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Journal of Applied Mathematics
VolumeΒ 2011, Article IDΒ 261237, 6 pages
http://dx.doi.org/10.1155/2011/261237
Research Article

An Optimal Double Inequality between Seiffert and Geometric Means

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

Received 30 June 2011; Accepted 14 October 2011

Academic Editor: J. C.Β Butcher

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

For 𝛼,π›½βˆˆ(0,1/2) we prove that the double inequality 𝐺(π›Όπ‘Ž+(1βˆ’π›Ό)𝑏,𝛼𝑏+(1βˆ’π›Ό)π‘Ž)<𝑃(π‘Ž,𝑏)<𝐺(π›½π‘Ž+(1βˆ’π›½)𝑏,𝛽𝑏+(1βˆ’π›½)π‘Ž) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if βˆšπ›Όβ‰€(1βˆ’1βˆ’4/πœ‹2)/2 and βˆšπ›½β‰₯(3βˆ’3)/6. Here, 𝐺(π‘Ž,𝑏) and 𝑃(π‘Ž,𝑏) denote the geometric and Seiffert means of two positive numbers π‘Ž and 𝑏, respectively.

1. Introduction

For π‘Ž,𝑏>0 with π‘Žβ‰ π‘ the Seiffert mean 𝑃(π‘Ž,𝑏) was introduced by Seiffert [1] as follows:𝑃(π‘Ž,𝑏)=π‘Žβˆ’π‘βˆš4arctan.π‘Ž/π‘βˆ’πœ‹(1.1)

Recently, the bivariate mean values have been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert mean can be found in the literature [1–9].

Let 𝐻(π‘Ž,𝑏)=2π‘Žπ‘/(π‘Ž+𝑏), √𝐺(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, 𝐢(π‘Ž,𝑏)=(π‘Ž2+𝑏2)/(π‘Ž+𝑏), 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, contraharmonic, and 𝑝th power means of two different positive numbers π‘Ž and 𝑏, 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)=𝑀1(π‘Ž,𝑏)<𝐢(π‘Ž,𝑏)<max{π‘Ž,𝑏}(1.2) for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

For all π‘Ž,𝑏>0 with π‘Žβ‰ π‘, Seiffert [1] established that 𝐿(π‘Ž,𝑏)<𝑃(π‘Ž,𝑏)<𝐼(π‘Ž,𝑏); Jagers [4] proved that 𝑀1/2(π‘Ž,𝑏)<𝑃(π‘Ž,𝑏)<𝑀2/3(π‘Ž,𝑏) and 𝑀2/3(π‘Ž,𝑏) is the best possible upper power mean bound for the Seiffert mean 𝑃(π‘Ž,𝑏); Seiffert [7] established that 𝑃(π‘Ž,𝑏)>𝐴(π‘Ž,𝑏)𝐺(π‘Ž,𝑏)/𝐿(π‘Ž,𝑏) and 𝑃(π‘Ž,𝑏)>2𝐴(π‘Ž,𝑏)/πœ‹; SΓ‘ndor [6] presented that √(𝐴(π‘Ž,𝑏)+𝐺(π‘Ž,𝑏))/2<𝑃(π‘Ž,𝑏)<𝐴(π‘Ž,𝑏)(𝐴(π‘Ž,𝑏)+𝐺(π‘Ž,𝑏))/2 and 3√𝐴2(π‘Ž,𝑏)𝐺(π‘Ž,𝑏)<𝑃(π‘Ž,𝑏)<(𝐺(π‘Ž,𝑏)+2𝐴(π‘Ž,𝑏))/3; HΓ€stΓΆ [3] proved that 𝑃(π‘Ž,𝑏)>𝑀log2/logπœ‹(π‘Ž,𝑏) and 𝑀log2/logπœ‹(π‘Ž,𝑏) is the best possible lower power mean bound for the Seiffert mean 𝑃(π‘Ž,𝑏).

Very recently, Wang and Chu [8] found the greatest value 𝛼 and the least value 𝛽 such that the double inequality 𝐴𝛼(π‘Ž,𝑏)𝐻1βˆ’π›Ό(π‘Ž,𝑏)<𝑃(π‘Ž,𝑏)<𝐴𝛽(π‘Ž,𝑏)𝐻1βˆ’π›½(π‘Ž,𝑏) holds for π‘Ž,𝑏>0 with π‘Žβ‰ π‘; For any π›Όβˆˆ(0,1), Chu et al. [10] presented the best possible bounds for 𝑃𝛼(π‘Ž,𝑏)𝐺1βˆ’π›Ό(π‘Ž,𝑏) in terms of the power mean; In [2] 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) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if 𝛼≀2/πœ‹ and 𝛽β‰₯5/6; Liu and Meng [5] proved that the inequalities 𝛼1𝐢(π‘Ž,𝑏)+1βˆ’π›Ό1𝐺(π‘Ž,𝑏)<𝑃(π‘Ž,𝑏)<𝛽1𝐢(π‘Ž,𝑏)+1βˆ’π›½1𝐺𝛼(π‘Ž,𝑏),2𝐢(π‘Ž,𝑏)+1βˆ’π›Ό2𝐻(π‘Ž,𝑏)<𝑃(π‘Ž,𝑏)<𝛽2𝐢(π‘Ž,𝑏)+1βˆ’π›½2𝐻(π‘Ž,𝑏)(1.3) hold for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if 𝛼1≀2/9, 𝛽1β‰₯1/πœ‹, 𝛼2≀1/πœ‹ and 𝛽2β‰₯5/12.

For fixed π‘Ž,𝑏>0 with π‘Žβ‰ π‘ and π‘₯∈[0,1/2], let 𝑔(π‘₯)=𝐺(π‘₯π‘Ž+(1βˆ’π‘₯)𝑏,π‘₯𝑏+(1βˆ’π‘₯)π‘Ž).(1.4)

Then it is not difficult to verify that 𝑔(π‘₯) is continuous and strictly increasing in [0,1/2]. Note that 𝑔(0)=𝐺(π‘Ž,𝑏)<𝑃(π‘Ž,𝑏) and 𝑔(1/2)=𝐴(π‘Ž,𝑏)>𝑃(π‘Ž,𝑏). Therefore, it is natural to ask what are the greatest value 𝛼 and least value 𝛽 in (0,1/2) such that the double inequality 𝐺(π›Όπ‘Ž+(1βˆ’π›Ό)𝑏,𝛼𝑏+(1βˆ’π›Ό)π‘Ž)<𝑃(π‘Ž,𝑏)<𝐺(π›½π‘Ž+(1βˆ’π›½)𝑏,𝛽𝑏+(1βˆ’π›½)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 the following Theorem 1.1.

Theorem 1.1. If 𝛼,π›½βˆˆ(0,1/2), then the double inequality 𝐺(π›Όπ‘Ž+(1βˆ’π›Ό)𝑏,𝛼𝑏+(1βˆ’π›Ό)π‘Ž)<𝑃(π‘Ž,𝑏)<𝐺(π›½π‘Ž+(1βˆ’π›½)𝑏,𝛽𝑏+(1βˆ’π›½)π‘Ž)(1.5) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘ if and only if βˆšπ›Όβ‰€(1βˆ’1βˆ’4/πœ‹2)/2 and βˆšπ›½β‰₯(3βˆ’3)/6.

2. Proof of Theorem 1.1

Proof of Theorem 1.1. Let βˆšπœ†=(1βˆ’1βˆ’4/πœ‹2)/2 and βˆšπœ‡=(3βˆ’3)/6. We first prove that inequalities 𝑃(π‘Ž,𝑏)>𝐺(πœ†π‘Ž+(1βˆ’πœ†)𝑏,πœ†π‘+(1βˆ’πœ†)π‘Ž),(2.1)𝑃(π‘Ž,𝑏)<𝐺(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 βˆšπ‘‘=π‘Ž/𝑏>1 and π‘βˆˆ(0,1/2), then from (1.1) one has =1log𝐺(π‘π‘Ž+(1βˆ’π‘)𝑏,𝑝𝑏+(1βˆ’π‘)π‘Ž)βˆ’log𝑃(π‘Ž,𝑏)2log𝑝𝑑2+(1βˆ’π‘)ξ€Έξ€·(1βˆ’π‘)𝑑2𝑑+π‘ξ€Έξ€»βˆ’log2βˆ’1.4arctanπ‘‘βˆ’πœ‹(2.3) Let 1𝑓(𝑑)=2log𝑝𝑑2+(1βˆ’π‘)ξ€Έξ€·(1βˆ’π‘)𝑑2𝑑+π‘ξ€Έξ€»βˆ’log2βˆ’1,4arctanπ‘‘βˆ’πœ‹(2.4) then simple computations lead to 𝑓(1)=0,(2.5)lim𝑑→+∞1𝑓(𝑑)=2[]𝑓log𝑝(1βˆ’π‘)+logπœ‹,(2.6)ξ…žπ‘‘ξ€·π‘‘(𝑑)=2ξ€Έ+1𝑑2ξ€Έξ€·βˆ’1(4arctanπ‘‘βˆ’πœ‹)𝑝𝑑2+(1βˆ’π‘)ξ€Έξ€·(1βˆ’π‘)𝑑2𝑓+𝑝1(𝑑),(2.7) where 𝑓14𝑑(𝑑)=2βˆ’1𝑝𝑑2+1βˆ’π‘ξ€Έξ€Ί(1βˆ’π‘)𝑑2ξ€»+𝑝𝑑𝑑2ξ€Έ+12π‘“βˆ’4arctan𝑑+πœ‹.(2.8)1(1)=0,(2.9)lim𝑑→+βˆžπ‘“1𝑓(𝑑)=+∞,(2.10)ξ…ž1(𝑑)=4𝑓2𝑑2𝑑2𝑑2ξ€Έ+14,(2.11) where 𝑓2(𝑑)=𝑝(1βˆ’π‘)𝑑5βˆ’(3π‘βˆ’2)(3π‘βˆ’1)𝑑4+2(5𝑝2βˆ’5𝑝+1)𝑑3+2(5𝑝2βˆ’5𝑝+1)𝑑2βˆ’(3π‘βˆ’2)(3π‘βˆ’1)𝑑+𝑝(1βˆ’π‘).
Note that 𝑓2(1)=0,(2.12)lim𝑑→+βˆžπ‘“2(𝑓𝑑)=+∞,(2.13)ξ…ž2(𝑑)=5𝑝(1βˆ’π‘)𝑑4βˆ’4(3π‘βˆ’2)(3π‘βˆ’1)𝑑3ξ€·+65𝑝2ξ€Έπ‘‘βˆ’5𝑝+12ξ€·+45𝑝2ξ€Έπ‘“βˆ’5𝑝+1π‘‘βˆ’(3π‘βˆ’2)(3π‘βˆ’1),(2.14)ξ…ž2(1)=0,(2.15)lim𝑑→+βˆžπ‘“ξ…ž2𝑓(𝑑)=+∞,(2.16)2ξ…žξ…ž(𝑑)=20𝑝(1βˆ’π‘)𝑑3βˆ’12(3π‘βˆ’2)(3π‘βˆ’1)𝑑2ξ€·+125𝑝2ξ€Έξ€·βˆ’5𝑝+1𝑑+45𝑝2ξ€Έπ‘“βˆ’5𝑝+1,(2.17)2ξ…žξ…ž(𝑑)=βˆ’86𝑝2ξ€Έβˆ’6𝑝+1,(2.18)lim𝑑→+βˆžπ‘“2ξ…žξ…žπ‘“(𝑑)=+∞,(2.19)3ξ…žξ…žξ…ž(𝑑)=60𝑝(1βˆ’π‘)𝑑2ξ€·βˆ’24(3π‘βˆ’2)(3π‘βˆ’1)𝑑+125𝑝2ξ€Έπ‘“βˆ’5𝑝+1,(2.20)2ξ…žξ…žξ…žξ€·(1)=βˆ’366𝑝2ξ€Έβˆ’6𝑝+1,(2.21)lim𝑑→+βˆžπ‘“2ξ…žξ…žξ…žπ‘“(𝑑)=+∞,(2.22)2(4)𝑓(𝑑)=120𝑝(1βˆ’π‘)π‘‘βˆ’24(3π‘βˆ’2)(3π‘βˆ’1),(2.23)2(4)ξ€·(1)=βˆ’487𝑝2ξ€Έβˆ’7𝑝+1,(2.24)lim𝑑→+βˆžπ‘“2(4)(𝑑)=+∞.(2.25)
We divide the proof into two cases.
Case 1 (βˆšπ‘=πœ†=(1βˆ’1βˆ’4/πœ‹2)/2). Then (2.6), (2.18), (2.21), and (2.24) become lim𝑑→+βˆžπ‘“π‘“(𝑑)=0,(2.26)2ξ…žξ…ž8ξ€·πœ‹(1)=βˆ’2ξ€Έβˆ’6πœ‹2𝑓<0,(2.27)2ξ…žξ…žξ…žξ€·πœ‹(1)=βˆ’362ξ€Έβˆ’6πœ‹2𝑓<0,(2.28)2(4)ξ€·πœ‹(1)=βˆ’482ξ€Έβˆ’7πœ‹2<0.(2.29)
From (2.23) we clearly see that 𝑓2(4)(𝑑) is strictly increasing in [1,+∞), then (2.25) and inequality (2.29) lead to the conclusion that there exists πœ†1>1 such that 𝑓2(4)(𝑑)<0 for π‘‘βˆˆ[1,πœ†1) and 𝑓2(4)(𝑑)>0 for π‘‘βˆˆ(πœ†1,+∞). Thus, 𝑓2ξ…žξ…žξ…ž(𝑑) is strictly decreasing in [1,πœ†1] and strictly increasing in [πœ†1,+∞).
It follows from (2.22) and inequality (2.28) together with the piecewise monotonicity of 𝑓2ξ…žξ…žξ…ž(𝑑) that there exists πœ†2>πœ†1>1 such that 𝑓2ξ…žξ…ž(𝑑) is strictly decreasing in [1,πœ†2] and strictly increasing in [πœ†2,+∞). Then (2.19) and inequality (2.27) lead to the conclusion that there exists πœ†3>πœ†2>1 such that π‘“ξ…ž2(𝑑) is strictly decreasing in [1,πœ†3] and strictly increasing in [πœ†3,+∞).
From (2.15) and (2.16) together with the piecewise monotonicity of π‘“ξ…ž2(𝑑) we know that there exists πœ†4>πœ†3>1 such that 𝑓2(𝑑) is strictly decreasing in [1,πœ†4] and strictly increasing in [πœ†4,+∞). Then (2.11)–(2.13) lead to the conclusion that there exists πœ†5>πœ†4>1 such that 𝑓1(𝑑) is strictly decreasing in √[1,πœ†5] and strictly increasing in [βˆšπœ†5,+∞).
It follows from (2.7)–(2.10) and the piecewise monotonicity of 𝑓1(𝑑) that there exists πœ†6>βˆšπœ†5>1 such that 𝑓(𝑑) is strictly decreasing in [1,πœ†6] and strictly increasing in [πœ†6,+∞).
Therefore, inequality (2.1) follows from (2.3)–(2.5) and the piecewise monotonicity of 𝑓(𝑑).

Case 2 (βˆšπ‘=πœ‡=(3βˆ’3)/6). Then (2.18), (2.21) and (2.24) become 𝑓2ξ…žξ…žπ‘“(1)=0,(2.30)2ξ…žξ…žξ…žπ‘“(1)=0,(2.31)2(4)(1)=8>0.(2.32)
From (2.23) we clearly see that 𝑓2(4)(𝑑) is strictly increasing in [1,+∞), then inequality (2.32) leads to the conclusion that 𝑓2ξ…žξ…žξ…ž(𝑑) is strictly increasing in [1,+∞).
Therefore, inequality (2.2) follows from (2.3)–(2.5), (2.7)–(2.9), (2.11), (2.12), (2.15), and inequalities (2.30) and (2.31) together with the monotonicity of 𝑓2ξ…žξ…žξ…ž(𝑑).
Next, we prove that βˆšπœ†=(1βˆ’1βˆ’4/πœ‹2)/2 is the best possible parameter such that inequality (2.1) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘. In fact, if √(1βˆ’1βˆ’4/πœ‹2)/2=πœ†<𝑝<1/2, then (2.6) leads to lim𝑑→+∞1𝑓(𝑑)=2[]log𝑝(1βˆ’π‘)+logπœ‹>0.(2.33)
Inequality (2.33) implies that there exists 𝑇=𝑇(𝑝)>1 such that 𝑓(𝑑)>0(2.34) for π‘‘βˆˆ(𝑇,+∞).
It follows from (2.3) and (2.4) together with inequality (2.34) that 𝑃(π‘Ž,𝑏)<𝐺(π‘π‘Ž+(1βˆ’π‘)𝑏,𝑝𝑏+(1βˆ’π‘)π‘Ž) for π‘Ž/π‘βˆˆ(𝑇2,+∞).
Finally, we prove that βˆšπœ‡=(3βˆ’3)/6 is the best possible parameter such that inequality (2.2) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘. In fact, if √0<𝑝<πœ‡=(3βˆ’3)/6, then from (2.18) we get 𝑓2ξ…žξ…ž(1)<0, which implies that there exists 𝛿>0 such that 𝑓2ξ…žξ…ž(𝑑)<0(2.35) for π‘‘βˆˆ[1,1+𝛿).
Therefore, 𝑃(π‘Ž,𝑏)>𝐺(π‘π‘Ž+(1βˆ’π‘)𝑏,𝑝𝑏+(1βˆ’π‘)π‘Ž) for π‘Ž/π‘βˆˆ(1,(1+𝛿)2) follows from (2.3)–(2.5), (2.7)–(2.9), (2.11), (2.12), and (2.15) together with inequality (2.35).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grant 11071069 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

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