Optimal Lehmer Mean Bounds for the Combinations of Identric and

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Aug 5, 2013 - Optimal Lehmer Mean Bounds for the Combinations of. Identric and Logarithmic Means. Xu-Hui Shen,1 Wei-Ming Gong,2 and Yu-Ming Chu2.
Hindawi Publishing Corporation Chinese Journal of Mathematics Volume 2013, Article ID 852516, 7 pages http://dx.doi.org/10.1155/2013/852516

Research Article Optimal Lehmer Mean Bounds for the Combinations of Identric and Logarithmic Means Xu-Hui Shen,1 Wei-Ming Gong,2 and Yu-Ming Chu2 1 2

College of Nursing, Huzhou Teachers College, Huzhou 313000, China School of Mathematics and Computation Science, Hunan City University, Yiyang 413000, China

Correspondence should be addressed to Yu-Ming Chu; [email protected] Received 11 July 2013; Accepted 5 August 2013 Academic Editors: M. Coppens, Y. Miao, and P.-y. Nie Copyright Β© 2013 Xu-Hui Shen 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. For any 𝛼 ∈ (0, 1), we answer the questions: what are the greatest values 𝑝 and πœ† and the least values π‘ž and πœ‡, such that the inequalities 𝐿 𝑝 (π‘Ž, 𝑏) < 𝐼𝛼 (π‘Ž, 𝑏)𝐿1βˆ’π›Ό (π‘Ž, 𝑏) < 𝐿 π‘ž (π‘Ž, 𝑏) and 𝐿 πœ† (π‘Ž, 𝑏) < 𝛼𝐼(π‘Ž, 𝑏) + (1 βˆ’ 𝛼)𝐿(π‘Ž, 𝑏) < 𝐿 πœ‡ (π‘Ž, 𝑏) hold for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏? Here, 𝐼(π‘Ž, 𝑏), 𝐿(π‘Ž, 𝑏), and 𝐿 𝑝 (π‘Ž, 𝑏) denote the identric, logarithmic, and 𝑝th Lehmer means of two positive numbers π‘Ž and 𝑏, respectively.

1. Introduction For π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏, the logarithmic mean 𝐿(π‘Ž, 𝑏) and identric mean 𝐼(π‘Ž, 𝑏) are defined by 𝐿 (π‘Ž, 𝑏) =

π‘Žβˆ’π‘ , log π‘Ž βˆ’ log 𝑏

(1)

1/(π‘βˆ’π‘Ž)

1 𝑏𝑏 𝐼 (π‘Ž, 𝑏) = ( π‘Ž ) 𝑒 π‘Ž

,

(2)

respectively. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for 𝐿 and 𝐼 can be found in the literature [1–19]. In [14, 17, 20], inequalities between 𝐿, 𝐼, and the classical arithmetic-geometric mean of Gauss are proved. The ratio of identric means leads to the weighted geometric mean 2

2

𝐼 (π‘Ž , 𝑏 ) 𝐼 (π‘Ž, 𝑏)

1/(π‘Ž+𝑏)

= (π‘Žπ‘Ž 𝑏𝑏 )

,

(3)

which has been investigated in [11, 13, 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 𝐿, which appears in a heat conduction

problem. A representation of 𝐿 as an infinite product and an iterative algorithm for computing the logarithmic mean as the common limit of two sequences of special geometric and arithmetic means are given in [5]. In [25, 26], it is shown that 𝐿 can be expressed in terms of Gauss’ hypergeometric function 2 𝐹1 . And in [26], the authors prove that the reciprocal of the logarithmic mean is strictly totally positive; that is, every 𝑛 Γ— 𝑛 determinant with elements 1/𝐿(π‘Žπ‘– , 𝑏𝑗 ), where 0 < π‘Ž1 < π‘Ž2 < β‹… β‹… β‹… < π‘Žπ‘› and 0 < 𝑏1 < 𝑏2 < β‹… β‹… β‹… < 𝑏𝑛 , is positive for all 𝑛 β‰₯ 1. For 𝑝 ∈ R, the power mean 𝑀𝑝 (π‘Ž, 𝑏) and Lehmer mean 𝐿 𝑝 (π‘Ž, 𝑏) of order 𝑝 of two positive numbers π‘Ž and 𝑏 with π‘Ž =ΜΈ 𝑏 are defined by 1/𝑝

𝑀𝑝 (π‘Ž, 𝑏) = ( 𝐿 𝑝 (π‘Ž, 𝑏) =

π‘Žπ‘ + 𝑏𝑝 ) 2

π‘Žπ‘+1 + 𝑏𝑝+1 . π‘Žπ‘ + 𝑏𝑝

,

(4) (5)

It is well known that 𝑀𝑝 (π‘Ž, 𝑏) and 𝐿 𝑝 (π‘Ž, 𝑏) are strictly increasing with respect to 𝑝 ∈ R for fixed π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏. The main properties for 𝑀𝑝 and 𝐿 𝑝 are given in [27–32]. Let 𝐴(π‘Ž, 𝑏) = (π‘Ž + 𝑏)/2, 𝐺(π‘Ž, 𝑏) = βˆšπ‘Žπ‘, and 𝐻(π‘Ž, 𝑏) = 2π‘Žπ‘/(π‘Ž+𝑏) be the arithmetic, geometric, and harmonic means

2

Chinese Journal of Mathematics

of two positive numbers π‘Ž and 𝑏, respectively. Then it is well known that min {π‘Ž, 𝑏} < 𝐻 (π‘Ž, 𝑏) = 𝐿 βˆ’1 (π‘Ž, 𝑏) = π‘€βˆ’1 (π‘Ž, 𝑏) < 𝐺 (π‘Ž, 𝑏) = 𝐿 βˆ’1/2 (π‘Ž, 𝑏) = 𝑀0 (π‘Ž, 𝑏) < 𝐿 (π‘Ž, 𝑏) < 𝐼 (π‘Ž, 𝑏) < 𝐴 (π‘Ž, 𝑏)

(6)

= 𝐿 0 (π‘Ž, 𝑏) = 𝑀1 (π‘Ž, 𝑏) < max {π‘Ž, 𝑏}

(6) 9𝑝3 + 33𝑝2 + 51𝑝 + 14 > 0.

1 18𝑝2 + 21𝑝 + 8 > 18𝑝2 + (βˆ’ ) Γ— 21 + 8 3

(2) Let (8)

𝑔1 (𝑝) = 270𝑝3 + 567𝑝2 + 384𝑝 + 104.

1 𝑔1 (βˆ’ ) = 29 > 0, 3 𝑔1σΈ€  (𝑝) = 6 (135𝑝2 + 189𝑝 + 64)

𝛼𝐴 (π‘Ž, 𝑏) + (1 βˆ’ 𝛼) 𝐺 (π‘Ž, 𝑏) (9)

1 > 6 [135𝑝 + 189 Γ— (βˆ’ ) + 64] 3 = 6 (135𝑝2 + 1) > 0.

for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏 if and only if 𝛼 ≀ 2/3, 𝛽 β‰₯ 2/𝑒 = 0.73575 . . ., and 𝑝 ≀ log 2/(1 + log 2) = 0.40938 . . .. The following sharp upper and lower Lehmer mean bounds for 𝐿, 𝐼, (𝐿𝐼)1/2 , and (𝐿 + 𝐼)/2 are presented in [34]:

Therefore, Lemma 1(2) follows from (14) and (15). (3) Let

𝐿 βˆ’1/3 (π‘Ž, 𝑏) < 𝐿 (π‘Ž, 𝑏) < 𝐿 0 (π‘Ž, 𝑏) ,

(10)

π‘€βˆ’1/6 (π‘Ž, 𝑏) < 𝐼 (π‘Ž, 𝑏) < 𝐿 0 (π‘Ž, 𝑏) ,

(11)

𝐿 βˆ’1/4 (π‘Ž, 𝑏) < 𝐼1/2 (π‘Ž, 𝑏) 𝐿1/2 (π‘Ž, 𝑏) < 𝐿 0 (π‘Ž, 𝑏) , (12)

for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏. The purpose of this paper is to present the best possible upper and lower Lehmer mean bounds of the product 𝐼𝛼 (π‘Ž, 𝑏)𝐿1βˆ’π›Ό (π‘Ž, 𝑏) and the sum 𝛼𝐼(π‘Ž, 𝑏) + (1 βˆ’ 𝛼)𝐿(π‘Ž, 𝑏) for any 𝛼 ∈ (0, 1) and all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏.

𝑔2 (𝑝) = 1134𝑝4 + 3429𝑝3 + 3879𝑝2 + 1950𝑝 + 448. (16) Then simple computations yield 1 𝑔2 (βˆ’ ) = 116 > 0, 3 𝑔2σΈ€  (𝑝) = 4536𝑝3 + 10287𝑝2 + 7758𝑝 + 1950, 1 𝑔2σΈ€  (βˆ’ ) = 339 > 0, 3

(17)

𝑔2σΈ€ σΈ€  (𝑝) = 13608𝑝2 + 20574𝑝 + 7758 1 > 13608𝑝2 + 20574 Γ— (βˆ’ ) + 7758 3 = 13608𝑝2 + 900 > 0.

2. Lemmas In order to establish our main result, we need several lemmas, which we present in this section. Lemma 1. If 𝑝 β‰₯ βˆ’1/3, then the following statements are true: 2

(1) 18𝑝 + 21𝑝 + 8 > 0; 3

(15)

2

1 𝑀𝑝 (π‘Ž, 𝑏) < (𝐼 (π‘Ž, 𝑏) + 𝐿 (π‘Ž, 𝑏)) 2

1 (𝐼 (π‘Ž, 𝑏) + 𝐿 (π‘Ž, 𝑏)) < 𝐿 0 (π‘Ž, 𝑏) 2

(14)

Then simple computations lead to

for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏. Alzer and Qiu [19] proved that

𝐿 βˆ’1/4 (π‘Ž, 𝑏)
0.

(7)

1 (𝐼 (π‘Ž, 𝑏) + 𝐿 (π‘Ž, 𝑏)) < 𝑀1/2 (π‘Ž, 𝑏) 2

< 𝐼 (π‘Ž, 𝑏) < 𝛽𝐴 (π‘Ž, 𝑏) + (1 βˆ’ 𝛽) 𝐺 (π‘Ž, 𝑏) ,

(5) 5346𝑝6 + 18225𝑝5 + 27297𝑝4 + 21996𝑝3 + 9672𝑝2 + 2264𝑝 + 320 > 0;

2

𝑀0 (π‘Ž, 𝑏) < 𝐿 (π‘Ž, 𝑏) < 𝑀1/3 (π‘Ž, 𝑏) , 𝑀0 (π‘Ž, 𝑏) < √𝐼 (π‘Ž, 𝑏) 𝐿 (π‘Ž, 𝑏) < 𝑀1/2 (π‘Ž, 𝑏) ,

(4) 1458𝑝5 +4941𝑝4 +6903𝑝3 +4809𝑝2 +1714𝑝+320 > 0;

Proof. (1) We clearly see that

for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏. The following sharp bounds for 𝐿, 𝐼, (𝐿𝐼)1/2 , and (𝐿 + 𝐼)/2 in terms of power means are proved in [2–4, 6, 8, 9, 33]: 𝑀2/3 (π‘Ž, 𝑏) < 𝐼 (π‘Ž, 𝑏) < 𝑀log 2 (π‘Ž, 𝑏) ,

(3) 1134𝑝4 + 3429𝑝3 + 3879𝑝2 + 1950𝑝 + 448 > 0;

2

(2) 270𝑝 + 567𝑝 + 384𝑝 + 104 > 0;

Therefore, Lemma 1(3) follows from (16) and (17). (4) Let 𝑔3 (𝑝) = 1458𝑝5 + 4941𝑝4 + 6903𝑝3 + 4809𝑝2 + 1714𝑝 + 320.

(18)

Chinese Journal of Mathematics

3

Then simple computations lead to

(6) Let

1 247 𝑔3 (βˆ’ ) = > 0, 3 3

𝑔5 (𝑝) = 9𝑝3 + 33𝑝2 + 51𝑝 + 14.

𝑔3σΈ€  (𝑝) = 7290𝑝4 + 19764𝑝3 + 20709𝑝2

Then we have

+ 9618𝑝 + 1714,

1 1 𝑔5 (βˆ’ ) = > 0, 3 3

1 𝑔3σΈ€  (βˆ’ ) = 167 > 0, 3 𝑔3σΈ€ σΈ€ 

3

2

(𝑝) = 29160𝑝 + 59292𝑝 + 41418𝑝 + 9618, 𝑔3σΈ€ σΈ€ 

𝑔5σΈ€  (𝑝) = 27𝑝2 + 66𝑝 + 51 (19)

1 (βˆ’ ) = 1320 > 0, 3

= 87480𝑝2 + 1890 > 0.

Proof. Let β„Ž1 (𝑑) = β„ŽσΈ€  (𝑑)/𝑑, β„Ž2 (𝑑) = 𝑑4βˆ’3𝑝 β„Ž1σΈ€ σΈ€  (𝑑), β„Ž3 (𝑑) = β„Ž2σΈ€ σΈ€ σΈ€  (𝑑)/𝑑, and β„Ž4 (𝑑) = 𝑑4βˆ’3𝑝 β„Ž3σΈ€ σΈ€ σΈ€  (𝑑)/[9𝑝(𝑝 + 1)(3𝑝 + 1)(3𝑝 + 2)2 (6𝑝 + 1)]; then elaborated computations lead to

Therefore, Lemma 1(4) follows from (18) and (19). (5) Let 𝑔4 (𝑝) = 5346𝑝6 + 18225𝑝5 + 27297𝑝4 + 21996𝑝3 + 9672𝑝2 + 2264𝑝 + 320.

(20)

β„Ž (1) = 0,

(32)

β„Ž1 (𝑑) = 6 (𝑝 + 1) (6𝑝 + 1) 𝑑6𝑝+4 βˆ’ 6 (3𝑝 + 2) 𝑑6𝑝+2

Then simple computations yield 1 284 𝑔4 (βˆ’ ) = > 0, 3 3 4

(21)

+ 65988𝑝2 + 19344𝑝 + 2264, 1 𝑔4σΈ€  (βˆ’ ) = 97 > 0, 3 𝑔4σΈ€ σΈ€  (𝑝) = 160380𝑝4 + 364500𝑝3 + 327564𝑝2 + 131976𝑝 + 19344, 1 𝑔4σΈ€ σΈ€  (βˆ’ ) = 228 > 0, 3

(22)

𝑔4(4) (𝑝) = 1944 (990𝑝2 + 1125𝑝 + 337) .

βˆ’ 3𝑝 (3𝑝 + 5) 𝑑3𝑝+3 βˆ’ 3 (𝑝 + 1) (3𝑝 + 4) 𝑑3𝑝+2

(33)

+ 3 (𝑝 + 1) (3𝑝 + 2) 𝑑3𝑝 + 3𝑝 (3𝑝 + 1) 𝑑3π‘βˆ’1 (23)

(24)

(25)

𝑔4σΈ€ σΈ€ σΈ€  (𝑝) = 108 (5940𝑝3 + 10125𝑝2 + 6066𝑝 + 1222) , (26) 1 (βˆ’ ) = 11340 > 0, 3

βˆ’ 6 (2𝑝 + 1) (3𝑝 + 2) 𝑑6𝑝+1 + 3 (𝑝 + 2) (3𝑝 + 1) 𝑑3𝑝+4

3

(𝑝) = 32076𝑝 + 91125𝑝 + 109188𝑝

𝑔4σΈ€ σΈ€ σΈ€ 

1 > 27𝑝2 βˆ’ 66 Γ— ( ) + 51 = 27𝑝2 + 29 > 0. 3

Lemma 2. Suppose that β„Ž(𝑑) = (6𝑝 + 1)𝑑6𝑝+6 βˆ’ 3𝑑6𝑝+4 βˆ’ 2(3𝑝 + 2)𝑑6𝑝+3 +(3𝑝+1)𝑑3𝑝+6 βˆ’3𝑝𝑑3𝑝+5 βˆ’3(𝑝+1)𝑑3𝑝+4 +3(𝑝+1)𝑑3𝑝+2 +3𝑝𝑑3𝑝+1 βˆ’ (3𝑝 + 1)𝑑3𝑝 + 2(3𝑝 + 2)𝑑3 + 3𝑑2 βˆ’ 6𝑝 βˆ’ 1. If 𝑝 ∈ (βˆ’1/3, βˆ’1/6), then β„Ž(𝑑) > 0 for 𝑑 > 1.

1 > 87480𝑝 + 41418 + 118584 Γ— (βˆ’ ) 3 2

5

(31)

Therefore, Lemma 1(6) follows from (30) and (31).

𝑔3σΈ€ σΈ€ σΈ€  (𝑝) = 87480𝑝2 + 118584𝑝 + 41418

𝑔4σΈ€ 

(30)

(27) (28)

βˆ’ 3𝑝 (3𝑝 + 1) 𝑑3π‘βˆ’2 + 6 (3𝑝 + 2) 𝑑 + 6, β„Ž1 (1) = 0, β„Ž1σΈ€  (𝑑) = 12 (𝑝 + 1) (3𝑝 + 2) (6𝑝 + 1) 𝑑6𝑝+3 βˆ’ 12 (3𝑝 + 1) (3𝑝 + 2) 𝑑6𝑝+1 βˆ’ 6 (2𝑝 + 1) (3𝑝 + 2) (6𝑝 + 1) 𝑑6𝑝 + 3 (𝑝 + 2) (3𝑝 + 1) (3𝑝 + 4) 𝑑3𝑝+3 βˆ’ 9𝑝 (𝑝 + 1) (3𝑝 + 5) 𝑑3𝑝+2 βˆ’ 3 (𝑝 + 1) (3𝑝 + 2) (3𝑝 + 4) 𝑑3𝑝+1

It follows from (28) and the discriminant Ξ” = (1125)2 βˆ’ 4 Γ— 990 Γ— 337 = βˆ’68895 < 0 of the quadratic function 𝑓(𝑝) = 990𝑝2 + 1125𝑝 + 337 that

+ 9𝑝 (𝑝 + 1) (3𝑝 + 2) 𝑑3π‘βˆ’1

𝑔4(4) (𝑝) > 0.

βˆ’ 3𝑝 (3𝑝 βˆ’ 2) (3𝑝 + 1) 𝑑3π‘βˆ’3

(29)

Therefore, Lemma 1(5) follows from (20)–(27) and (29).

(34)

+ 3𝑝 (3𝑝 βˆ’ 1) (3𝑝 + 1) 𝑑3π‘βˆ’2

+ 6 (3𝑝 + 2) ,

(35)

4

Chinese Journal of Mathematics β„Ž1σΈ€  (1) = 0,

(36)

Γ— (3𝑝 + 4) (3𝑝 + 5) (6𝑝 + 1) 𝑑3𝑝+2

β„Ž2 (𝑑) = 36 (𝑝 + 1) (2𝑝 + 1) (3𝑝 + 2) (6𝑝 + 1) 𝑑3𝑝+6

βˆ’ 36 (𝑝 + 1) (3𝑝 + 1) (3𝑝 + 2)

βˆ’ 12 (3𝑝 + 1) (3𝑝 + 2) (6𝑝 + 1) 𝑑3𝑝+4

βˆ’ 108𝑝 (𝑝 + 1) (2𝑝 + 1) (3𝑝 + 1) (3𝑝 + 2)

+ 9 (𝑝 + 1) (𝑝 + 2) (3𝑝 + 1) (3𝑝 + 4) 𝑑6

3π‘βˆ’1

(37)

βˆ’ 3 (𝑝 + 1) (3𝑝 + 1) (3𝑝 + 2) (3𝑝 + 4) 𝑑4

(38)

β„Ž2σΈ€  (𝑑) = 108 (𝑝 + 1) (𝑝 + 2) (2𝑝 + 1) (3𝑝 + 2) (6𝑝 + 1) 𝑑3𝑝+5

Γ— (3𝑝 + 1) (3𝑝 + 4) 𝑑2

Γ— (1134𝑝4 + 3429𝑝3 + 3879𝑝2 + 1950𝑝 + 448) , (44) β„Ž3σΈ€  (𝑑) = 108 (𝑝 + 1) (𝑝 + 2) (2𝑝 + 1) (3𝑝 + 2)

βˆ’ 12 (3𝑝 + 1) (3𝑝 + 2) (3𝑝 + 4) (6𝑝 + 1) 𝑑3𝑝+3

Γ— (3𝑝 + 4) (3𝑝 + 5) (6𝑝 + 1) 𝑑3𝑝+1

βˆ’ 108𝑝 (𝑝 + 1) (2𝑝 + 1) (3𝑝 + 2) (6𝑝 + 1) 𝑑3𝑝+2

βˆ’ 108𝑝 (𝑝 + 1) (3𝑝 + 1) (3𝑝 + 2)

2

2

Γ— (3𝑝 + 4) (6𝑝 + 1) 𝑑3π‘βˆ’1

+ 54 (𝑝 + 1) (𝑝 + 2) (3𝑝 + 1) (3𝑝 + 4) 𝑑5

βˆ’ 108𝑝 (𝑝 + 1) (2𝑝 + 1) (3𝑝 βˆ’ 1) (3𝑝 + 1) (3𝑝 + 2)

βˆ’ 45𝑝 (𝑝 + 1) (3𝑝 + 2) (3𝑝 + 5) 𝑑4

2

Γ— (6𝑝 + 1) 𝑑3π‘βˆ’2 + 2160 (𝑝 + 1) (𝑝 + 2)

βˆ’ 12 (𝑝 + 1) (3𝑝 + 1) (3𝑝 + 2) (3𝑝 + 4) 𝑑3

Γ— (3𝑝 + 1) (3𝑝 + 4) 𝑑

+ 18𝑝 (𝑝 + 1) (3𝑝 βˆ’ 1) (3𝑝 + 2) 𝑑

βˆ’ 540𝑝 (𝑝 + 1) (3𝑝 + 2) (3𝑝 + 5) ,

+ 3𝑝 (3𝑝 βˆ’ 1) (3𝑝 βˆ’ 2) (3𝑝 + 1) ,

(45) (39)

β„Ž2σΈ€ σΈ€ 

+ 1080 (𝑝 + 1) (𝑝 + 2)

β„Ž3 (1) = 36 (3𝑝 + 1) (𝑝 + 1)

βˆ’ 9𝑝 (𝑝 βˆ’ 1) (3𝑝 βˆ’ 2) (3𝑝 + 1) ,

β„Ž2σΈ€ 

(43)

βˆ’ 72 (𝑝 + 1) (3𝑝 + 1) (3𝑝 + 2) (3𝑝 + 4) ,

+ 3𝑝 (3𝑝 βˆ’ 2) (3𝑝 βˆ’ 1) (3𝑝 + 1) 𝑑

β„Ž2 (1) = 12 (3𝑝 + 1) (18𝑝 + 21𝑝 + 8) ,

Γ— (6𝑝 + 1) 𝑑

2

βˆ’ 540𝑝 (𝑝 + 1) (3𝑝 + 2) (3𝑝 + 5) 𝑑

+ 9𝑝 (𝑝 + 1) (3𝑝 βˆ’ 1) (3𝑝 + 2) 𝑑2

2

2

Γ— (3𝑝 + 4) (6𝑝 + 1) 𝑑3𝑝

βˆ’ 36𝑝 (2𝑝 + 1) (3𝑝 + 2) (6𝑝 + 1) 𝑑3𝑝+3

βˆ’ 9𝑝 (𝑝 + 1) (3𝑝 + 2) (3𝑝 + 5) 𝑑5

β„Ž3 (𝑑) = 108 (𝑝 + 1) (𝑝 + 2) (2𝑝 + 1) (3𝑝 + 2)

2

(1) = 12 (3𝑝 + 1) (9𝑝 + 7) (18𝑝 + 21𝑝 + 8) ,

(40)

β„Ž3σΈ€  (1) = 108 (𝑝 + 1) (3𝑝 + 1) Γ— (1458𝑝5 + 4941𝑝4 + 6903𝑝3

(𝑑) = 108 (𝑝 + 1) (𝑝 + 2) (2𝑝 + 1)

(46)

+4809𝑝2 + 1714𝑝 + 320) ,

Γ— (3𝑝 + 2) (3𝑝 + 5) (6𝑝 + 1) 𝑑3𝑝+4

β„Ž3σΈ€ σΈ€  (𝑑) = 108 (𝑝 + 1) (𝑝 + 2) (2𝑝 + 1) (3𝑝 + 1) 2

βˆ’ 36 (𝑝 + 1) (3𝑝 + 1) (3𝑝 + 2)

Γ— (3𝑝 + 2) (3𝑝 + 4) (3𝑝 + 5) (6𝑝 + 1) 𝑑3𝑝

Γ— (3𝑝 + 4) (6𝑝 + 1) 𝑑3𝑝+2

βˆ’ 108𝑝 (𝑝 + 1) (3𝑝 βˆ’ 1) (3𝑝 + 1) (3𝑝 + 2) (3𝑝 + 4)

2

2

βˆ’ 108𝑝 (𝑝 + 1) (2𝑝 + 1) (3𝑝 + 2) (6𝑝 + 1) 𝑑3𝑝+1

Γ— (6𝑝 + 1) 𝑑3π‘βˆ’2 βˆ’ 108𝑝 (𝑝 + 1) (2𝑝 + 1) (3𝑝 βˆ’ 2)

+ 270 (𝑝 + 1) (𝑝 + 2) (3𝑝 + 1) (3𝑝 + 4) 𝑑4

Γ— (3𝑝 βˆ’ 1) (3𝑝 + 1) (3𝑝 + 2) (6𝑝 + 1) 𝑑3π‘βˆ’3

βˆ’ 180𝑝 (𝑝 + 1) (3𝑝 + 2) (3𝑝 + 5) 𝑑3

+ 2160 (𝑝 + 1) (𝑝 + 2) (3𝑝 + 1) (3𝑝 + 4) ,

2

(47)

βˆ’ 36 (𝑝 + 1) (3𝑝 + 1) (3𝑝 + 2) (3𝑝 + 4) 𝑑2

β„Ž3σΈ€ σΈ€  (1) = 108 (𝑝 + 1) (3𝑝 + 1)

+ 18𝑝 (𝑝 + 1) (3𝑝 βˆ’ 1) (3𝑝 + 2) , (41) β„Ž2σΈ€ σΈ€  (1) = 36 (3𝑝 + 1) (𝑝 + 1) (270𝑝3 + 567𝑝2 + 384𝑝 + 104), (42)

Γ— (5346𝑝6 + 18225𝑝5 + 27297𝑝4 + 21996𝑝3 +9672𝑝2 + 2264𝑝 + 320) , (48)

Chinese Journal of Mathematics

5

β„Ž4 (𝑑) = 36 (𝑝 + 2) (2𝑝 + 1) (3𝑝 + 4) (3𝑝 + 5) 𝑑3

Then simple computations lead to

βˆ’ 12 (1 βˆ’ 3𝑝) (2 βˆ’ 3𝑝) (3𝑝 + 4) 𝑑

𝑓 (1) = 0,

+ 36 (1 βˆ’ 𝑝) (1 βˆ’ 3𝑝) (2 βˆ’ 3𝑝) (2𝑝 + 1)

𝑓󸀠 (𝑑) =

> 36 (𝑝 + 2) (2𝑝 + 1) (3𝑝 + 4) (3𝑝 + 5) 𝑑 (49)

βˆ’ 12 (1 βˆ’ 3𝑝) (2 βˆ’ 3𝑝) (3𝑝 + 4) 𝑑 + 36 (1 βˆ’ 𝑝) (1 βˆ’ 3𝑝) (2 βˆ’ 3𝑝) (2𝑝 + 1)

β„Ž (𝑑) , 𝑑 (𝑑 + 1) (𝑑2 + 1) (𝑑3𝑝 + 1) (𝑑3𝑝+3 + 1)

𝑓 (𝑑) > 0

It follows from Lemma 1(6) and (49) that (50)

for 𝑑 > 1. From Lemma 1(1)–(5) and (38)–(48), we clearly see that β„Ž2 (1) > 0, β„Ž2σΈ€  (1) > 0, (51)

β„Ž3 (1) > 0,

𝛼

𝛼

1βˆ’π›Ό

Lemma 3. Inequality [𝑀2/3 (π‘Ž, 𝑏)] [𝐿 βˆ’1/3 (π‘Ž, 𝑏)] > 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) holds for any 𝛼 ∈ (0, 1) and all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏. Proof. Without loss of generality, we assume that π‘Ž > 𝑏. Let 𝑑 = (π‘Ž/𝑏)1/3 > 1 and 𝑝 = (𝛼 βˆ’ 2)/6; then 𝑝 ∈ (βˆ’1/3, βˆ’1/6), 𝛼 = 6𝑝 + 2, and from (1) and (4) we have [𝑀2/3 (π‘Ž, 𝑏)] [𝐿 βˆ’1/3 (π‘Ž, 𝑏)]

(𝑑 + 1)6𝑝+1

βˆ’29𝑝+3 𝑑6𝑝+1 (𝑑3𝑝+3 + 1) ])

+ [𝑀2/3 (π‘Ž, 𝑏)] [𝐿 βˆ’1/3 (π‘Ž, 𝑏)]

1βˆ’π›Ό

βˆ’ 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) . (57)

Therefore, 𝐼𝛼 (π‘Ž, 𝑏)𝐿1βˆ’π›Ό (π‘Ž, 𝑏) > 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) follows from (7) and (10) together with Lemma 3 and (57). Next, we prove that 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) and 𝐿 0 (π‘Ž, 𝑏) are the best possible lower and upper Lehmer mean bounds for the product 𝐼𝛼 (π‘Ž, 𝑏)𝐿1βˆ’π›Ό (π‘Ž, 𝑏). For any 0 < πœ€ < 1 and π‘₯ > 0, from (1) and (2) we have 𝐼𝛼 (1, π‘₯) 𝐿1βˆ’π›Ό (1, π‘₯) π‘₯ β†’ +∞ 𝐿 βˆ’πœ€ (1, π‘₯) = lim [

(52)

π‘₯ β†’ +∞

π‘₯βˆ’πœ€ + 1 𝛼((log π‘₯)/(π‘₯βˆ’1)βˆ’1) 𝑒 π‘₯πœ€βˆ’1 + 1

Γ—(1 βˆ’

βˆ’1

Γ— (29𝑝+3 𝑑6𝑝+1 (𝑑3𝑝 + 1)) .

π‘₯πœ€ 1 1βˆ’π›Ό ] = +∞, ) 1βˆ’π›Ό π‘₯ (log π‘₯)

𝐿 (π›Όβˆ’2)/6+πœ€ (1, 1 + π‘₯) βˆ’ 𝐼𝛼 (1, 1 + π‘₯) 𝐿1βˆ’π›Ό (1, 1 + π‘₯)

Let 𝑓 (𝑑) = log (𝑑3𝑝 + 1) + (3𝑝 + 2) log (𝑑2 + 1) + (6𝑝 + 1) log (𝑑 + 1) βˆ’ (6𝑝 + 1) log 𝑑 3𝑝+3

βˆ’ log (𝑑

+ 1) βˆ’ 3 (3𝑝 + 1) log 2.

1βˆ’π›Ό

lim

βˆ’ 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏)

3𝑝+2

= (𝑏 [(𝑑3𝑝 + 1) (𝑑2 + 1)

Proof. Inequality 𝐼𝛼 (π‘Ž, 𝑏)𝐿1βˆ’π›Ό (π‘Ž, 𝑏) < 𝐿 0 (π‘Ž, 𝑏) follows directly from (10) and (11). For the other inequality, we note that

𝛼

(1) > 0.

1βˆ’π›Ό

Theorem 4. Inequality 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) < 𝐼𝛼 (π‘Ž, 𝑏)𝐿1βˆ’π›Ό (π‘Ž, 𝑏) < 𝐿 0 (π‘Ž, 𝑏) holds for any 𝛼 ∈ (0, 1) and all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏, and 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) and 𝐿 0 (π‘Ž, 𝑏) are the best possible lower and upper Lehmer mean bounds for the product 𝐼𝛼 (π‘Ž, 𝑏)𝐿1βˆ’π›Ό (π‘Ž, 𝑏).

= 𝐼𝛼 (π‘Ž, 𝑏) 𝐿1βˆ’π›Ό (π‘Ž, 𝑏) βˆ’ [𝑀2/3 (π‘Ž, 𝑏)] [𝐿 βˆ’1/3 (π‘Ž, 𝑏)]

Therefore, Lemma 2 follows from 9𝑝(𝑝 + 1)(3𝑝 + 1)(3𝑝 + 2)2 (6𝑝+1) > 0 and (32)–(36) together with (50) and (51).

𝛼

3. Main Results

𝐼𝛼 (π‘Ž, 𝑏) 𝐿1βˆ’π›Ό (π‘Ž, 𝑏) βˆ’ 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏)

β„Ž3σΈ€  (1) > 0, β„Ž3σΈ€ σΈ€ 

(56)

for 𝑑 > 1. Therefore, Lemma 3 follows from (52) and (53) together with (56).

+ 36 (1 βˆ’ 𝑝) (1 βˆ’ 3𝑝) (2 βˆ’ 3𝑝) (2𝑝 + 1) .

β„Ž2σΈ€ σΈ€  (1) > 0,

(55)

where β„Ž(𝑑) is defined as in Lemma 2. From (54) and (55) together with Lemma 2, we clearly see that

= 24 (3𝑝 + 4) (9𝑝3 + 33𝑝2 + 51𝑝 + 14) 𝑑

β„Ž4 (𝑑) > 0

(54)

=

(1 + π‘₯)(𝛼+4)/6+πœ€ + 1 βˆ’ 𝑒𝛼[(1+π‘₯) log(1+π‘₯)/π‘₯βˆ’1] (1 + π‘₯)(π›Όβˆ’2)/6+πœ€ + 1

(53) Γ—

π‘₯1βˆ’π›Ό 𝛼. [log (1 + π‘₯)]

(58)

6

Chinese Journal of Mathematics

Letting π‘₯ β†’ 0 and making use of Taylor expansion, one has

=

(1 + π‘₯)(𝛼+4)/6+πœ€ + 1 βˆ’ 𝑒𝛼[(1+π‘₯) log(1+π‘₯)/π‘₯βˆ’1] (1 + π‘₯)(π›Όβˆ’2)/6+πœ€ + 1 Γ—

Letting π‘₯ β†’ 0 and making use of Taylor expansion, one has (1 βˆ’ 𝛼) π‘₯ (1 + π‘₯)(𝛼+4)/6+πœ€ + 1 βˆ’ 𝛼𝑒(1+π‘₯) log(1+π‘₯)/π‘₯βˆ’1 βˆ’ (π›Όβˆ’2)/6+πœ€ log (1 + π‘₯) +1 (1 + π‘₯)

𝛼 1 1 π‘₯ + 𝛼 ( 𝛼 βˆ’ ) π‘₯2 + π‘œ (π‘₯2 )] 2 8 6

1 𝛼 1 1 1 Γ— [1 + ( βˆ’ ) π‘₯ + ( 𝛼2 βˆ’ 𝛼 βˆ’ ) π‘₯2 + π‘œ (π‘₯2 )] 2 2 8 24 12 1 𝛼 1 πœ€ =1+ π‘₯+( βˆ’ + ) π‘₯2 2 24 12 4 1 𝛼 1 βˆ’ [1 + π‘₯ + ( βˆ’ ) π‘₯2 ] + π‘œ (π‘₯2 ) 2 24 12 πœ€ 2 = π‘₯ + π‘œ (π‘₯2 ) . 4

(59)

imply that for any 0 < πœ€ < 1, > 1 and 𝛿1 = 𝛿1 (𝛼, πœ€) > 0, > 𝐿 βˆ’πœ€ (1, π‘₯) for π‘₯ ∈ (𝑋1 , +∞) 𝐼𝛼 (1, 1 + π‘₯)𝐿1βˆ’π›Ό (1, 1 + π‘₯) for

Proof. Inequality 𝛼𝐼(π‘Ž, 𝑏) + (1 βˆ’ 𝛼)𝐿(π‘Ž, 𝑏) < 𝐿 0 (π‘Ž, 𝑏) follows directly from (10) and (11), and inequality 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) < 𝛼𝐼(π‘Ž, 𝑏) + (1 βˆ’ 𝛼)𝐿(π‘Ž, 𝑏) follows from Theorem 4. Next, we prove that 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) and 𝐿 0 (π‘Ž, 𝑏) are the best possible lower and upper Lehmer mean bounds for the sum 𝛼𝐼(π‘Ž, 𝑏) + (1 βˆ’ 𝛼)𝐿(π‘Ž, 𝑏). For any 0 < πœ€ < 1 and π‘₯ > 0, from (1) and (2), we have

lim

𝛼𝐼 (1, π‘₯) + (1 βˆ’ 𝛼) 𝐿 (1, π‘₯) 𝐿 βˆ’πœ€ (1, π‘₯)

= lim [ π‘₯ β†’ +∞

π‘₯βˆ’πœ€ + 1 π‘₯πœ€βˆ’1 + 1

Γ— (𝛼𝑒(log π‘₯)/(π‘₯βˆ’1) π‘₯πœ€ +

1 𝛼 1 πœ€ =1+ π‘₯+( βˆ’ + ) π‘₯2 + π‘œ (π‘₯2 ) 2 24 12 4 1 1 βˆ’ 𝛼 [1 + π‘₯ βˆ’ π‘₯2 + π‘œ (π‘₯2 )] 2 24

(62)

1 1 βˆ’ (1 βˆ’ 𝛼) [1 + π‘₯ βˆ’ π‘₯2 + π‘œ (π‘₯2 )] 2 12 πœ€ = π‘₯2 + π‘œ (π‘₯2 ) . 4

Theorem 5. Inequality 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) < 𝛼𝐼(π‘Ž, 𝑏) + (1 βˆ’ 𝛼)𝐿(π‘Ž, 𝑏) < 𝐿 0 (π‘Ž, 𝑏) holds for any 𝛼 ∈ (0, 1) and all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏, and 𝐿 (π›Όβˆ’2)/6 (π‘Ž, 𝑏) and 𝐿 0 (π‘Ž, 𝑏) are the best possible lower and upper Lehmer mean bounds for the sum 𝛼𝐼(π‘Ž, 𝑏) + (1 βˆ’ 𝛼)𝐿(π‘Ž, 𝑏).

π‘₯ β†’ +∞

(1 βˆ’ 𝛼) π‘₯ . log (1 + π‘₯) (61)

1 𝛼 1 πœ€ =1+ π‘₯+( βˆ’ + ) π‘₯2 + π‘œ (π‘₯2 ) 2 24 12 4

Equations (58) and (59) there exist 𝑋1 = 𝑋1 (𝛼, πœ€) such that 𝐼𝛼 (1, π‘₯)𝐿1βˆ’π›Ό (1, π‘₯) and 𝐿 (π›Όβˆ’2)/6+πœ€ (1, 1 + π‘₯) > π‘₯ ∈ (0, 𝛿1 ).

(1 + π‘₯)(𝛼+4)/6+πœ€ + 1 βˆ’ 𝛼𝑒(1+π‘₯) log(1+π‘₯)/π‘₯βˆ’1 (π›Όβˆ’2)/6+πœ€ + 1 + π‘₯) (1 βˆ’

π‘₯1βˆ’π›Ό 𝛼 [log (1 + π‘₯)]

βˆ’ [1 +

𝐿 (π›Όβˆ’2)/6+πœ€ (1, 1 + π‘₯) βˆ’ 𝛼𝐼 (1, 1 + π‘₯) βˆ’ (1 βˆ’ 𝛼) 𝐿 (1, 1 + π‘₯)

(1 βˆ’ 𝛼) (1 βˆ’ 1/π‘₯) πœ€ π‘₯ )] log π‘₯

= +∞, (60)

Equations (60)-(62) imply that for any 0 < πœ€ < 1, there exist 𝑋2 = 𝑋2 (𝛼, πœ€) > 1 and 𝛿2 = 𝛿2 (𝛼, πœ€) > 0, such that 𝛼𝐼(1, π‘₯) + (1 βˆ’ 𝛼)𝐿(1, π‘₯) > 𝐿 βˆ’πœ€ (1, π‘₯) for π‘₯ ∈ (𝑋2 , +∞) and 𝐿 (π›Όβˆ’2)/6+πœ€ (1, 1 + π‘₯) > 𝛼𝐼(1, 1 + π‘₯) + (1 βˆ’ 𝛼)𝐿(1, 1 + π‘₯) for π‘₯ ∈ (0, 𝛿2 ).

Acknowledgments This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307 and the Natural Science Foundation of Zhejiang Province under Grants LY13H070004 and LY13A010004.

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