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Journal of Applied Mathematics
VolumeΒ 2011Β (2011), Article IDΒ 618929, 9 pages
http://dx.doi.org/10.1155/2011/618929
Research Article

Optimal Inequalities between Harmonic, Geometric, Logarithmic, and Arithmetic-Geometric Means

Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China

Received 19 July 2011; Accepted 4 September 2011

Academic Editor: LaurentΒ Gosse

Copyright Β© 2011 Yu-Ming Chu and Miao-Kun Wang. 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 find the least values 𝑝, π‘ž, and 𝑠 in (0, 1/2) such that the inequalities 𝐻(π‘π‘Ž+(1βˆ’π‘)𝑏, 𝑝𝑏+(1βˆ’π‘)π‘Ž)>AG(π‘Ž,𝑏), 𝐺(π‘žπ‘Ž+(1βˆ’π‘ž)𝑏,π‘žπ‘+(1βˆ’π‘ž)π‘Ž)>AG(π‘Ž,𝑏), and 𝐿(π‘ π‘Ž+(1βˆ’π‘ )𝑏,𝑠𝑏+(1βˆ’π‘ )π‘Ž)>AG(π‘Ž,𝑏) hold for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘, respectively. Here AG(π‘Ž,𝑏),𝐻(π‘Ž,𝑏),𝐺(π‘Ž,𝑏), and 𝐿(π‘Ž,𝑏) denote the arithmetic-geometric, harmonic, geometric, and logarithmic means of two positive numbers π‘Ž and 𝑏, respectively.

1. Introduction

The classical arithmetic-geometric mean AG(π‘Ž,𝑏) of two positive real numbers π‘Ž and 𝑏 is defined as the common limit of sequences {π‘Žπ‘›} and {𝑏𝑛}, which are given byπ‘Ž0=π‘Ž,𝑏0π‘Ž=𝑏,𝑛+1=π‘Žπ‘›+𝑏𝑛2,𝑏𝑛+1=βˆšπ‘Žπ‘›π‘π‘›.(1.1)

Let 𝐻(π‘Ž,𝑏)=2π‘Žπ‘/(π‘Ž+𝑏),β€‰βˆšπΊ(π‘Ž,𝑏)=π‘Žπ‘, 𝐿(π‘Ž,𝑏)=(π‘Žβˆ’π‘)/(logπ‘Žβˆ’log𝑏),𝐼(π‘Ž,𝑏)=(1/𝑒)(𝑏𝑏/π‘Žπ‘Ž)1/(π‘βˆ’π‘Ž), 𝐴(π‘Ž,𝑏)=(π‘Ž+𝑏)/2, and 𝑀𝑝(π‘Ž,𝑏)=[(π‘Žπ‘+𝑏𝑝)/2]1/𝑝(𝑝≠0) and 𝑀0√(π‘Ž,𝑏)=π‘Žπ‘ be the harmonic, geometric, logarithmic, identric, arithmetic, and 𝑝-th power means of two positive numbers π‘Ž and 𝑏 with π‘Žβ‰ π‘, respectively. Then it is well known that min{π‘Ž,𝑏}<𝐻(π‘Ž,𝑏)=π‘€βˆ’1(π‘Ž,𝑏)<𝐺(π‘Ž,𝑏)=𝑀0(π‘Ž,𝑏)<𝐿(π‘Ž,𝑏)<𝐼(π‘Ž,𝑏)<𝐴(π‘Ž,𝑏)=π‘€βˆ’1(π‘Ž,𝑏)<max{π‘Ž,𝑏}(1.2) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

Recently, the inequalities for means have been the subject of intensive research. In particular, many remarkable inequalities for arithmetic-geometric mean can be found in the literature [1–9].

Carlson and Vuorinen [2], and Bracken [9] proved that 𝐿(π‘Ž,𝑏)<AG(π‘Ž,𝑏)(1.3) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

In [3], Vamanamurthy and Vuorinen established the following inequalities: AG(π‘Ž,𝑏)<𝐼(π‘Ž,𝑏)<𝐴(π‘Ž,𝑏),AG(π‘Ž,𝑏)<𝑀1/2πœ‹(π‘Ž,𝑏),AG(π‘Ž,𝑏)<2𝐿𝑀(π‘Ž,𝑏),1ξ€·π‘Ž(π‘Ž,𝑏)<AG2,𝑏2ξ€ΈAG(π‘Ž,𝑏)<𝑀2(π‘Ž,𝑏)(1.4) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

We recall the Gauss identity [6, 7]ξ€·AG1,π‘Ÿξ…žξ€Έπœ‹π’¦(π‘Ÿ)=2(1.5) for π‘Ÿβˆˆ[0,1) and π‘Ÿξ…ž=√1βˆ’π‘Ÿ2. As usual, 𝒦 and β„° denote the complete elliptic integrals [8] given byξ€œπ’¦(π‘Ÿ)=0πœ‹/2ξ€·1βˆ’π‘Ÿ2sin2πœƒξ€Έβˆ’1/2πœ‹π‘‘πœƒ=2βˆžξ“π‘›=0(1/2,𝑛)2(𝑛!)2π‘Ÿ2𝑛,π’¦ξ…ž(ξ€·π‘Ÿπ‘Ÿ)=π’¦ξ…žξ€Έ,β„°ξ€œ(π‘Ÿ)=0πœ‹/2ξ€·1βˆ’π‘Ÿ2sin2πœƒξ€Έ1/2πœ‹π‘‘πœƒ=2βˆžξ“π‘›=0(βˆ’1/2,𝑛)(1/2,𝑛)(𝑛!)2π‘Ÿ2𝑛,β„°ξ…žξ€·π‘Ÿ(π‘Ÿ)=β„°ξ…žξ€Έ,(1.6)

where (π‘Ž,0)=1 for π‘Žβ‰ 0, and ∏(π‘Ž,𝑛)=π‘›βˆ’1π‘˜=0(π‘Ž+π‘˜).

For fixed π‘Ž,𝑏>0 with π‘Žβ‰ π‘ and π‘₯∈[0,1/2], let𝑓1𝑓(π‘₯)=𝐻(π‘₯π‘Ž+(1βˆ’π‘₯)𝑏,π‘₯𝑏+(1βˆ’π‘₯)π‘Ž),(1.7)2(𝑓π‘₯)=𝐺(π‘₯π‘Ž+(1βˆ’π‘₯)𝑏,π‘₯𝑏+(1βˆ’π‘₯)π‘Ž),(1.8)3(π‘₯)=𝐿(π‘₯π‘Ž+(1βˆ’π‘₯)𝑏,π‘₯𝑏+(1βˆ’π‘₯)π‘Ž).(1.9)

Then it is not difficult to verify that 𝑓1(π‘₯),𝑓2(π‘₯), and 𝑓3(π‘₯) are continuous and strictly increasing in [0,1/2], respectively. Note that 𝑓1(0)=𝐻(π‘Ž,𝑏)<AG(π‘Ž,𝑏)<𝑓1(1/2)=𝐴(π‘Ž,𝑏), 𝑓2(0)=𝐺(π‘Ž,𝑏)<AG(π‘Ž,𝑏)<𝑓2(1/2) = 𝐴(π‘Ž,𝑏) and 𝑓3(0)=𝐿(π‘Ž,𝑏)<AG(π‘Ž,𝑏)<𝑓3(1/2)=𝐴(π‘Ž,𝑏).

Therefore, it is natural to ask what are the least values 𝑝, π‘ž, and 𝑠 in (0,1/2) such that the inequalities 𝐻(π‘π‘Ž+(1βˆ’π‘)𝑏,𝑝𝑏+(1βˆ’π‘)π‘Ž)>AG(π‘Ž,𝑏), 𝐺(π‘žπ‘Ž+(1βˆ’π‘ž)𝑏,π‘žπ‘ + (1βˆ’π‘ž)π‘Ž)>AG(π‘Ž,𝑏), and 𝐿(π‘ π‘Ž+(1βˆ’π‘ )𝑏,𝑠𝑏+(1βˆ’π‘ )π‘Ž)>AG(π‘Ž,𝑏) hold for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘, respectively. The main purpose of this paper is to answer these questions. Our main results are Theorems 1.1–1.3.

Theorem 1.1. If π‘βˆˆ(0,1/2), then inequality 𝐻(π‘π‘Ž+(1βˆ’π‘)𝑏,𝑝𝑏+(1βˆ’π‘)π‘Ž)>𝐴𝐺(π‘Ž,𝑏)(1.10) holds for all π‘Ž,b>0 with π‘Žβ‰ π‘ if and only if 𝑝β‰₯1/4.

Theorem 1.2. If π‘žβˆˆ(0,1/2), then inequality 𝐺(π‘žπ‘Ž+(1βˆ’π‘ž)𝑏,π‘žπ‘+(1βˆ’π‘ž)π‘Ž)>𝐴𝐺(π‘Ž,𝑏)(1.11) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if βˆšπ‘žβ‰₯1/2βˆ’2/4.

Theorem 1.3. If π‘ βˆˆ(0,1/2), then inequality 𝐿(π‘ π‘Ž+(1βˆ’π‘ )𝑏,𝑠𝑏+(1βˆ’π‘ )π‘Ž)>𝐴𝐺(π‘Ž,𝑏)(1.12) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if βˆšπ‘ β‰₯1/2βˆ’3/4.

2. Lemmas

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

For 0<π‘Ÿ<1, the following derivative formulas were presented in [6, Appendix E, pp. 474-475]: 𝑑𝒦=π‘‘π‘Ÿβ„°βˆ’π‘Ÿξ…ž2π’¦π‘Ÿπ‘Ÿξ…ž2,𝑑ℰ=π‘‘π‘Ÿβ„°βˆ’π’¦π‘Ÿ,π‘‘ξ‚€β„°βˆ’π‘Ÿξ…ž2π’¦ξ‚π‘‘π‘Ÿ=π‘Ÿπ’¦,𝑑(π’¦βˆ’β„°)=π‘‘π‘Ÿπ‘Ÿβ„°π‘Ÿξ…ž2,𝒦2√(2.1)π‘Ÿξƒͺ1+π‘Ÿ=(1+π‘Ÿ)𝒦(π‘Ÿ).(2.2)

The following Lemma 2.1 can be found in [6, Theorem 3.21(7) and Exercise 3.43(4)].

Lemma 2.1. (1)  (1+π‘Ÿξ…ž2)β„°(π‘Ÿ)βˆ’2π‘Ÿξ…ž2𝒦(π‘Ÿ) is strictly increasing from (0,1) onto (0,1);
(2)  ℰ(π‘Ÿ)/π‘Ÿξ…ž1/2 is strictly increasing from (0,1) onto (πœ‹/2,+∞).

Lemma 2.2. Inequality 2πœ‹π’¦ξ‚™(π‘Ÿ)11βˆ’2π‘Ÿ2>1(2.3) holds for all π‘Ÿβˆˆ(0,1).

Proof. Let 2𝑓(π‘Ÿ)=logπœ‹ξ‚™π’¦(π‘Ÿ)11βˆ’2π‘Ÿ2ξƒ­.(2.4) Then simple computations lead to 𝑓𝑓(0)=0,(2.5)ξ…žβ„°(π‘Ÿ)=(π‘Ÿ)βˆ’π‘Ÿξ…ž2𝒦(π‘Ÿ)π‘Ÿπ‘Ÿξ…ž2βˆ’π‘Ÿπ’¦(π‘Ÿ)2βˆ’π‘Ÿ2=ξ‚€1+π‘Ÿξ…ž2ℰ(π‘Ÿ)βˆ’2π‘Ÿξ…ž2𝒦(π‘Ÿ)π‘Ÿπ‘Ÿξ…ž2ξ€·2βˆ’π‘Ÿ2ξ€Έ.𝒦(π‘Ÿ)(2.6)
It follows from Lemma 2.1 (1) and (2.6) that π‘“ξ…ž(π‘Ÿ)>0 for π‘Ÿβˆˆ(0,1), which implies that 𝑓(π‘Ÿ) is strictly increasing in (0,1).
Therefore, inequality (2.3) follows from (2.4) and (2.5) together with the monotonicity of 𝑓(π‘Ÿ).

Lemma 2.3. Inequality 2√3πœ‹ξƒ©βˆšπ‘Ÿπ’¦(π‘Ÿ)>log2+3π‘Ÿβˆš2βˆ’ξƒͺ3π‘Ÿ(2.7) holds for all π‘Ÿβˆˆ(0,1).

Proof. Let 2βˆšπ‘”(π‘Ÿ)=3πœ‹ξƒ©βˆšπ‘Ÿπ’¦(π‘Ÿ)βˆ’log2+3π‘Ÿβˆš2βˆ’ξƒͺ3π‘Ÿ.(2.8) Then simple computations lead to 𝑔𝑔(0)=0,(2.9)ξ…ž2√(π‘Ÿ)=3πœ‹ξƒ©π’¦(π‘Ÿ)+π‘Ÿβ„°(π‘Ÿ)βˆ’π‘Ÿξ…ž2𝒦(π‘Ÿ)π‘Ÿπ‘Ÿβ€²2ξƒͺβˆ’4√34βˆ’3π‘Ÿ2=2√3πœ‹ξ€·1+3π‘Ÿξ…ž2ξ€Έξ‚΅1+3π‘Ÿξ…ž2π‘Ÿξ…ž3/2β„°(π‘Ÿ)π‘Ÿξ…ž1/2ξ‚Άβˆ’2πœ‹.(2.10)
Clearly the function π‘Ÿβ†’(1+3π‘Ÿ2)/π‘Ÿ3/2 is strictly decreasing from (0,1) onto (4,+∞). Then (2.10) and Lemma 2.1 (2) lead to the conclusion that π‘”ξ…ž(π‘Ÿ)>0 for π‘Ÿβˆˆ(0,1). Thus, 𝑔(π‘Ÿ) is strictly increasing in (0,1).
Therefore, inequality (2.7) follows from (2.8) and (2.9) together with the monotonicity of 𝑔(π‘Ÿ).

3. Proof of Theorems 1.1–1.3

Proof of Theorem 1.1. Let πœ†=1/4, then from the monotonicity of the function 𝑓1(π‘₯)=𝐻(π‘₯π‘Ž+(1βˆ’π‘₯)𝑏,π‘₯𝑏+(1βˆ’π‘₯)π‘Ž) in [0,1/2] we know that to prove inequality (1.10) we only need to prove that AG(π‘Ž,𝑏)<𝐻(πœ†π‘Ž+(1βˆ’πœ†)𝑏,πœ†π‘+(1βˆ’πœ†)π‘Ž)(3.1) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.
From (1.1) and (1.7) we clearly see that both AG(π‘Ž,𝑏) and 𝐻(πœ†π‘Ž+(1βˆ’πœ†)𝑏,πœ†π‘+(1βˆ’πœ†)π‘Ž) are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that π‘Ž=1>𝑏. Let 𝑑=π‘βˆˆ(0,1) and π‘Ÿ=(1βˆ’π‘‘)/(1+𝑑), then from (1.5) we have 𝐻(πœ†π‘Ž+(1βˆ’πœ†)𝑏,πœ†π‘+(1βˆ’πœ†)π‘Ž)βˆ’AG(π‘Ž,𝑏)=(𝑑+3)(3𝑑+1)βˆ’πœ‹8(1+𝑑).2𝒦(𝑑′)(3.2)
Let 𝐹(𝑑)=(𝑑+3)(3𝑑+1)βˆ’πœ‹8(1+𝑑)2𝒦(π‘‘ξ…ž).(3.3) Then making use of (2.2) we get 𝐹(𝑑)=(2+π‘Ÿ)(2βˆ’π‘Ÿ)βˆ’πœ‹4(1+π‘Ÿ)=πœ‹2(1+π‘Ÿ)𝒦(π‘Ÿ)𝐹8(1+π‘Ÿ)𝒦(π‘Ÿ)1(π‘Ÿ),(3.4) where 𝐹1(π‘Ÿ)=(2/πœ‹)(4βˆ’π‘Ÿ2)𝒦(π‘Ÿ)βˆ’4. Note that 𝐹1(π‘Ÿ)=βˆžξ“π‘›=0(1/2,𝑛)2(𝑛!)2π‘Ÿ2𝑛4βˆ’π‘Ÿ2ξ€Έβˆ’4=4π‘Ÿ2βˆžξ“π‘›=0(1/2,𝑛+1)2[(]𝑛+1)!2π‘Ÿ2π‘›βˆ’π‘Ÿ2βˆžξ“π‘›=0(1/2,𝑛)2(𝑛!)2π‘Ÿ2𝑛=βˆžξ“π‘›=0(1/2,𝑛)2[](𝑛+1)!2ξ€·3𝑛2ξ€Έπ‘Ÿ+2𝑛2(𝑛+1)>0.(3.5)
Therefore, inequality (3.1) follows from (3.2)–(3.5).

Next, we prove that the parameter 𝑝=πœ†=1/4 is the best possible parameter in (0,1/2) such that inequality (1.10) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

Since for 0<𝑝<1/2 and small π‘₯>0,πœ‹AG(1,1βˆ’π‘₯)=ξ‚€βˆš2𝒦2π‘₯βˆ’π‘₯21=1βˆ’21π‘₯βˆ’π‘₯162ξ€·π‘₯+π‘œ3ξ€Έ,1(3.6)𝐻(𝑝(1βˆ’π‘₯)+1βˆ’π‘,(1βˆ’π‘)(1βˆ’π‘₯)+𝑝)=1βˆ’2ξ‚€π‘₯+βˆ’π‘21+π‘βˆ’4π‘₯2ξ€·π‘₯+π‘œ3ξ€Έ.(3.7)

It follows from (3.6) and (3.7) that inequality AG(1,1βˆ’π‘₯)≀𝐻(𝑝(1βˆ’π‘₯)+1βˆ’π‘,(1βˆ’π‘)(1βˆ’π‘₯)+𝑝) holds for small π‘₯ only 𝑝β‰₯1/4.

Remark 3.1. For 0<𝑝<1/2 and π‘₯>0, one has limπ‘₯β†’0𝐻(𝑝π‘₯+1βˆ’π‘,(1βˆ’π‘)π‘₯+𝑝)AG(1,π‘₯)=limπ‘₯β†’04[]𝑝π‘₯+1βˆ’π‘][(1βˆ’π‘)π‘₯+𝑝𝒦π‘₯(1+π‘₯)πœ‹ξ…žξ€Έ=+∞.(3.8)
Equation (3.8) implies that there does not exist π‘βˆˆ(0,1/2) such that AG(1,π‘₯)>𝐻(𝑝π‘₯+1βˆ’π‘,(1βˆ’π‘)π‘₯+𝑝) for all π‘₯∈(0,1).

Proof of Theorem 1.2. Let βˆšπœ‡=1/2βˆ’2/4, then from the monotonicity of the function 𝑓2(π‘₯)=𝐺(π‘₯π‘Ž+(1βˆ’π‘₯)𝑏,π‘₯𝑏+(1βˆ’π‘₯)π‘Ž) in [0,1/2] we know that to prove inequality (1.11) we only need to prove that AG(π‘Ž,𝑏)<𝐺(πœ‡π‘Ž+(1βˆ’πœ‡)𝑏,πœ‡π‘+(1βˆ’πœ‡)π‘Ž)(3.9) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.
From (1.1) and (1.8) we clearly see that both AG(π‘Ž,𝑏) and 𝐺(πœ‡π‘Ž+(1βˆ’πœ‡)𝑏,πœ‡π‘+(1βˆ’πœ‡)π‘Ž) are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that π‘Ž=1>𝑏. Let 𝑑=π‘βˆˆ(0,1) and π‘Ÿ=(1βˆ’π‘‘)/(1+𝑑), then from (1.5) we have √𝐺(πœ‡π‘Ž+(1βˆ’πœ‡)𝑏,πœ‡π‘+(1βˆ’πœ‡)π‘Ž)βˆ’AG(π‘Ž,𝑏)=[]βˆ’πœ‹πœ‡+(1βˆ’πœ‡)𝑑][πœ‡π‘‘+(1βˆ’πœ‡)2𝒦(π‘‘ξ…ž).(3.10)
Let √𝐺(𝑑)=[]βˆ’πœ‹πœ‡+(1βˆ’πœ‡)𝑑][πœ‡π‘‘+(1βˆ’πœ‡)2𝒦(π‘‘ξ…ž).(3.11) Then making use of (2.2) we have πœ‹πΊ(𝑑)=22(1+π‘Ÿ)𝒦(π‘Ÿ)πœ‹ξ‚™π’¦(π‘Ÿ)11βˆ’2π‘Ÿ2ξƒ­βˆ’1.(3.12)
Therefore, inequality (3.9) follows from (3.10)–(3.12) together with Lemma 2.2.

Next, we prove that the parameter βˆšπ‘ž=πœ‡=1/2βˆ’2/4 is the best possible parameter in (0,1/2) such that inequality (1.11) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

Since for 0<π‘ž<1/2 and small π‘₯>0,1𝐺(π‘ž(1βˆ’π‘₯)+1βˆ’π‘ž,(1βˆ’π‘ž)(1βˆ’π‘₯)+π‘ž)=1βˆ’21π‘₯+8ξ€·βˆ’4π‘ž2ξ€Έπ‘₯+4π‘žβˆ’12ξ€·π‘₯+π‘œ3ξ€Έ.(3.13)

It follows from (3.6) and (3.13) that inequality AG(1,1βˆ’π‘₯)≀𝐺(π‘ž(1βˆ’π‘₯)+1βˆ’π‘ž,(1βˆ’π‘ž)(1βˆ’π‘₯)+π‘ž) holds for small π‘₯ only βˆšπ‘žβ‰₯1/2βˆ’2/4.

Remark 3.2. For 0<π‘ž<1/2 and π‘₯>0, one has limπ‘₯β†’0𝐺(π‘žπ‘₯+1βˆ’π‘ž,(1βˆ’π‘ž)π‘₯+π‘ž)AG(1,π‘₯)=limπ‘₯β†’02πœ‹βˆš[(]𝒦π‘₯π‘žπ‘₯+1βˆ’π‘ž][1βˆ’π‘ž)π‘₯+π‘žξ…žξ€Έ=+∞.(3.14)
Equation (3.14) implies that there does not exist π‘žβˆˆ(0,1/2) such that AG(1,π‘₯)>𝐺(π‘žπ‘₯+1βˆ’π‘ž,(1βˆ’π‘ž)π‘₯+π‘ž) for all π‘₯∈(0,1).

Proof of Theorem 1.3. Let βˆšπ›½=1/2βˆ’3/4, then from the monotonicity of 𝑓3(π‘₯)=𝐿(π‘₯π‘Ž+(1βˆ’π‘₯)𝑏,π‘₯𝑏+(1βˆ’π‘₯)π‘Ž) in [0,1/2] we know that to prove inequality (1.12) we only need to prove that AG(π‘Ž,𝑏)<𝐿(π›½π‘Ž+(1βˆ’π›½)𝑏,𝛽𝑏+(1βˆ’π›½)π‘Ž)(3.15) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.
From (1.1) and (1.9) we clearly see that both AG(π‘Ž,𝑏) and 𝐿(π›½π‘Ž+(1βˆ’π›½)𝑏,𝛽𝑏+(1βˆ’π›½)π‘Ž) are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that π‘Ž=1>𝑏. Let 𝑑=π‘βˆˆ(0,1) and π‘Ÿ=(1βˆ’π‘‘)/(1+𝑑), then from (1.5) one has =√𝐿(π›½π‘Ž+(1βˆ’π›½)𝑏,𝛽𝑏+(1βˆ’π›½)π‘Ž)βˆ’AG(π‘Ž,𝑏)3(1βˆ’π‘‘)√2log2βˆ’3ξ‚βˆšπ‘‘+2+3ξ‚ξ€Ήβˆšξ‚€ξ‚€2+3ξ‚βˆšπ‘‘+2βˆ’3βˆ’πœ‹ξ‚ξ‚„2𝒦(π‘‘ξ…ž).(3.16)
Let √𝐽(𝑑)=3(1βˆ’π‘‘)√2log2βˆ’3ξ‚βˆšπ‘‘+2+3ξ‚ξ€Ήβˆšξ‚€ξ‚€2+3ξ‚βˆšπ‘‘+2βˆ’3βˆ’πœ‹ξ‚ξ‚„2𝒦(π‘‘ξ…ž).(3.17) Then from (2.2) we get πœ‹π½(𝑑)=2√(1+π‘Ÿ)𝒦(π‘Ÿ)logξ‚€ξ‚€2+ξ‚ξ€Ήξ‚€βˆš3π‘Ÿ2βˆ’3π‘Ÿξ‚ξ‚π‘”(π‘Ÿ),(3.18) where 𝑔(π‘Ÿ) is defined as in Lemma 2.3.
Therefore, inequality (3.15) follows from (3.16)–(3.18) together with Lemma 2.3.

Next, we prove that the parameter βˆšπ‘ =𝛽=1/2βˆ’3/4 is the best possible parameter in (0,1/2) such that inequality (1.12) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

Since for 0<𝑠<1/2 and small π‘₯>0,1𝐿(𝑠(1βˆ’π‘₯)+1βˆ’π‘ ,(1βˆ’π‘ )(1βˆ’π‘₯)+𝑠)=1βˆ’21π‘₯+ξ€·12βˆ’4𝑠2ξ€Έπ‘₯+4π‘ βˆ’12ξ€·π‘₯+π‘œ3ξ€Έ.(3.19)

It follows from (3.6) and (3.19) that inequality AG(1,1βˆ’π‘₯)≀𝐿(𝑠(1βˆ’π‘₯)+1βˆ’π‘ ,(1βˆ’π‘ )(1βˆ’π‘₯)+𝑠) holds for small π‘₯ only βˆšπ‘ β‰₯1/2βˆ’3/4.

Remark 3.3. For 0<𝑠<1/2 and π‘₯>0, one has limπ‘₯β†’0𝐿(𝑠π‘₯+1βˆ’π‘ ,(1βˆ’π‘ )π‘₯+𝑠)AG(1,π‘₯)=limπ‘₯β†’02πœ‹π’¦ξ€·π‘₯ξ…žξ€Έ(1βˆ’2𝑠)(1βˆ’π‘₯)[(]log𝑠π‘₯+1βˆ’π‘ )/((1βˆ’π‘ )π‘₯+𝑠)=+∞.(3.20)
Equation (3.20) implies that there exist no values π‘ βˆˆ(0,1/2) such that AG(1,π‘₯)>𝐿(𝑠π‘₯+1βˆ’π‘ ,(1βˆ’π‘ )π‘₯+𝑠) for all π‘₯∈(0,1).

Acknowledgments

The authors wish to thank the anonymous referees for their careful reading of the paper and their fruitful comments and suggestions. This research was supported by the Natural Science Foundation of China under Grant 11071069, and Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

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