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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 480689, 8 pages
http://dx.doi.org/10.1155/2012/480689
Research Article

Sharp Bounds by the Generalized Logarithmic Mean for the Geometric Weighted Mean of the Geometric and Harmonic Means

1School of Distance Education, Huzhou Broadcast and TV University, Huzhou 313000, China
2School of Mathematics Science, Anhui University, Hefei 230039, China

Received 29 January 2012; Revised 19 February 2012; Accepted 12 March 2012

Academic Editor: Yuri Sotskov

Copyright © 2012 Wei-Mao Qian and Bo-Yong Long. 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 present sharp upper and lower generalized logarithmic mean bounds for the geometric weighted mean of the geometric and harmonic means.

1. Introduction

For 𝑝 the generalized logarithmic mean 𝐿𝑝(𝑎,𝑏) of two positive numbers 𝑎 and 𝑏 is defined by 𝐿𝑝𝑎(𝑎,𝑏)=𝑎,𝑎=𝑏,𝑝+1𝑏𝑝+1(𝑝+1)(𝑎𝑏)1/𝑝1,𝑝0,𝑝1,𝑎𝑏,𝑒𝑏𝑏𝑎𝑎1/(𝑏𝑎),𝑝=0,𝑎𝑏,𝑏𝑎log𝑏log𝑎,𝑝=1,𝑎𝑏.(1.1)

It is well-known that 𝐿𝑝(𝑎,𝑏) is continuous and strictly increasing with respect to 𝑝 for fixed 𝑎 and 𝑏 with 𝑎𝑏. In the recent past, the generalized logarithmic mean has been the subject of intensive research. In particular, many remarkable inequalities for 𝐿𝑝 can be found in the literature [123]. The generalized logarithmic mean has applications in convex function, economics, physics, and even in meteorology [2427]. In [26] the authors study a variant of Jensen’s functional equation involving 𝐿𝑝, which appear in a heat conduction problem. Let 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2,𝐼(𝑎,𝑏)=(1/𝑒)(𝑏𝑏/𝑎𝑎)1/(𝑏𝑎), 𝐿(𝑎,𝑏)=(𝑏𝑎)/(log𝑏log𝑎), 𝐺(𝑎,𝑏)=𝑎𝑏, and 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏) be the arithmetic, identric, logarithmic, geometric, and harmonic means of two positive numbers 𝑎 and 𝑏 with 𝑎𝑏, respectively. Then it is well known thatmin{𝑎,𝑏}<𝐻(𝑎,𝑏)<𝐺(𝑎,𝑏)=𝐿2(𝑎,𝑏)<𝐿(𝑎,𝑏)=𝐿1(𝑎,𝑏)<𝐼(𝑎,𝑏)=𝐿0(𝑎,𝑏)<𝐴(𝑎,𝑏)=𝐿1(𝑎,𝑏)<max{𝑎,𝑏}.(1.2)

In [2830], the authors present bounds for 𝐿 and 𝐼 in terms of 𝐺 and 𝐴.

Proposition 1.1. For all positive real numbers 𝑎 and 𝑏 with 𝑎𝑏, one has 𝐴1/3(𝑎,𝑏)𝐺2/31(𝑎,𝑏)<𝐿(𝑎,𝑏)<32𝐴(𝑎,𝑏)+31𝐺(𝑎,𝑏),32𝐺(𝑎,𝑏)+3𝐴(𝑎,𝑏)<𝐼(𝑎,𝑏).(1.3)

The proof of the following Proposition 1.2 can be found in [31].

Proposition 1.2. For all positive real numbers 𝑎 and 𝑏 with 𝑎𝑏, we have 𝐺(𝑎,𝑏)𝐴(𝑎,𝑏)<1𝐿(𝑎,𝑏)𝐼(𝑎,𝑏)<21(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))<2(𝐺(𝑎,𝑏)+𝐴(𝑎,𝑏)).(1.4)

For 𝑟 the 𝑟th power mean 𝑀𝑟(𝑎,𝑏) of two positive numbers 𝑎 and 𝑏 is defined by𝑀𝑟𝑎(𝑎,𝑏)=𝑟+𝑏𝑟21/𝑟,𝑟0,𝑎𝑏,𝑟=0.(1.5)

The main properties of these means are given in [32]. Several authors discussed the relationship of certain means to 𝑀𝑟. The following sharp bounds for 𝐿, 𝐼, (𝐼𝐿)1/2, and (𝐼+𝐿)/2 in terms of power means are proved in [31, 3337].

Proposition 1.3. For all positive real numbers 𝑎 and 𝑏 with 𝑎𝑏 one has 𝑀0(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝑀1/3(𝑎,𝑏),𝑀2/3(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝑀log2𝑀(𝑎,𝑏),0(𝑎,𝑏)<𝐼1/2(𝑎,𝑏)𝐿1/2(𝑎,𝑏)<𝑀1/21(𝑎,𝑏),2[]𝐼(𝑎,𝑏)+𝐿(𝑎,𝑏)<𝑀1/2(𝑎,𝑏).(1.6)

The following three results were established by Alzer and Qiu in [38].

Proposition 1.4. The inequalities 𝛼𝐴(𝑎,𝑏)+(1𝛼)𝐺(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝛽𝐴(𝑎,𝑏)+(1𝛽)𝐺(𝑎,𝑏)(1.7) hold for all positive real numbers 𝑎 and 𝑏 with 𝑎𝑏 if and only if 2𝛼32,𝛽𝑒=0.73575.(1.8)

Proposition 1.5. Let 𝑎 and 𝑏 be real numbers with 𝑎𝑏. If 0<𝑎,𝑏𝑒, then []𝐺(𝑎,𝑏)𝐴(𝑎,𝑏)<[]𝐿(𝑎,𝑏)𝐼(𝑎,𝑏)<[]𝐴(𝑎,𝑏)𝐺(𝑎,𝑏).(1.9) And, if 𝑎,𝑏𝑒, then []𝐴(𝑎,𝑏)𝐺(𝑎,𝑏)<[]𝐼(𝑎,𝑏)𝐿(𝑎,𝑏)<[]𝐺(𝑎,𝑏)𝐴(𝑎,𝑏).(1.10)

Proposition 1.6. For all positive real numbers 𝑎 and 𝑏 with 𝑎𝑏, one has 𝑀𝑐1(𝑎,𝑏)<2(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))(1.11) with the best possible parameter 𝑐=log2/(1+log2)=0.40938

In [39] the authors presented inequalities between the generalized logarithmic mean and the product 𝐴𝛼(𝑎,𝑏)𝐺𝛽(𝑎,b)𝐻𝛾(𝑎,𝑏) for all 𝑎,𝑏>0 with 𝑎𝑏 and 𝛼,𝛽>0 with 𝛼+𝛽<1.

It is the aim of this paper to give a solution to the problem: for 𝛼(0,1), what are the greatest value 𝑝 and the least value 𝑞, such that the inequality𝐿𝑝(𝑎,𝑏)𝐺𝛼(𝑎,𝑏)𝐻1𝛼(𝑎,𝑏)𝐿𝑞(𝑎,𝑏)(1.12) holds for all 𝑎,𝑏>0?

2. Main Result

Theorem 2.1. For 𝛼(0,1) and all 𝑎,𝑏>0, one has the following:(1)𝐿3𝛼5(𝑎,𝑏)=𝐺𝛼(𝑎,𝑏)𝐻1𝛼(𝑎,𝑏)=𝐿(2/𝛼)(𝑎,𝑏) for 𝛼=2/3,(2)𝐿3𝛼5(𝑎,𝑏)𝐺𝛼(𝑎,𝑏)𝐻1𝛼(𝑎,𝑏)𝐿(2/𝛼)(𝑎,𝑏) for 0<𝛼<2/3, and 𝐿3𝛼5(𝑎,𝑏)𝐺𝛼(𝑎,𝑏)𝐻1𝛼(𝑎,𝑏)𝐿(2/𝛼)(𝑎,𝑏) for 2/3<𝛼<1, with equality if and only if 𝑎=𝑏, and the parameters 3𝛼5 and 2/𝛼 in each inequality cannot be improved.

Proof. (1) If 𝛼=2/3 and 𝑎=𝑏, then (1.1) implies that 𝐿3𝛼5(𝑎,𝑏)=𝐺𝛼(𝑎,𝑏)𝐻1𝛼(𝑎,𝑏)=𝐿(2/𝛼)(𝑎,𝑏)=𝑎.
If 𝛼=2/3 and 𝑎𝑏, then (1.1) leads to 𝐿3𝛼5(𝑎,𝑏)=𝐿2/𝛼(𝑎,𝑏)=𝐿3𝑎(𝑎,𝑏)=2𝑏22(𝑏𝑎)1/3=(𝑎𝑏)1/32𝑎𝑏𝑎+𝑏1/3=𝐺2/3(𝑎,𝑏)𝐻1/3(𝑎,𝑏)=𝐺𝛼(𝑎,𝑏)𝐻1𝛼(𝑎,𝑏).(2.1)
(2) If 𝑎=𝑏, then from (1.1) we clearly see that 𝐿3𝛼5(𝑎,𝑏)=𝐺𝛼(𝑎,𝑏)𝐻1𝛼(𝑎,𝑏)=𝐿(2/𝛼)(𝑎,𝑏)=𝑎 for any 𝛼(0,1).
If 𝑎𝑏, without loss of generality, we assume 𝑎>𝑏. Let 𝑎/𝑏=𝑡>1 and 𝑓(𝑡)=log𝐿3𝛼5𝐺(𝑎,𝑏)log𝛼(𝑎,𝑏)𝐻1𝛼.(𝑎,𝑏)(2.2) Then (1.1) and simple computations yield 1𝑓(𝑡)=𝑡3𝛼5log3𝛼41𝛼(3𝛼4)(𝑡1)2log𝑡(1𝛼)log2𝑡,1+𝑡lim𝑡1+𝑓𝑓(𝑡)=0,(2.3)𝑡(𝑡)=43𝛼𝑡𝑡2𝑡143𝛼1𝑔(𝑡),(2.4) where 𝑔(𝑡)=(2𝛼/2)𝑡3𝛼2((2𝛼)(23𝛼)/53𝛼)𝑡3𝛼3+((1𝛼)(23𝛼)/2(53𝛼))𝑡3𝛼4((1𝛼)(23𝛼)/2(53𝛼))𝑡2+((2𝛼)(23𝛼)/(53𝛼))𝑡(2𝛼)/2, 𝑔𝑔(1)=0,(𝑡)=(2𝛼)(3𝛼2)2𝑡3𝛼33(2𝛼)(23𝛼)(𝛼1)𝑡53𝛼3𝛼4+(1𝛼)(23𝛼)(3𝛼4)𝑡2(53𝛼)3𝛼5(1𝛼)(23𝛼)𝑡+(53𝛼)(2𝛼)(23𝛼),𝑔(53𝛼)𝑔(1)=0,(𝑡)=3(2𝛼)(3𝛼2)(𝛼1)2𝑡3𝛼43(2𝛼)(23𝛼)(𝛼1)(3𝛼4)𝑡53𝛼3𝛼5(1𝛼)(23𝛼)(3𝛼4)2𝑡3𝛼6(1𝛼)(23𝛼),𝑔(53𝛼)(2.5)𝑔(1)=0,(2.6)3(𝑡)=2(1𝛼)(2𝛼)(43𝛼)(3𝛼2)𝑡3𝛼7(𝑡1)2.(2.7)
If 0<𝛼<2/3, then (2.7) implies 𝑔(𝑡)<0(2.8) for 𝑡>1.
From (2.3)–(2.6) and (2.8) we know that 𝑓(𝑡)>0 for 𝑡>1.
If 2/3<𝛼<1, then (2.7) leads to 𝑔(𝑡)>0(2.9) for 𝑡>1. Therefore 𝑓(𝑡)<0 for 𝑡>1 follows from (2.3)–(2.6) and (2.9).
Let (𝑡)=log𝐿(2/𝛼)𝐺(𝑎,𝑏)log𝛼(𝑎,𝑏)𝐻1𝛼(𝑎,𝑏)(2.10) for 𝑡=𝑎/𝑏>1; then (1.1) and elementary calculations lead to 𝛼(𝑡)=2𝑡log(𝛼2)/𝛼1𝛼((𝛼2)/𝛼)(𝑡1)2log𝑡(1𝛼)log2𝑡,1+𝑡lim𝑡1+(𝑡)=0,(2.11)𝑡(𝑡)=(2𝛼)/𝛼𝑡𝑡2𝑡1(2𝛼)/𝛼1𝑣(𝑡),(2.12) where 𝑣(𝑡)=((2𝛼)/2)𝑡(3𝛼2)/𝛼+((3𝛼2)/2)𝑡(2𝛼2)/𝛼((3𝛼2)/2)𝑡(2𝛼)/2, 𝑣𝑣(1)=0,(𝑡)=(2𝛼)(3𝛼2)𝑡2𝛼(2𝛼2)/𝛼+(3𝛼2)(𝛼1)𝛼𝑡(𝛼2)/𝛼3𝛼22,𝑣(2.13)𝑣(1)=0,(2.14)(𝑡)=(2𝛼)(1𝛼)(23𝛼)𝛼2𝑡2/𝛼(𝑡1).(2.15)
If 𝛼(0,2/3), then (2.15) implies 𝑣(𝑡)>0(2.16) for 𝑡>1.
From (2.11)–(2.14) and (2.16) we know that (𝑡)<0 for 𝑡>1.
If 𝛼(2/3,1), then (2.15) leads to 𝑣(𝑡)<0(2.17) for 𝑡>1. Therefore, (𝑡)>0 for 𝑡>1 follows from (2.11)–(2.14) and (2.17).
Next, we prove that the parameters (2/𝛼) and 3𝛼5 in either case cannot be improved. The proof is divided into two cases.
Case 1 (𝛼(0,2/3)). For any 𝜖>0 and 𝑥(0,1), from (1.1) one has 𝐺𝛼(1,1+𝑥)𝐻1𝛼(1,1+𝑥)53𝛼+𝜖𝐿3𝛼5𝜖(1,1+𝑥)53𝛼+𝜖=𝑓1(𝑥)(1+𝑥/2)(1𝛼)(53𝛼+𝜖)(1+𝑥)43𝛼+𝜖,1(2.18) where 𝑓1(𝑥)=(1+𝑥)(1𝛼/2)(53𝛼+𝜖)[(1+𝑥)43𝛼+𝜖1](43𝛼+𝜖)𝑥(1+𝑥)43𝛼+𝜖(1+𝑥/2)(1𝛼)(53𝛼+𝜖).
Let 𝑥0; making use of the Taylor expansion, we get 𝑓1(𝑥)=𝜖(43𝛼+𝜖)(53𝛼+𝜖)𝑥243𝑥+𝑜3.(2.19)
Equations (2.18) and (2.19) imply that for any 𝛼(0,2/3) and 𝜖>0 there exists 𝛿=𝛿(𝜖,𝛼)(0,1), such that 𝐿3𝛼5𝜖(1,1+𝑥)<𝐺𝛼(1,1+𝑥)𝐻1𝛼(1,1+𝑥) for 𝑥(0,𝛿).
On the other hand, for any 𝜖(0,(2/𝛼)1) we have 𝐿(2/𝛼)+𝜖(1,𝑡)𝐺𝛼(1,𝑡)𝐻1𝛼(1,𝑡)=𝑡𝛼/(2𝜖𝛼)1𝑡2/𝛼+𝜖+1(2/𝛼𝜖1)(11/𝑡)𝛼/(2𝜖𝛼)𝑡𝜖𝛼2/2(2𝜖𝛼)2𝑡1+𝑡1𝛼,lim𝑡+1𝑡2/𝛼+𝜖+1(2/𝛼𝜖1)(11/𝑡)𝛼/(2𝜖𝛼)𝑡𝜖𝛼2/2(2𝜖𝛼)2𝑡1+𝑡1𝛼=2𝛼𝜖1𝛼/(2𝜖𝛼)>0.(2.20)
From (2.20) we know that for any 𝛼(0,2/3) and 𝜖(0,2/𝛼1) there exists 𝑇=𝑇(𝜖,𝛼)>1, such that 𝐿2/𝛼+𝜖(1,𝑡)>𝐺𝛼(1,𝑡)𝐻1𝛼(1,𝑡) for 𝑡(𝑇,).
Case 2 (𝛼(2/3,1)). For any 𝜖(0,43𝛼) and 𝑥(0,1), from (1.1) one has 𝐿3𝛼5+𝜖(1,1+𝑥)53𝛼𝜖𝐺𝛼(1,1+𝑥)𝐻1𝛼(1,1+𝑥)53𝛼𝜖=𝑓2(𝑥)(1+𝑥/2)(1𝛼)(53𝛼𝜖)(1+𝑥)43𝛼𝜖,1(2.21) where 𝑓2(𝑥)=(43𝛼𝜖)𝑥(1+𝑥)43𝛼𝜖(1+𝑥/2)(1𝛼)(53𝛼𝜖)(1+𝑥)(1𝛼/2)(53𝛼𝜖)[(1+𝑥)43𝛼𝜖1].
Let 𝑥0; making use of the Taylor expansion, we have 𝑓2𝜖(𝑥)=24(43𝛼𝜖)(53𝛼𝜖)𝑥3𝑥+𝑜3.(2.22)
Equations (2.21) and (2.22) imply that for any 𝛼(2/3,1) and 𝜖(0,43𝛼) there exists 𝛿=𝛿(𝜖,𝛼)(0,1), such that 𝐿3𝛼5+𝜖(1,1+𝑥)>𝐺𝛼(1,1+𝑥)𝐻1𝛼(1,1+𝑥) for 𝑥(0,𝛿).
On the other hand, for any 𝜖>0, we have 𝐺𝛼(1,𝑡)𝐻1𝛼(1,𝑡)𝐿(2/𝛼)𝜖(1,𝑡)=𝑡𝛼/22𝑡1+𝑡1𝛼𝑡𝜖𝛼2/2(2+𝜖𝛼)1𝑡(2/𝛼+𝜖1)(2/𝛼+𝜖1)(11/𝑡)𝛼/(2+𝜖𝛼),lim𝑡+2𝑡1+𝑡1𝛼𝑡𝜖𝛼2/2(2+𝜖𝛼)1𝑡(2/𝛼+𝜖1)(2/𝛼+𝜖1)(11/𝑡)𝛼/(2+𝜖𝛼)=21𝛼>0.(2.23)
From (2.23) we know that for any 𝛼(2/3,1) and 𝜖>0 there exists 𝑇=𝑇(𝜖,𝛼)>1, such that 𝐿(2/𝛼)𝜖(1,𝑡)<𝐺𝛼(1,𝑡)𝐻1𝛼(1,𝑡) for 𝑡(𝑇,).

Acknowledgment

This work was supported by the Natural Science Foundation of Zhejiang Broad-cast and TV University under Grant XKT-09G21.

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