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Abstract and Applied Analysis
VolumeΒ 2011Β (2011), Article IDΒ 657935, 7 pages
http://dx.doi.org/10.1155/2011/657935
Research Article

A Sharp Double Inequality between Harmonic and Identric Means

1Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
2Department of Mathematics, Hangzhou Normal University, Hangzhou 310012, China

Received 31 May 2011; Accepted 6 August 2011

Academic Editor: OndΕ™ejΒ DoΕ‘lΓ½

Copyright Β© 2011 Yu-Ming Chu et al. 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 greatest value 𝑝 and the least value π‘ž in (0,1/2) such that the double inequality 𝐻(π‘π‘Ž+(1βˆ’π‘)𝑏,𝑝𝑏+(1βˆ’π‘)π‘Ž)<𝐼(π‘Ž,𝑏)<𝐻(π‘žπ‘Ž+(1βˆ’π‘ž)𝑏,π‘žπ‘+(1βˆ’π‘ž)π‘Ž) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘. Here, 𝐻(π‘Ž,𝑏), and 𝐼(π‘Ž,𝑏) denote the harmonic and identric means of two positive numbers π‘Ž and 𝑏, respectively.

1. Introduction

The classical harmonic mean 𝐻(π‘Ž,𝑏) and identric mean 𝐼(π‘Ž,𝑏) of two positive numbers π‘Ž and 𝑏 are defined by𝐻(π‘Ž,𝑏)=2π‘Žπ‘,⎧βŽͺ⎨βŽͺ⎩1π‘Ž+𝑏(1.1)𝐼(π‘Ž,𝑏)=π‘’ξ‚΅π‘π‘π‘Žπ‘Žξ‚Ά1/(π‘βˆ’π‘Ž),π‘Žβ‰ π‘,π‘Ž,π‘Ž=𝑏,(1.2) respectively. Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for 𝐻 and 𝐼 can be found in the literature [1–17].

Let 𝑀𝑝(π‘Ž,𝑏)=[(π‘Žπ‘+𝑏𝑝)/2]1/𝑝, 𝐿(π‘Ž,𝑏)=(π‘Žβˆ’π‘)/(logπ‘Žβˆ’log𝑏), √𝐺(π‘Ž,𝑏)=π‘Žπ‘, 𝐴(π‘Ž,𝑏)=(π‘Ž+𝑏)/2, and βˆšπ‘ƒ(π‘Ž,𝑏)=(π‘Žβˆ’π‘)/[4arctan(π‘Ž/𝑏)βˆ’πœ‹] be the 𝑝th power, logarithmic, geometric, arithmetic, and Seiffert 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)<𝑃(π‘Ž,𝑏)<𝐼(π‘Ž,𝑏)<𝐴(π‘Ž,𝑏)=𝑀1(π‘Ž,𝑏)<max{π‘Ž,𝑏}(1.3) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

Long and Chu [18] answered the question: what are the greatest value 𝑝 and the least value π‘ž such that 𝑀𝑝(π‘Ž,𝑏)<𝐴𝛼(π‘Ž,𝑏)𝐺𝛽(π‘Ž,𝑏)𝐻1βˆ’π›Όβˆ’π›½(π‘Ž,𝑏)<π‘€π‘ž(π‘Ž,𝑏) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ and 𝛼,𝛽>0 with 𝛼+𝛽<1.

In [19], the authors proved 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)(1.4) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if 𝛼≀2/πœ‹ and 𝛽β‰₯5/6.

The following sharp bounds for 𝐼,(𝐿𝐼)1/2, and (𝐿+𝐼)/2 in terms of power means are presented in [20]: 𝑀2/3(π‘Ž,𝑏)<𝐼(π‘Ž,𝑏)<𝑀log2(π‘Ž,𝑏),𝑀0√(π‘Ž,𝑏)<𝐿(π‘Ž,𝑏)𝐼(π‘Ž,𝑏)<𝑀1/2𝑀(π‘Ž,𝑏),log2/(1+log2)(π‘Ž,𝑏)<𝐿(π‘Ž,𝑏)+𝐼(π‘Ž,𝑏)2<𝑀1/2(π‘Ž,𝑏)(1.5)

for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

Alzer and Qiu [21] proved that the inequalities 𝛼𝐴(π‘Ž,𝑏)+(1βˆ’π›Ό)𝐺(π‘Ž,𝑏)<𝐼(π‘Ž,𝑏)<𝛽𝐴(π‘Ž,𝑏)+(1βˆ’π›½)𝐺(π‘Ž,𝑏)(1.6) hold for all positive real numbers π‘Ž and 𝑏 with π‘Žβ‰ π‘ if and only if 𝛼≀2/3 and 𝛽β‰₯2/𝑒=0.73575, and so forth.

For fixed π‘Ž,𝑏>0 with π‘Žβ‰ π‘ and π‘₯∈[0,1/2], let 𝑓(π‘₯)=𝐻(π‘₯π‘Ž+(1βˆ’π‘₯)𝑏,π‘₯𝑏+(1βˆ’π‘₯)π‘Ž).(1.7)

Then it is not difficult to verify that 𝑓(π‘₯) is continuous and strictly increasing in [0,1/2]. Note that 𝑓(0)=𝐻(π‘Ž,𝑏)<𝐼(π‘Ž,𝑏) and 𝑓(1/2)=𝐴(π‘Ž,𝑏)>𝐼(π‘Ž,𝑏). Therefore, it is natural to ask what are the greatest value 𝑝 and the least value π‘ž in (0,1/2) such that the double inequality 𝐻(π‘π‘Ž+(1βˆ’π‘)𝑏,𝑝𝑏+(1βˆ’π‘)π‘Ž)<𝐼(π‘Ž,𝑏)<𝐻(π‘žπ‘Ž+(1βˆ’π‘ž)𝑏,π‘žπ‘+(1βˆ’π‘ž)π‘Ž) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘. The main purpose of this paper is to answer these questions. Our main result is Theorem 1.1.

Theorem 1.1. If 𝑝,π‘žβˆˆ(0,1/2), then the double inequality 𝐻(π‘π‘Ž+(1βˆ’π‘)𝑏,𝑝𝑏+(1βˆ’π‘)π‘Ž)<𝐼(π‘Ž,𝑏)<𝐻(π‘žπ‘Ž+(1βˆ’π‘ž)𝑏,π‘žπ‘+(1βˆ’π‘ž)π‘Ž)(1.8) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if βˆšπ‘β‰€(1βˆ’1βˆ’2/𝑒)/2 and βˆšπ‘žβ‰₯(6βˆ’6)/12.

2. Proof of Theorem 1.1

Proof of Theorem 1.1. Let βˆšπœ†=(6βˆ’6)/12 and βˆšπœ‡=(1βˆ’1βˆ’2/𝑒)/2. Then from the monotonicity of the function 𝑓(π‘₯)=𝐻(π‘₯π‘Ž+(1βˆ’π‘₯)𝑏,π‘₯𝑏+(1βˆ’π‘₯)π‘Ž) in [0,1/2] we know that to prove inequality (1.8) we only need to prove that inequalities 𝐼(π‘Ž,𝑏)<𝐻(πœ†π‘Ž+(1βˆ’πœ†)𝑏,πœ†π‘+(1βˆ’πœ†)π‘Ž),(2.1)𝐼(π‘Ž,𝑏)>𝐻(πœ‡π‘Ž+(1βˆ’πœ‡)𝑏,πœ‡π‘+(1βˆ’πœ‡)π‘Ž),(2.2) hold for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.
Without loss of generality, we assume that π‘Ž>𝑏. Let 𝑑=π‘Ž/𝑏>1 and π‘Ÿβˆˆ(0,1/2), then from (1.1) and (1.2) one has ξ€½log𝐻(π‘Ÿπ‘Ž+(1βˆ’π‘Ÿ)𝑏,π‘Ÿπ‘+(1βˆ’π‘Ÿ)π‘Ž)βˆ’log𝐼(π‘Ž,𝑏)=logπ‘Ÿ(1βˆ’π‘Ÿ)𝑑2+ξ€Ίπ‘Ÿ2+(1βˆ’π‘Ÿ)2𝑑+π‘Ÿ(1βˆ’π‘Ÿ)βˆ’log(𝑑+1)βˆ’π‘‘logπ‘‘π‘‘βˆ’1+1+log2.(2.3)
Let 𝑔(𝑑)=logπ‘Ÿ(1βˆ’π‘Ÿ)𝑑2+ξ€Ίπ‘Ÿ2+(1βˆ’π‘Ÿ)2ξ€»ξ€Ύβˆ’π‘‘+π‘Ÿ(1βˆ’π‘Ÿ)βˆ’log(𝑑+1)𝑑logπ‘‘π‘‘βˆ’1+1+log2.(2.4) Then simple computations lead to 𝑔(1)=0,(2.5)lim𝑑→+∞[]𝑔𝑔(𝑑)=logπ‘Ÿ(1βˆ’π‘Ÿ)+1+log2,(2.6)ξ…ž(𝑔𝑑)=1(𝑑)(π‘‘βˆ’1)2,(2.7) where 𝑔1(𝑑)=logπ‘‘βˆ’(π‘‘βˆ’1)ξ€Ίξ€·2π‘Ÿ2ξ€Έπ‘‘βˆ’2π‘Ÿ+12+4π‘Ÿ(1βˆ’π‘Ÿ)𝑑+2π‘Ÿ2ξ€»βˆ’2π‘Ÿ+1ξ€Ί(𝑑+1)π‘Ÿ(1βˆ’π‘Ÿ)𝑑2+ξ€·2π‘Ÿ2ξ€Έξ€»,π‘”βˆ’2π‘Ÿ+1𝑑+π‘Ÿ(1βˆ’π‘Ÿ)(2.8)1(1)=0,(2.9)lim𝑑→+βˆžπ‘”1𝑔(𝑑)=+∞,(2.10)ξ…ž1𝑔(𝑑)=2(𝑑)𝑑(𝑑+1)2ξ€Ίπ‘Ÿ(1βˆ’π‘Ÿ)𝑑2+ξ€·2π‘Ÿ2ξ€Έξ€»βˆ’2π‘Ÿ+1𝑑+π‘Ÿ(1βˆ’π‘Ÿ)2,(2.11) where 𝑔2(𝑑)=π‘Ÿ2(1βˆ’π‘Ÿ)2𝑑6+ξ€·2π‘Ÿ4βˆ’4π‘Ÿ3βˆ’2π‘Ÿ2𝑑+4π‘Ÿβˆ’15βˆ’ξ€·17π‘Ÿ4βˆ’34π‘Ÿ3+25π‘Ÿ2ξ€Έπ‘‘βˆ’8π‘Ÿ+14ξ€·+47π‘Ÿ4βˆ’14π‘Ÿ3+13π‘Ÿ2ξ€Έπ‘‘βˆ’6π‘Ÿ+13βˆ’ξ€·17π‘Ÿ4βˆ’34π‘Ÿ3+25π‘Ÿ2ξ€Έπ‘‘βˆ’8π‘Ÿ+12+ξ€·2π‘Ÿ4βˆ’4π‘Ÿ3βˆ’2π‘Ÿ2ξ€Έ+4π‘Ÿβˆ’1𝑑+π‘Ÿ2(1βˆ’π‘Ÿ)2,𝑔(2.12)2(1)=0,(2.13)lim𝑑→+βˆžπ‘”2𝑔(𝑑)=+∞,(2.14)ξ…ž2(𝑑)=6π‘Ÿ2(1βˆ’π‘Ÿ)2𝑑5ξ€·+52π‘Ÿ4βˆ’4π‘Ÿ3βˆ’2π‘Ÿ2𝑑+4π‘Ÿβˆ’14ξ€·βˆ’417π‘Ÿ4βˆ’34π‘Ÿ3+25π‘Ÿ2ξ€Έπ‘‘βˆ’8π‘Ÿ+13ξ€·+127π‘Ÿ4βˆ’14π‘Ÿ3+13π‘Ÿ2ξ€Έπ‘‘βˆ’6π‘Ÿ+12ξ€·βˆ’217π‘Ÿ4βˆ’34π‘Ÿ3+25π‘Ÿ2ξ€Έπ‘‘βˆ’8π‘Ÿ+1+2π‘Ÿ4βˆ’4π‘Ÿ3βˆ’2π‘Ÿ2𝑔+4π‘Ÿβˆ’1,(2.15)ξ…ž2(1)=0,(2.16)lim𝑑→+βˆžπ‘”ξ…ž2𝑔(𝑑)=+∞,(2.17)2ξ…žξ…ž(𝑑)=30π‘Ÿ2(1βˆ’π‘Ÿ)2𝑑4ξ€·+202π‘Ÿ4βˆ’4π‘Ÿ3βˆ’2π‘Ÿ2𝑑+4π‘Ÿβˆ’13ξ€·βˆ’1217π‘Ÿ4βˆ’34π‘Ÿ3+25π‘Ÿ2ξ€Έπ‘‘βˆ’8π‘Ÿ+12ξ€·+247π‘Ÿ4βˆ’14π‘Ÿ3+13π‘Ÿ2ξ€Έξ€·βˆ’6π‘Ÿ+1π‘‘βˆ’217π‘Ÿ4βˆ’34π‘Ÿ3+25π‘Ÿ2ξ€Έ,π‘”βˆ’8π‘Ÿ+1(2.18)2ξ…žξ…žξ€·(1)=βˆ’224π‘Ÿ2ξ€Έβˆ’24π‘Ÿ+5,(2.19)lim𝑑→+βˆžπ‘”2ξ…žξ…žπ‘”(𝑑)=+∞,(2.20)2ξ…žξ…žξ…ž(𝑑)=120π‘Ÿ2(1βˆ’π‘Ÿ)2𝑑3ξ€·+602π‘Ÿ4βˆ’4π‘Ÿ3βˆ’2π‘Ÿ2𝑑+4π‘Ÿβˆ’12ξ€·βˆ’2417π‘Ÿ4βˆ’34π‘Ÿ3+25π‘Ÿ2ξ€Έξ€·βˆ’8π‘Ÿ+1𝑑+247π‘Ÿ4βˆ’14π‘Ÿ3+13π‘Ÿ2ξ€Έ,π‘”βˆ’6π‘Ÿ+1(2.21)2ξ…žξ…žξ…žξ€·(1)=βˆ’1224π‘Ÿ2ξ€Έβˆ’24π‘Ÿ+5,(2.22)limπ‘‘β†’βˆžπ‘”2ξ…žξ…žξ…ž(𝑔𝑑)=∞,(2.23)2(4)(𝑑)=360π‘Ÿ2(1βˆ’π‘Ÿ)2𝑑2ξ€·+1202π‘Ÿ4βˆ’4π‘Ÿ3βˆ’2π‘Ÿ2𝑑+4π‘Ÿβˆ’1βˆ’2417π‘Ÿ4βˆ’34π‘Ÿ3+25π‘Ÿ2ξ€Έ,π‘”βˆ’8π‘Ÿ+1(2.24)2(4)ξ€·(1)=484π‘Ÿ4βˆ’8π‘Ÿ3βˆ’10π‘Ÿ2ξ€Έ+14π‘Ÿβˆ’3,(2.25)lim𝑑→+βˆžπ‘”2(4)𝑔(𝑑)=+∞,(2.26)2(5)(𝑑)=720π‘Ÿ2(1βˆ’π‘Ÿ)2𝑑+1202π‘Ÿ4βˆ’4π‘Ÿ3βˆ’2π‘Ÿ2𝑔+4π‘Ÿβˆ’1,(2.27)2(5)ξ€·(1)=1208π‘Ÿ4βˆ’16π‘Ÿ3+4π‘Ÿ2ξ€Έ.+4π‘Ÿβˆ’1(2.28)
We divide the proof into two cases. Case 1 (βˆšπ‘Ÿ=πœ†=(6βˆ’6)/12). Then (2.19), (2.22), (2.25), and (2.28) lead to 𝑔2ξ…žξ…žπ‘”(1)=0,(2.29)2ξ…žξ…žξ…žπ‘”(1)=0,(2.30)2(4)(1)=133𝑔>0,(2.31)2(5)(1)=653>0.(2.32)
From (2.27) we clearly see that 𝑔2(5)(𝑑) is strictly increasing in [1,+∞), then inequality (2.32) leads to the conclusion that 𝑔2(5)(𝑑)>0 for π‘‘βˆˆ[1,+∞), hence 𝑔2(4)(𝑑) is strictly increasing in [1,+∞).
It follows from inequality (2.31) and the monotonicity of 𝑔2(4)(𝑑) that 𝑔2ξ…žξ…žξ…ž(𝑑) is strictly increasing in [1,+∞). Then (2.30) implies that 𝑔2ξ…žξ…žξ…ž(𝑑)>0 for π‘‘βˆˆ[1,+∞), so 𝑔2ξ…žξ…ž(𝑑) is strictly increasing in [1,+∞).
From (2.29) and the monotonicity of 𝑔2ξ…žξ…ž(𝑑) we clearly see that π‘”ξ…ž2(𝑑) is strictly increasing in [1,+∞).
From (2.5), (2.7), (2.9), (2.11), (2.13), (2.16), and the monotonicity of π‘”ξ…ž2(𝑑) we conclude that 𝑔(𝑑)>0(2.33) for π‘‘βˆˆ(1,+∞).
Therefore, inequality (2.1) follows from (2.3) and (2.4) together with inequality (2.33).

Case 2 (βˆšπ‘Ÿ=πœ‡=(1βˆ’1βˆ’2/𝑒)/2). Then (2.19), (2.22), (2.25), and (2.28) lead to 𝑔2ξ…žξ…ž2(1)=βˆ’π‘’π‘”(5π‘’βˆ’12)<0,(2.34)2ξ…žξ…žξ…ž(1)=βˆ’12𝑒𝑔(5π‘’βˆ’12)<0,(2.35)2(4)(1)=βˆ’48𝑒2ξ€·3𝑒2ξ€Έπ‘”βˆ’7π‘’βˆ’1<0,(2.36)2(5)(1)=120𝑒2ξ€·2+2π‘’βˆ’π‘’2ξ€Έ>0.(2.37) From (2.27) and (2.37) we know that 𝑔2(4)(𝑑) is strictly increasing in [1,+∞). Then (2.26) and (2.36) lead to the conclusion that there exists 𝑑1>1 such that 𝑔2(4)(𝑑)<0 for π‘‘βˆˆ[1,𝑑1) and 𝑔2(4)(𝑑)>0 for π‘‘βˆˆ(𝑑1,+∞), hence 𝑔2ξ…žξ…žξ…ž(𝑑) is strictly decreasing in [1,𝑑1] and strictly increasing in [𝑑1,+∞).
It follows from (2.23) and (2.35) together with the piecewise monotonicity of 𝑔2ξ…žξ…žξ…ž(𝑑) that there exists 𝑑2>𝑑1>1 such that 𝑔2ξ…žξ…ž(𝑑) is strictly decreasing in [1,𝑑2] and strictly increasing in [𝑑2,+∞). Then (2.20) and (2.34) lead to the conclusion that there exists 𝑑3>𝑑2>1 such that π‘”ξ…ž2(𝑑) is strictly decreasing in [1,𝑑3] and strictly increasing in [𝑑3,+∞).
From (2.16) and (2.17) together with the piecewise monotonicity of π‘”ξ…ž2(𝑑) we clearly see that there exists 𝑑4>𝑑3>1 such that π‘”ξ…ž2(𝑑)<0 for π‘‘βˆˆ(1,𝑑4) and π‘”ξ…ž2(𝑑)>0 for π‘‘βˆˆ(𝑑4,+∞). Therefore, 𝑔2(𝑑) is strictly decreasing in [1,𝑑4] and strictly increasing in [𝑑4,+∞). Then (2.11)–(2.14) lead to the conclusion that there exists 𝑑5>𝑑4>1 such that 𝑔1(𝑑) is strictly decreasing in [1,𝑑5] and strictly increasing in [𝑑5,+∞).
It follows from (2.7)–(2.10) and the piecewise monotonicity of 𝑔1(𝑑) that there exists 𝑑6>𝑑5>1 such that 𝑔(𝑑) is strictly decreasing in [1,𝑑6] and strictly increasingin [𝑑6,+∞).
Note that (2.6) becomes lim𝑑→+∞[]𝑔(𝑑)=logπ‘Ÿ(1βˆ’π‘Ÿ)+1+log2=0(2.38) for βˆšπ‘Ÿ=πœ‡=(1βˆ’1βˆ’2/𝑒)/2.
From (2.5) and (2.38) together with the piecewise monotonicity of 𝑔(𝑑) we clearly see that 𝑔(𝑑)<0(2.39) for π‘‘βˆˆ(1,+∞).
Therefore, inequality (2.2) follows from (2.3) and (2.4) together with inequality (2.39).
Next, we prove that the parameter βˆšπœ†=(6βˆ’6)/12 is the best possible parameter in (0,1/2) such that inequality (2.1) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘. In fact, if βˆšπ‘Ÿ<πœ†=(6βˆ’6)/12, then (2.19) leads to 𝑔2ξ…žξ…ž(1)=βˆ’2(24π‘Ÿ2βˆ’24π‘Ÿ+5)<0. From the continuity of 𝑔2ξ…žξ…ž(𝑑) we know that there exists 𝛿>0 such that 𝑔2ξ…žξ…ž(𝑑)<0(2.40) for π‘‘βˆˆ(1,1+𝛿).
It follows from (2.3)–(2.5), (2.7), (2.9), (2.11), (2.13), and (2.16) that 𝐼(π‘Ž,𝑏)>𝐻(π‘Ÿπ‘Ž+(1βˆ’π‘Ÿ)𝑏,π‘Ÿπ‘+(1βˆ’π‘Ÿ)π‘Ž) for π‘Ž/π‘βˆˆ(1,1+𝛿).
Finally, we prove that the parameter βˆšπœ‡=(1βˆ’1βˆ’2/𝑒)/2 is the best possible parameter in (0,1/2) such that inequality (2.2) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘. In fact, if √(1βˆ’1βˆ’2/𝑒)/2=πœ‡<π‘Ÿ<1/2, then (2.6) leads to lim𝑑→+βˆžπ‘”(𝑑)>0. Hence, there exists 𝑇>1 such that 𝑔(𝑑)>0(2.41) for π‘‘βˆˆ(𝑇,+∞).
Therefore, 𝐻(π‘Ÿπ‘Ž+(1βˆ’π‘Ÿ)𝑏,π‘Ÿπ‘+(1βˆ’π‘Ÿ)π‘Ž)>𝐼(π‘Ž,𝑏) for π‘Ž/π‘βˆˆ(𝑇,+∞), follows from (2.3) and (2.4) together with inequality (2.41).

Acknowledgments

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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