆ’βˆž,(2.5)ξ…ž1(𝑑)=βˆ’4𝛼(𝑝+1)24(𝑝+2)𝑑+(1βˆ’π›Ό)𝑝(𝑝+1)(2βˆ’π‘)=βˆ’27(1βˆ’π›Ό)2[]16𝛼(7βˆ’4𝛼)𝑑+(4π›Όβˆ’1)(4𝛼+5)<0(2.6) for π‘‘βˆˆ[1,+∞).
Inequality (2.6) implies that 𝑔1(𝑑) is strictly decreasing in [1,+∞). Then (2.4) and (2.5) lead to the conclusion that there exists πœ†1>1 such that 𝑔1(𝑑)>0 for π‘‘βˆˆ[1,πœ†1) and 𝑔1(𝑑)<0 for π‘‘βˆˆ(πœ†1,+∞). It follows from (2.3) that 𝑔(𝑑) is strictly increasing in [1,πœ†1] and strictly decreasing in [πœ†1,+∞).
Therefore, Lemma 2.1 follows from (2.1) and (2.2) together with the piecewise monotonicity of 𝑔(𝑑).

Lemma 2.2. Let π›Όβˆˆ(1/4,1),𝑝=(1βˆ’4𝛼)/3∈(βˆ’1,0), and β„Ž(𝑑)=βˆ’(1βˆ’π›Ό)(𝑝+1)(𝑝+2)2(𝑝+3)𝑑𝑝+(𝑝+1)(𝑝3βˆ’π›Όπ‘3βˆ’19𝛼𝑝2+3𝑝2βˆ’34𝛼𝑝+2π‘βˆ’8𝛼)π‘‘π‘βˆ’1+(1βˆ’π›Ό)𝑝(𝑝3βˆ’8𝑝2βˆ’π‘+4)π‘‘π‘βˆ’2+(1βˆ’π›Ό)(1βˆ’π‘)(𝑝3+5𝑝2βˆ’14𝑝+4)π‘‘π‘βˆ’3+4(1βˆ’π›Ό)(7βˆ’4𝑝)βˆ’4𝑝(1βˆ’π›Ό)π‘‘βˆ’1+4𝛼(1+𝑝)π‘‘βˆ’2. Then β„Ž(𝑑)>0 for π‘‘βˆˆ[1,+∞).

Proof. Let β„Ž1(𝑑)=𝑑3βˆ’π‘β„Ž(𝑑).(2.7) Then simple computations lead to β„Ž1(1)=16ξ€·27(1βˆ’π›Ό)80𝛼2ξ€Έβ„Ž+110π›Όβˆ’1>0,(2.8)ξ…ž1(𝑑)=βˆ’3(1βˆ’π›Ό)(𝑝+1)(𝑝+2)2(𝑝+3)𝑑2×𝑝+2(𝑝+1)3βˆ’π›Όπ‘3βˆ’19𝛼𝑝2+3𝑝2ξ€Έξ€·π‘βˆ’34𝛼𝑝+2π‘βˆ’8𝛼𝑑+(1βˆ’π›Ό)𝑝3βˆ’8𝑝2ξ€Έβˆ’π‘+4+4(1βˆ’π›Ό)(7βˆ’4𝑝)(3βˆ’π‘)𝑑2βˆ’π‘βˆ’4𝑝(1βˆ’π›Ό)(2βˆ’π‘)𝑑1βˆ’π‘ξ€·+4𝛼1βˆ’π‘2ξ€Έπ‘‘βˆ’π‘,β„Žξ…ž1(1)=32(ξ€·271βˆ’π›Ό)βˆ’16𝛼3+38𝛼2ξ€Έβ„Ž+176π›Όβˆ’9>0,(2.9)1ξ…žξ…ž(𝑑)=βˆ’6(1βˆ’π›Ό)(𝑝+1)(𝑝+2)2×𝑝(𝑝+3)𝑑+2(𝑝+1)3βˆ’π›Όπ‘3βˆ’19𝛼𝑝2+3𝑝2ξ€Έβˆ’34𝛼𝑝+2π‘βˆ’8𝛼+4(1βˆ’π›Ό)(7βˆ’4𝑝)(3βˆ’π‘)(2βˆ’π‘)𝑑1βˆ’π‘βˆ’4𝑝(1βˆ’π›Ό)(2βˆ’π‘)(1βˆ’π‘)π‘‘βˆ’π‘ξ€·βˆ’4𝛼𝑝1βˆ’π‘2ξ€Έπ‘‘βˆ’π‘βˆ’1,β„Ž1ξ…žξ…ž8(1)=ξ€·81(1βˆ’π›Ό)βˆ’128𝛼4+896𝛼3+288𝛼2ξ€Έβ„Ž+5294π›Όβˆ’437>0,(2.10)1ξ…žξ…žξ…ž(𝑑)=βˆ’6(1βˆ’π›Ό)(𝑝+1)(𝑝+2)2(𝑝+3)+4(1βˆ’π›Ό)(7βˆ’4𝑝)(3βˆ’π‘)(2βˆ’π‘)(1βˆ’π‘)π‘‘βˆ’π‘+4𝑝2(1βˆ’π›Ό)(2βˆ’π‘)(1βˆ’π‘)π‘‘βˆ’π‘βˆ’1+4𝛼𝑝(1+𝑝)2(1βˆ’π‘)π‘‘βˆ’π‘βˆ’2,β„Ž1ξ…žξ…žξ…ž8(1)=ξ€·81(1βˆ’π›Ό)576𝛼4+3872𝛼3+660𝛼2ξ€Έβ„Ž+6612π›Όβˆ’785>0,(2.11)1(4)(𝑑)=βˆ’4𝑝(1βˆ’π‘)π‘‘βˆ’π‘βˆ’3β„Ž2(𝑑),(2.12) where β„Ž2(𝑑)=(1βˆ’π›Ό)(7βˆ’4𝑝)(3βˆ’π‘)(2βˆ’π‘)𝑑2+(1βˆ’π›Ό)𝑝(2βˆ’π‘)(𝑝+1)𝑑+𝛼(𝑝+1)2β„Ž(𝑝+2),24(1)=ξ€·27(1βˆ’π›Ό)96𝛼3+232𝛼2ξ€Έβ„Ž+388𝛼+175>0,ξ…ž2(𝑑)=2(1βˆ’π›Ό)(7βˆ’4𝑝)(3βˆ’π‘)(2βˆ’π‘)𝑑+(1βˆ’π›Ό)𝑝(2βˆ’π‘)(𝑝+1)β‰₯β„Žξ…ž24(1)=9ξ€·(1βˆ’π›Ό)(5+4𝛼)12𝛼2ξ€Έ+31𝛼+23>0(2.13) for π‘‘βˆˆ[1,+∞).
From inequalities (2.13) we clearly see that β„Ž2(𝑑)>0 for π‘‘βˆˆ[1,+∞). Then (2.12) leads to the conclusion that β„Ž1ξ…žξ…žξ…ž(𝑑) is strictly increasing in [1,+∞).
Therefore, Lemma 2.2 follows from (2.7)–(2.11) and the monotonicity of β„Ž1ξ…žξ…žξ…ž(𝑑).

Lemma 2.3. Let π›Όβˆˆ(0,1),𝑝=(1βˆ’4𝛼)/3,πœ†0√=3345/80βˆ’11/16=0.00903…, and 𝑓(𝑑)=2𝛼(1βˆ’π‘‘π‘+1)𝑑log2𝑑+(1βˆ’π›Ό)(1+π‘‘π‘βˆ’1)(1+𝑑)2𝑑log𝑑+(1βˆ’π›Ό)(1+𝑑)2(1βˆ’π‘‘)(𝑑𝑝+1). Then the following two statements are true:(1)if π›Όβˆˆ(1/4,1), then 𝑓(𝑑)>0 for π‘‘βˆˆ(1,+∞);(2)if π›Όβˆˆ(0,πœ†0], then 𝑓(𝑑)<0 for π‘‘βˆˆ(1,+∞).

Proof. Let 𝑓1(𝑑)=π‘‘βˆ’π‘π‘“ξ…žξ…ž(𝑑), 𝑓2(𝑑)=𝑑𝑝+2𝑓1β€²(𝑑), 𝑓3(𝑑)=𝑑4βˆ’π‘π‘“2ξ…žξ…žξ…ž(𝑑), 𝑓4(𝑑)=𝑑𝑝+2𝑓3ξ…žξ…žξ…ž(𝑑) and 𝑓5(𝑑)=𝑑4βˆ’π‘π‘“4ξ…žξ…žξ…ž(𝑑). Then simple computations lead to 𝑓𝑓(1)=0,(2.14)ξ…žξ€Ί(𝑑)=2𝛼1βˆ’(𝑝+2)𝑑𝑝+1ξ€»log2𝑑+ξ€Ί(𝑝+2βˆ’π›Όπ‘βˆ’6𝛼)𝑑𝑝+1+2(1βˆ’π›Ό)(𝑝+1)𝑑𝑝+(1βˆ’π›Ό)π‘π‘‘π‘βˆ’1+3(1βˆ’π›Ό)𝑑2ξ€»ξ€Ί+4(1βˆ’π›Ό)𝑑+3𝛼+1logπ‘‘βˆ’(1βˆ’π›Ό)(𝑝+3)𝑑𝑝+2+(𝑝+1)𝑑𝑝+1βˆ’(𝑝+3)π‘‘π‘βˆ’(𝑝+1)π‘‘π‘βˆ’1+2𝑑2ξ€»,π‘“βˆ’2ξ…žπ‘“(1)=0,(2.15)1(𝑑)=βˆ’2𝛼(𝑝+1)(𝑝+2)log2𝑑+𝑝2βˆ’π›Όπ‘2ξ€Έ+3π‘βˆ’11π›Όπ‘βˆ’14𝛼+2+2(1βˆ’π›Ό)𝑝(𝑝+1)π‘‘βˆ’1βˆ’(1βˆ’π›Ό)𝑝(1βˆ’π‘)π‘‘βˆ’2+6(1βˆ’π›Ό)𝑑1βˆ’π‘+4(1βˆ’π›Ό)π‘‘βˆ’π‘+4π›Όπ‘π‘‘βˆ’1βˆ’π‘ξ€»ξ€·π‘logπ‘‘βˆ’(1βˆ’π›Ό)(𝑝+2)(𝑝+3)𝑑+(1βˆ’π›Ό)2𝑑+5𝑝+2βˆ’1𝑝+(1βˆ’π›Ό)2𝑑+π‘βˆ’1βˆ’2βˆ’(1βˆ’π›Ό)𝑑1βˆ’π‘+4(1βˆ’π›Ό)π‘‘βˆ’π‘+(1+3𝛼)π‘‘βˆ’1βˆ’π‘βˆ’(1βˆ’π›Ό)𝑝2π‘“βˆ’(1βˆ’π›Ό)π‘βˆ’5𝛼+1,1𝑓(1)=0,(2.16)2ξ€Ί(𝑑)=βˆ’4𝛼(𝑝+1)(𝑝+2)𝑑𝑝+1+2(1βˆ’π›Ό)𝑝(𝑝+1)π‘‘π‘βˆ’2(1βˆ’π›Ό)𝑝(1βˆ’π‘)π‘‘π‘βˆ’1βˆ’6(1βˆ’π›Ό)(1βˆ’π‘)𝑑2ξ€»+4(1βˆ’π›Ό)𝑝𝑑+4𝛼(1+𝑝)logπ‘‘βˆ’(1βˆ’π›Ό)(𝑝+2)(𝑝+3)𝑑𝑝+2+𝑝2βˆ’π›Όπ‘2𝑑+3π‘βˆ’11π›Όπ‘βˆ’14𝛼+2𝑝+1+𝑝(1βˆ’π›Ό)2ξ€Έπ‘‘βˆ’3π‘βˆ’2π‘βˆ’ξ€·π‘(1βˆ’π›Ό)2𝑑+3π‘βˆ’2π‘βˆ’1+(1βˆ’π›Ό)(𝑝+5)𝑑2𝑓+4(1βˆ’π›Ό)(1βˆ’π‘)𝑑+π›Όβˆ’3π›Όπ‘βˆ’π‘βˆ’1,2(𝑓1)=0,(2.17)ξ…ž2(𝑑)=βˆ’4𝛼(𝑝+1)2(𝑝+2)𝑑𝑝+2(1βˆ’π›Ό)𝑝2(𝑝+1)π‘‘π‘βˆ’1+2(1βˆ’π›Ό)𝑝(1βˆ’π‘)2π‘‘π‘βˆ’2]βˆ’12(1βˆ’π›Ό)(1βˆ’π‘)𝑑+4(1βˆ’π›Ό)𝑝logπ‘‘βˆ’(1βˆ’π›Ό)(𝑝+2)2(𝑝+3)𝑑𝑝+1𝑝+(𝑝+1)2βˆ’π›Όπ‘2ξ€Έπ‘‘βˆ’15𝛼𝑝+3π‘βˆ’22𝛼+2𝑝𝑝+(1βˆ’π›Ό)𝑝2ξ€Έπ‘‘βˆ’5π‘βˆ’4π‘βˆ’1𝑝+(1βˆ’π›Ό)(1βˆ’π‘)2𝑑+5π‘βˆ’2π‘βˆ’2+4(1βˆ’π›Ό)(4βˆ’π‘)π‘‘βˆ’4𝛼(1+𝑝)π‘‘βˆ’1𝑓+4(1βˆ’π›Ό)(1βˆ’2𝑝),ξ…ž2(𝑓1)=0,(2.18)2ξ…žξ…žξ€Ί(𝑑)=βˆ’4𝛼𝑝(𝑝+1)2(𝑝+2)π‘‘π‘βˆ’1βˆ’2(1βˆ’π›Ό)𝑝2ξ€·1βˆ’π‘2ξ€Έπ‘‘π‘βˆ’2βˆ’2(1βˆ’π›Ό)𝑝(1βˆ’π‘)2Γ—(2βˆ’π‘)π‘‘π‘βˆ’3ξ€»βˆ’12(1βˆ’π›Ό)(1βˆ’π‘)logπ‘‘βˆ’(1βˆ’π›Ό)(𝑝+1)(𝑝+2)2Γ—(𝑝+3)𝑑𝑝𝑝+(𝑝+1)3βˆ’π›Όπ‘3βˆ’19𝛼𝑝2+3𝑝2ξ€Έπ‘‘βˆ’34𝛼𝑝+2π‘βˆ’8π›Όπ‘βˆ’1+𝑝(1βˆ’π›Ό)𝑝3βˆ’8𝑝2ξ€Έπ‘‘βˆ’π‘+4π‘βˆ’2+𝑝(1βˆ’π›Ό)(1βˆ’π‘)3+5𝑝2ξ€Έπ‘‘βˆ’14𝑝+4π‘βˆ’3βˆ’4(1βˆ’π›Ό)π‘π‘‘βˆ’1+4𝛼(1+𝑝)π‘‘βˆ’2+4(1βˆ’π›Ό)(7βˆ’4𝑝).(2.19)
(1) If π›Όβˆˆ(1/4,1), then from (2.19) we note that 𝑓2ξ…žξ…ž(𝑑)=𝑔(𝑑)log𝑑+β„Ž(𝑑),(2.20) where 𝑔(𝑑) and β„Ž(𝑑) are defined as in Lemmas 2.1 and 2.2, respectively.
Lemmas 2.1 and 2.2 together with (2.20) imply that 𝑓2β€²(𝑑) is strictly increasing in [1,+∞). Therefore, 𝑓(𝑑)>0 for π‘‘βˆˆ(1,+∞) follows from (2.14)–(2.18) and the monotonicity of 𝑓2β€²(𝑑).
(2) If π›Όβˆˆ(0,πœ†0], then from (2.19) we have 𝑓2ξ…žξ…ž(1)=16ξ€·27(1βˆ’π›Ό)80𝛼2ξ€Έ=+110π›Όβˆ’11280ξ€·27(1βˆ’π›Ό)π›Όβˆ’πœ†03βˆšπ›Ό+345+8011ξƒͺ𝑓16≀0,(2.21)3ξ€Ί(𝑑)=4𝛼𝑝(1βˆ’π‘)(𝑝+1)2(𝑝+2)𝑑2βˆ’2(1βˆ’π›Ό)𝑝2ξ€·1βˆ’π‘2ξ€Έ(2βˆ’π‘)π‘‘βˆ’2(1βˆ’π›Ό)𝑝(1βˆ’π‘)2ξ€»(2βˆ’π‘)(3βˆ’π‘)logπ‘‘βˆ’(1βˆ’π›Ό)𝑝(𝑝+1)(𝑝+2)2(𝑝+3)𝑑3𝑝+(𝑝+1)4βˆ’π›Όπ‘4βˆ’22𝛼𝑝3+2𝑝3βˆ’27𝛼𝑝2βˆ’π‘2ξ€Έπ‘‘βˆ’2𝑝+18𝛼𝑝+8𝛼2𝑝+(1βˆ’π›Ό)𝑝4βˆ’12𝑝3+15𝑝2𝑝+8π‘βˆ’8𝑑+(1βˆ’π›Ό)(1βˆ’π‘)4+4𝑝3βˆ’35𝑝2ξ€Έ+50π‘βˆ’12+12(1βˆ’π›Ό)(1βˆ’π‘)𝑑3βˆ’π‘+4𝑝(1βˆ’π›Ό)𝑑2βˆ’π‘βˆ’8𝛼(1+𝑝)𝑑1βˆ’π‘,𝑓3(1)=32ξ€·81(1βˆ’π›Ό)(1βˆ’4𝛼)80𝛼2ξ€Έ=+110π›Όβˆ’110240ξ‚€181(1βˆ’π›Ό)4ξ‚ξ€·βˆ’π›Όπ›Όβˆ’πœ†03βˆšπ›Ό+345+8011ξƒͺ𝑓16≀0,(2.22)ξ…ž3ξ€Ί(𝑑)=8𝛼𝑝(1βˆ’π‘)(𝑝+1)2(𝑝+2)π‘‘βˆ’2(1βˆ’π›Ό)𝑝2ξ€·1βˆ’π‘2ξ€Έξ€»(2βˆ’π‘)logπ‘‘βˆ’3(1βˆ’π›Ό)𝑝(𝑝+1)(𝑝+2)2Γ—ξ€·(𝑝+3)π‘‘βˆ’2(𝑝+1)3𝛼𝑝4βˆ’π‘4+26𝛼𝑝3βˆ’2𝑝3+25𝛼𝑝2+𝑝2ξ€Έπ‘‘ξ€·π‘βˆ’22𝛼𝑝+2π‘βˆ’8π›Όβˆ’(1βˆ’π›Ό)𝑝4+8𝑝3βˆ’17𝑝2ξ€Έβˆ’4𝑝+8βˆ’2(1βˆ’π›Ό)𝑝(1βˆ’π‘)2(2βˆ’π‘)(3βˆ’π‘)π‘‘βˆ’1+12(1βˆ’π›Ό)(1βˆ’π‘)(3βˆ’π‘)𝑑2βˆ’π‘+4(1βˆ’π›Ό)𝑝(2βˆ’π‘)𝑑1βˆ’π‘ξ€·βˆ’8𝛼1βˆ’π‘2ξ€Έπ‘‘βˆ’π‘,π‘“ξ…ž38(1)=ξ€·243(1βˆ’π›Ό)3328𝛼4+128𝛼3βˆ’7248𝛼2ξ€Έ<8+7118π›Όβˆ’167ξ€·243(1βˆ’π›Ό)3328πœ†40+128πœ†30+7118πœ†0ξ€Έ<8βˆ’167ξ€Ί243(1βˆ’π›Ό)3328Γ—(0.01)4+128Γ—(0.01)3𝑓+7118Γ—0.01βˆ’167<0,(2.23)3ξ…žξ…ž(𝑑)=8𝛼𝑝(1βˆ’π‘)(𝑝+1)2(𝑝+2)logπ‘‘βˆ’6(1βˆ’π›Ό)𝑝(𝑝+1)(𝑝+2)2(𝑝+3)π‘‘βˆ’2(1βˆ’π›Ό)𝑝2ξ€·1βˆ’π‘2ξ€Έ(2βˆ’π‘)π‘‘βˆ’1+2(1βˆ’π›Ό)𝑝(1βˆ’π‘)2(2βˆ’π‘)(3βˆ’π‘)π‘‘βˆ’2+12(1βˆ’π›Ό)(1βˆ’π‘)(2βˆ’π‘)(3βˆ’π‘)𝑑1βˆ’π‘+4(1βˆ’π›Ό)𝑝(1βˆ’π‘)(2βˆ’π‘)π‘‘βˆ’π‘ξ€·+8𝛼𝑝1βˆ’π‘2ξ€Έπ‘‘βˆ’1βˆ’π‘Γ—ξ€·βˆ’2(𝑝+1)7𝛼𝑝4βˆ’π‘4+34𝛼𝑝3βˆ’2𝑝3+21𝛼𝑝2+𝑝2ξ€Έ,π‘“βˆ’30𝛼𝑝+2π‘βˆ’8𝛼3ξ…žξ…ž8(1)=ξ€·243(1βˆ’π›Ό)(7βˆ’4𝛼)256𝛼4βˆ’64𝛼3βˆ’1152𝛼2ξ€Έ<8+2066π›Όβˆ’53ξ€·243(1βˆ’π›Ό)(7βˆ’4𝛼)256πœ†40+2066πœ†0ξ€Έ<8βˆ’53[ξ€·243(1βˆ’π›Ό)7βˆ’4𝛼)256Γ—(0.01)4𝑓+2066Γ—0.01βˆ’53<0,(2.24)4(𝑑)=βˆ’6(1βˆ’π›Ό)𝑝(𝑝+1)(𝑝+2)2(𝑝+3)𝑑𝑝+2+8𝛼𝑝(1βˆ’π‘)(𝑝+1)2(𝑝+2)𝑑𝑝+1+2(1βˆ’π›Ό)𝑝2ξ€·1βˆ’π‘2ξ€Έ(2βˆ’π‘)π‘‘π‘βˆ’4(1βˆ’π›Ό)𝑝(1βˆ’π‘)2(2βˆ’π‘)(3βˆ’π‘)π‘‘π‘βˆ’1+12(1βˆ’π›Ό)(1βˆ’π‘)2(2βˆ’π‘)(3βˆ’π‘)𝑑2βˆ’4(1βˆ’π›Ό)𝑝2(1βˆ’π‘)(2βˆ’π‘)π‘‘βˆ’8𝛼𝑝(1βˆ’π‘)(1+𝑝)2,𝑓48(1)=(ξ€·2431βˆ’π›Ό)βˆ’1024𝛼4+21952𝛼3βˆ’10968𝛼2ξ€Έ<8+13474π›Όβˆ’835ξ€·243(1βˆ’π›Ό)21952πœ†30+13474πœ†0ξ€Έ<8βˆ’835ξ€Ί243(1βˆ’π›Ό)21952Γ—(0.01)3𝑓+13474Γ—0.01βˆ’835<0,(2.25)ξ…ž4(𝑑)=βˆ’6(1βˆ’π›Ό)𝑝(𝑝+1)(𝑝+2)3(𝑝+3)𝑑𝑝+1+8𝛼𝑝(1βˆ’π‘)(𝑝+1)3(𝑝+2)𝑑𝑝+2(1βˆ’π›Ό)𝑝3ξ€·1βˆ’π‘2ξ€Έ(2βˆ’π‘)π‘‘π‘βˆ’1+4(1βˆ’π›Ό)𝑝(1βˆ’π‘)3(2βˆ’π‘)(3βˆ’π‘)π‘‘π‘βˆ’2+24(1βˆ’π›Ό)(1βˆ’π‘)2(2βˆ’π‘)(3βˆ’π‘)π‘‘βˆ’4(1βˆ’π›Ό)𝑝2(𝑓1βˆ’π‘)(2βˆ’π‘),ξ…ž48(1)=ξ€·729(1βˆ’π›Ό)(7βˆ’4𝛼)βˆ’1024𝛼4+21952𝛼3βˆ’10968𝛼2ξ€Έ<8+13474π›Όβˆ’835(ξ€·7291βˆ’π›Ό)(7βˆ’4𝛼)21952πœ†30+13474πœ†0ξ€Έ<8βˆ’835ξ€Ί729(1βˆ’π›Ό)(7βˆ’4𝛼)21952Γ—(0.01)3𝑓+13474Γ—0.01βˆ’835<0,(2.26)4ξ…žξ…ž(𝑑)=βˆ’6(1βˆ’π›Ό)𝑝(𝑝+1)2(𝑝+2)3(𝑝+3)𝑑𝑝+8𝛼𝑝2(1βˆ’π‘)(𝑝+1)3(𝑝+2)π‘‘π‘βˆ’1βˆ’2(1βˆ’π›Ό)𝑝3(1+𝑝)(1βˆ’π‘)2(2βˆ’π‘)π‘‘π‘βˆ’2βˆ’4(1βˆ’π›Ό)𝑝(1βˆ’π‘)3(2βˆ’π‘)2(3βˆ’π‘)π‘‘π‘βˆ’3+24(1βˆ’π›Ό)(1βˆ’π‘)2(𝑓2βˆ’π‘)(3βˆ’π‘),4ξ…žξ…ž(1)=32ξ€·2187(1βˆ’π›Ό)βˆ’4096𝛼6+136320𝛼5βˆ’241440𝛼4+383672𝛼3βˆ’209850𝛼2ξ€Έ<+100113π›Όβˆ’725532ξ€·2187(1βˆ’π›Ό)136320πœ†50+383672πœ†30+100113πœ†0ξ€Έ<βˆ’725532ξ€Ί2187(1βˆ’π›Ό)136320Γ—(0.01)5+383672Γ—(0.01)3𝑓+100113Γ—0.01βˆ’7255<0,(2.27)5(𝑑)=βˆ’6(1βˆ’π›Ό)𝑝2(𝑝+1)2(𝑝+2)3(𝑝+3)𝑑3βˆ’8𝛼𝑝2(1βˆ’π‘)2(𝑝+1)3(𝑝+2)𝑑2+2(1βˆ’π›Ό)𝑝3(1+𝑝)(1βˆ’π‘)2(2βˆ’π‘)2𝑑+4(1βˆ’π›Ό)𝑝(1βˆ’π‘)3(2βˆ’π‘)2(3βˆ’π‘)2,𝑓5(1)=32ξ€·6561(1βˆ’π›Ό)(1βˆ’4𝛼)βˆ’4096𝛼6+173568𝛼5βˆ’190368𝛼4+439136𝛼3βˆ’191370𝛼2ξ€Έ<+96723π›Όβˆ’866532ξ€·6561(1βˆ’π›Ό)(1βˆ’4𝛼)173568πœ†50+439136πœ†30+96723πœ†0ξ€Έ<βˆ’866532ξ€Ί6561(1βˆ’π›Ό)(1βˆ’4𝛼)173568Γ—(0.01)5+439136Γ—(0.01)3𝑓+96723Γ—0.01βˆ’8665<0,(2.28)ξ…ž5(𝑑)=βˆ’18(1βˆ’π›Ό)𝑝2(𝑝+1)2(𝑝+2)3(𝑝+3)𝑑2βˆ’16𝛼𝑝2(1βˆ’π‘)2(𝑝+1)3(𝑝+2)𝑑+2(1βˆ’π›Ό)𝑝3(1+𝑝)(1βˆ’π‘)2(2βˆ’π‘)2,π‘“ξ…ž5(1)=326561(1βˆ’π›Ό)2(1βˆ’4𝛼)2ξ€·βˆ’17408𝛼4+69920𝛼3βˆ’119136𝛼2ξ€Έ<+95282π›Όβˆ’30845326561(1βˆ’π›Ό)(1βˆ’4𝛼)2ξ€·69920πœ†30+95282πœ†0ξ€Έ<βˆ’30845326561(1βˆ’π›Ό)(1βˆ’4𝛼)2ξ€·69920Γ—(0.01)3𝑓+95282Γ—0.01βˆ’30845<0,(2.29)5ξ…žξ…ž(𝑑)=βˆ’36(1βˆ’π›Ό)𝑝2(𝑝+1)2(𝑝+2)3(𝑝+3)π‘‘βˆ’16𝛼𝑝2(1βˆ’π‘)2(𝑝+1)3𝑓(𝑝+2),5ξ…žξ…ž(1)=1286561(1βˆ’π›Ό)3(1βˆ’4𝛼)2ξ€·(7βˆ’4𝛼)160𝛼3βˆ’1856𝛼2ξ€Έ<+3370π›Όβˆ’22051286561(1βˆ’π›Ό)3(1βˆ’4𝛼)2ξ€·(7βˆ’4𝛼)160πœ†30+3370πœ†0ξ€Έ<βˆ’22051286561(1βˆ’π›Ό)3(1βˆ’4𝛼)2ξ€·(7βˆ’4𝛼)160Γ—(0.01)3𝑓+3370Γ—0.01βˆ’2205<0,5ξ…žξ…žξ…ž(𝑑)=βˆ’36(1βˆ’π›Ό)𝑝2(𝑝+1)2(𝑝+2)3(𝑝+3)=βˆ’128(7295βˆ’2𝛼)(1βˆ’4𝛼)2(1βˆ’π›Ό)3(7βˆ’4𝛼)3<0.(2.30)
Inequalities (2.30) imply that 𝑓5β€²(𝑑) is strictly decreasing in [1,+∞). Then (2.29) leads to the conclusion that 𝑓5(𝑑) is strictly decreasing in [1,+∞).
It follows from (2.28) and the monotonicity of 𝑓5(𝑑) that 𝑓4ξ…žξ…ž(𝑑) is strictly decreasing in [1,+∞). Then inequalities (2.25)–(2.27) lead to the conclusion that 𝑓4(𝑑)<0 for π‘‘βˆˆ[1,+∞). Thus, 𝑓3ξ…žξ…ž(𝑑) is strictly decreasing in [1,+∞).
From inequalities (2.22)–(2.24) and the monotonicity of 𝑓3ξ…žξ…ž(𝑑) we clearly see that 𝑓3(𝑑)<0 for π‘‘βˆˆ(1,+∞). Thus, 𝑓2ξ…žξ…ž(𝑑) is strictly decreasing in [1,+∞).
It follows from (2.17) and (2.18) and inequality (2.21) together with the monotonicity of 𝑓2ξ…žξ…ž(𝑑) that 𝑓2(𝑑)<0 for π‘‘βˆˆ(1,+∞), which implies that 𝑓1(𝑑) is strictly decreasing in [1,+∞).
Therefore, 𝑓(𝑑)<0 for π‘‘βˆˆ(1,+∞) follows from (2.14)–(2.16) and the monotonicity of 𝑓1(𝑑).

Lemma 2.4. 3𝑑4βˆ’4𝑑(2𝑑2βˆ’π‘‘+2)logπ‘‘βˆ’3>0 for 𝑑>1.

Proof. Let 𝐽(𝑑)=3𝑑4ξ€·βˆ’4𝑑2𝑑2ξ€Έβˆ’π‘‘+2logπ‘‘βˆ’3.(2.31) Then simple computations lead to 𝐽𝐽(1)=0,ξ…žξ€·(𝑑)=43𝑑3βˆ’2𝑑2ξ€Έξ€·+π‘‘βˆ’2βˆ’83𝑑2ξ€Έπ½βˆ’π‘‘+1log𝑑,ξ…žπ½(1)=0,ξ…žξ…ž4(𝑑)=𝑑𝐽1(𝑑),(2.32) where 𝐽1(𝑑)=9𝑑3βˆ’10𝑑2+3π‘‘βˆ’2βˆ’2(6π‘‘βˆ’1)𝑑log𝑑, 𝐽1𝐽(1)=0,(2.33)ξ…ž1(𝑑)=27𝑑2π½βˆ’32𝑑+5βˆ’2(12π‘‘βˆ’1)log𝑑,ξ…ž1(𝐽1)=0,(2.34)1ξ…žξ…ž2(𝑑)=𝑑𝐽2(𝑑),(2.35) where 𝐽2(𝑑)=27𝑑2βˆ’12𝑑logπ‘‘βˆ’28𝑑+1, 𝐽2𝐽(1)=0,ξ…ž2(𝑑)=54π‘‘βˆ’12logπ‘‘βˆ’40>0(2.36) for 𝑑>1.

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

3. Proof of Theorem 1.1

Proof of Theorem 1.1. For all π‘Ž,𝑏>0 with π‘Žβ‰ π‘, we first prove that 𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)>𝑀(1βˆ’4𝛼)/3(π‘Ž,𝑏)(3.1) for π›Όβˆˆ[1/4,1), 𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)<𝑀(1βˆ’4𝛼)/3(π‘Ž,𝑏)(3.2) for βˆšπ›Όβˆˆ(0,3345/80βˆ’11/16).
Without loss of generality, we assume that π‘Ž>𝑏,𝑑=π‘Ž/𝑏>1 and 𝑝=(1βˆ’4𝛼)/3. We divide the proof into two cases.
Case 1 (𝛼=1/4). Let √π‘₯=βˆšπ‘‘=π‘Ž/𝑏>1. Then we clearly see that 𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)βˆ’π‘€(1βˆ’4𝛼)/31(π‘Ž,𝑏)=4[]βˆ’βˆšπ»(π‘Ž,𝑏)+3𝐿(π‘Ž,𝑏)=π‘ξ€Ίπ‘Žπ‘3π‘₯4ξ€·βˆ’4π‘₯2π‘₯2ξ€Έξ€»βˆ’π‘₯+2logπ‘₯βˆ’38ξ€·π‘₯2ξ€Έ.+1logπ‘₯(3.3) Therefore, inequality (3.1) follows from (3.3) and Lemma 2.4.
Case 2 (βˆšπ›Όβˆˆ(0,3345/80βˆ’11/16)βˆͺ(1/4,1)). Then we have 𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)βˆ’π‘€(1βˆ’4𝛼)/3(π‘Ž,𝑏)=𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)βˆ’π‘€π‘ξƒ¬(π‘Ž,𝑏)=𝑏2𝛼𝑑+𝑑+1(1βˆ’π›Ό)(π‘‘βˆ’1)βˆ’ξ‚΅π‘‘log𝑑𝑝+12ξ‚Ά1/𝑝.(3.4)
Let 𝐹(𝑑)=log2𝛼𝑑+𝑑+1(1βˆ’π›Ό)(π‘‘βˆ’1)ξ‚Ήβˆ’1log𝑑𝑝𝑑log𝑝+12ξ‚Ά.(3.5) Then simple computations lead to lim𝑑→1𝐹𝐹(𝑑)=0,ξ…ž(𝑑)=𝑓(𝑑)𝑑(𝑑+1)(𝑑𝑝𝑑+1)2𝛼𝑑log𝑑+(1βˆ’π›Ό)2,βˆ’1ξ€Έξ€»log𝑑(3.6) where 𝑓(𝑑) is defined as in Lemma 2.3.
If π›Όβˆˆ(1/4,1), then inequality (3.1) follows from (3.4)–(3.6) and Lemma 2.3(1). If βˆšπ›Όβˆˆ(0,3345/80βˆ’11/16), then inequality (3.2) follows from (3.4)–(3.6) and Lemma 2.3(2).
Next, we prove that the parameter (1βˆ’4𝛼)/3 in inequalities (3.1) and (3.2) is the best possible.
For any βˆšπ›Όβˆˆ(0,3345/80βˆ’11/16)βˆͺ(1/4,1), 𝑝≠0, and π‘₯>0, one has 𝛼𝐻(1,1+π‘₯)+(1βˆ’π›Ό)𝐿(1,1+π‘₯)βˆ’π‘€π‘(1,1+π‘₯)=𝑄(π‘₯)21/𝑝,(1+π‘₯/2)log(1+π‘₯)(3.7) where 𝑄(π‘₯)=21/𝑝𝛼(1+π‘₯)log(1+π‘₯)+21/𝑝(1βˆ’π›Ό)π‘₯(1+π‘₯/2)βˆ’(1+π‘₯/2)log(1+π‘₯)[1+(1+π‘₯)𝑝]1/𝑝.
Letting π‘₯β†’0 and making use of Taylor expansion, we get 2𝑄(π‘₯)=1/𝑝8ξ‚€1βˆ’4𝛼3π‘₯βˆ’π‘3ξ€·π‘₯+π‘œ3ξ€Έ.(3.8)
If π›Όβˆˆ[1/4,1) and 𝑝>(1βˆ’4𝛼)/3, then (3.7) and (3.8) imply that there exists 𝛿1=𝛿1(𝛼,𝑝)>0 such that 𝛼𝐻(1,1+π‘₯)+(1βˆ’π›Ό)𝐿(1,1+π‘₯)<𝑀𝑝(1,1+π‘₯) for π‘₯∈(0,𝛿1). If βˆšπ›Όβˆˆ(0,3345/80βˆ’11/16) and 𝑝<(1βˆ’4𝛼)/3, then (3.7) and (3.8) imply that there exists 𝛿2=𝛿2(𝛼,𝑝)>0 such that 𝛼𝐻(1,1+π‘₯)+(1βˆ’π›Ό)𝐿(1,1+π‘₯)>𝑀𝑝(1,1+π‘₯) for π‘₯∈(0,𝛿2).
Finally, we prove that there exist π‘Ž1,𝑏1,π‘Ž2,𝑏2>0 with π‘Ž1≠𝑏1 and π‘Ž2≠𝑏2 such that 𝛼𝐻(π‘Ž1,𝑏1)+(1βˆ’π›Ό)𝐿(π‘Ž1,𝑏1)<𝑀(1βˆ’4𝛼)/3(π‘Ž1,𝑏1) and 𝛼𝐻(π‘Ž2,𝑏2)+(1βˆ’π›Ό)𝐿(π‘Ž2,𝑏2)>𝑀(1βˆ’4𝛼)/3(π‘Ž2,𝑏2) for any 3√345/80βˆ’11/16<𝛼<1/4.
If 3√345/80βˆ’11/16<𝛼<1/4, then from the expression of 𝑓2ξ…žξ…ž(1) in (2.21) we clearly see that 𝑓2ξ…žξ…ž(1)>0, which leads to the conclusion that there exists πœ†>1 such that 𝑓2ξ…žξ…ž(𝑑)>0(3.9) for π‘‘βˆˆ[1,πœ†).
From (2.14)–(2.18) and inequality (3.9) we know that 𝑓(𝑑)>0(3.10) for π‘‘βˆˆ(1,πœ†). Equations (3.4)–(3.6) and inequality (3.10) lead to the conclusion that 𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)>𝑀(1βˆ’4𝛼)/3(π‘Ž,𝑏) for all π‘Ž/π‘βˆˆ(1,πœ†).
On the other hand, simple computations lead to limπ‘₯β†’+βˆžπ‘€(1βˆ’4𝛼)/3(1,π‘₯)𝛼𝐻(1,π‘₯)+(1βˆ’π›Ό)𝐿(1,π‘₯)=23/(4π›Όβˆ’1)limπ‘₯β†’+βˆžξ€·1+π‘₯(4π›Όβˆ’1)/3ξ€Έ3/(1βˆ’4𝛼)2𝛼/(π‘₯+1)+(1βˆ’π›Ό)(1βˆ’1/π‘₯)/logπ‘₯=+∞.(3.11)
Equation (3.11) implies that there exists 𝑋=𝑋(𝛼)>1 such that 𝛼𝐻(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐿(π‘Ž,𝑏)<𝑀(1βˆ’4𝛼)/3(π‘Ž,𝑏) for all π‘Ž/π‘βˆˆ(𝑋,+∞).

Acknowledgment

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

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