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 πΌ β (0, 1), we answer the questions: what are the greatest values π and π and the least values π and π, such that the inequalities πΏ π (π, π) < πΌπΌ (π, π)πΏ1βπΌ (π, π) < πΏ π (π, π) and πΏ π (π, π) < πΌπΌ(π, π) + (1 β πΌ)πΏ(π, π) < πΏ π (π, π) hold for all π, π > 0 with π =ΜΈ π? Here, πΌ(π, π), πΏ(π, π), and πΏ π (π, π) denote the identric, logarithmic, and πth Lehmer means of two positive numbers π and π, respectively.
1. Introduction For π, π > 0 with π =ΜΈ π, the logarithmic mean πΏ(π, π) and identric mean πΌ(π, π) are defined by πΏ (π, π) =
πβπ , log π β log π
(1)
1/(πβπ)
1 ππ πΌ (π, π) = ( π ) π π
,
(2)
respectively. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for πΏ and πΌ can be found in the literature [1β19]. In [14, 17, 20], inequalities between πΏ, πΌ, and the classical arithmetic-geometric mean of Gauss are proved. The ratio of identric means leads to the weighted geometric mean 2
2
πΌ (π , π ) πΌ (π, π)
1/(π+π)
= (ππ ππ )
,
(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 [22β24]. In [22], the authors study a variant of Jensenβs functional equation involving πΏ, which appears in a heat conduction
problem. A representation of πΏ 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 πΏ can be expressed in terms of Gaussβ hypergeometric function 2 πΉ1 . And in [26], the authors prove that the reciprocal of the logarithmic mean is strictly totally positive; that is, every π Γ π determinant with elements 1/πΏ(ππ , ππ ), where 0 < π1 < π2 < β
β
β
< ππ and 0 < π1 < π2 < β
β
β
< ππ , is positive for all π β₯ 1. For π β R, the power mean ππ (π, π) and Lehmer mean πΏ π (π, π) of order π of two positive numbers π and π with π =ΜΈ π are defined by 1/π
ππ (π, π) = ( πΏ π (π, π) =
ππ + ππ ) 2
ππ+1 + ππ+1 . ππ + ππ
,
(4) (5)
It is well known that ππ (π, π) and πΏ π (π, π) are strictly increasing with respect to π β R for fixed π, π > 0 with π =ΜΈ π. The main properties for ππ and πΏ π are given in [27β32]. Let π΄(π, π) = (π + π)/2, πΊ(π, π) = βππ, and π»(π, π) = 2ππ/(π+π) be the arithmetic, geometric, and harmonic means
2
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of two positive numbers π and π, respectively. Then it is well known that min {π, π} < π» (π, π) = πΏ β1 (π, π) = πβ1 (π, π) < πΊ (π, π) = πΏ β1/2 (π, π) = π0 (π, π) < πΏ (π, π) < πΌ (π, π) < π΄ (π, π)
(6)
= πΏ 0 (π, π) = π1 (π, π) < max {π, π}
(6) 9π3 + 33π2 + 51π + 14 > 0.
1 18π2 + 21π + 8 > 18π2 + (β ) Γ 21 + 8 3
(2) Let (8)
π1 (π) = 270π3 + 567π2 + 384π + 104.
1 π1 (β ) = 29 > 0, 3 π1σΈ (π) = 6 (135π2 + 189π + 64)
πΌπ΄ (π, π) + (1 β πΌ) πΊ (π, π) (9)
1 > 6 [135π + 189 Γ (β ) + 64] 3 = 6 (135π2 + 1) > 0.
for all π, π > 0 with π =ΜΈ π if and only if πΌ β€ 2/3, π½ β₯ 2/π = 0.73575 . . ., and π β€ log 2/(1 + log 2) = 0.40938 . . .. The following sharp upper and lower Lehmer mean bounds for πΏ, πΌ, (πΏπΌ)1/2 , and (πΏ + πΌ)/2 are presented in [34]:
Therefore, Lemma 1(2) follows from (14) and (15). (3) Let
πΏ β1/3 (π, π) < πΏ (π, π) < πΏ 0 (π, π) ,
(10)
πβ1/6 (π, π) < πΌ (π, π) < πΏ 0 (π, π) ,
(11)
πΏ β1/4 (π, π) < πΌ1/2 (π, π) πΏ1/2 (π, π) < πΏ 0 (π, π) , (12)
for all π, π > 0 with π =ΜΈ π. The purpose of this paper is to present the best possible upper and lower Lehmer mean bounds of the product πΌπΌ (π, π)πΏ1βπΌ (π, π) and the sum πΌπΌ(π, π) + (1 β πΌ)πΏ(π, π) for any πΌ β (0, 1) and all π, π > 0 with π =ΜΈ π.
π2 (π) = 1134π4 + 3429π3 + 3879π2 + 1950π + 448. (16) Then simple computations yield 1 π2 (β ) = 116 > 0, 3 π2σΈ (π) = 4536π3 + 10287π2 + 7758π + 1950, 1 π2σΈ (β ) = 339 > 0, 3
(17)
π2σΈ σΈ (π) = 13608π2 + 20574π + 7758 1 > 13608π2 + 20574 Γ (β ) + 7758 3 = 13608π2 + 900 > 0.
2. Lemmas In order to establish our main result, we need several lemmas, which we present in this section. Lemma 1. If π β₯ β1/3, then the following statements are true: 2
(1) 18π + 21π + 8 > 0; 3
(15)
2
1 ππ (π, π) < (πΌ (π, π) + πΏ (π, π)) 2
1 (πΌ (π, π) + πΏ (π, π)) < πΏ 0 (π, π) 2
(14)
Then simple computations lead to
for all π, π > 0 with π =ΜΈ π. Alzer and Qiu [19] proved that
πΏ β1/4 (π, π)
0.
(7)
1 (πΌ (π, π) + πΏ (π, π)) < π1/2 (π, π) 2
< πΌ (π, π) < π½π΄ (π, π) + (1 β π½) πΊ (π, π) ,
(5) 5346π6 + 18225π5 + 27297π4 + 21996π3 + 9672π2 + 2264π + 320 > 0;
2
π0 (π, π) < πΏ (π, π) < π1/3 (π, π) , π0 (π, π) < βπΌ (π, π) πΏ (π, π) < π1/2 (π, π) ,
(4) 1458π5 +4941π4 +6903π3 +4809π2 +1714π+320 > 0;
Proof. (1) We clearly see that
for all π, π > 0 with π =ΜΈ π. The following sharp bounds for πΏ, πΌ, (πΏπΌ)1/2 , and (πΏ + πΌ)/2 in terms of power means are proved in [2β4, 6, 8, 9, 33]: π2/3 (π, π) < πΌ (π, π) < πlog 2 (π, π) ,
(3) 1134π4 + 3429π3 + 3879π2 + 1950π + 448 > 0;
2
(2) 270π + 567π + 384π + 104 > 0;
Therefore, Lemma 1(3) follows from (16) and (17). (4) Let π3 (π) = 1458π5 + 4941π4 + 6903π3 + 4809π2 + 1714π + 320.
(18)
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3
Then simple computations lead to
(6) Let
1 247 π3 (β ) = > 0, 3 3
π5 (π) = 9π3 + 33π2 + 51π + 14.
π3σΈ (π) = 7290π4 + 19764π3 + 20709π2
Then we have
+ 9618π + 1714,
1 1 π5 (β ) = > 0, 3 3
1 π3σΈ (β ) = 167 > 0, 3 π3σΈ σΈ
3
2
(π) = 29160π + 59292π + 41418π + 9618, π3σΈ σΈ
π5σΈ (π) = 27π2 + 66π + 51 (19)
1 (β ) = 1320 > 0, 3
= 87480π2 + 1890 > 0.
Proof. Let β1 (π‘) = βσΈ (π‘)/π‘, β2 (π‘) = π‘4β3π β1σΈ σΈ (π‘), β3 (π‘) = β2σΈ σΈ σΈ (π‘)/π‘, and β4 (π‘) = π‘4β3π β3σΈ σΈ σΈ (π‘)/[9π(π + 1)(3π + 1)(3π + 2)2 (6π + 1)]; then elaborated computations lead to
Therefore, Lemma 1(4) follows from (18) and (19). (5) Let π4 (π) = 5346π6 + 18225π5 + 27297π4 + 21996π3 + 9672π2 + 2264π + 320.
(20)
β (1) = 0,
(32)
β1 (π‘) = 6 (π + 1) (6π + 1) π‘6π+4 β 6 (3π + 2) π‘6π+2
Then simple computations yield 1 284 π4 (β ) = > 0, 3 3 4
(21)
+ 65988π2 + 19344π + 2264, 1 π4σΈ (β ) = 97 > 0, 3 π4σΈ σΈ (π) = 160380π4 + 364500π3 + 327564π2 + 131976π + 19344, 1 π4σΈ σΈ (β ) = 228 > 0, 3
(22)
π4(4) (π) = 1944 (990π2 + 1125π + 337) .
β 3π (3π + 5) π‘3π+3 β 3 (π + 1) (3π + 4) π‘3π+2
(33)
+ 3 (π + 1) (3π + 2) π‘3π + 3π (3π + 1) π‘3πβ1 (23)
(24)
(25)
π4σΈ σΈ σΈ (π) = 108 (5940π3 + 10125π2 + 6066π + 1222) , (26) 1 (β ) = 11340 > 0, 3
β 6 (2π + 1) (3π + 2) π‘6π+1 + 3 (π + 2) (3π + 1) π‘3π+4
3
(π) = 32076π + 91125π + 109188π
π4σΈ σΈ σΈ
1 > 27π2 β 66 Γ ( ) + 51 = 27π2 + 29 > 0. 3
Lemma 2. Suppose that β(π‘) = (6π + 1)π‘6π+6 β 3π‘6π+4 β 2(3π + 2)π‘6π+3 +(3π+1)π‘3π+6 β3ππ‘3π+5 β3(π+1)π‘3π+4 +3(π+1)π‘3π+2 +3ππ‘3π+1 β (3π + 1)π‘3π + 2(3π + 2)π‘3 + 3π‘2 β 6π β 1. If π β (β1/3, β1/6), then β(π‘) > 0 for π‘ > 1.
1 > 87480π + 41418 + 118584 Γ (β ) 3 2
5
(31)
Therefore, Lemma 1(6) follows from (30) and (31).
π3σΈ σΈ σΈ (π) = 87480π2 + 118584π + 41418
π4σΈ
(30)
(27) (28)
β 3π (3π + 1) π‘3πβ2 + 6 (3π + 2) π‘ + 6, β1 (1) = 0, β1σΈ (π‘) = 12 (π + 1) (3π + 2) (6π + 1) π‘6π+3 β 12 (3π + 1) (3π + 2) π‘6π+1 β 6 (2π + 1) (3π + 2) (6π + 1) π‘6π + 3 (π + 2) (3π + 1) (3π + 4) π‘3π+3 β 9π (π + 1) (3π + 5) π‘3π+2 β 3 (π + 1) (3π + 2) (3π + 4) π‘3π+1
It follows from (28) and the discriminant Ξ = (1125)2 β 4 Γ 990 Γ 337 = β68895 < 0 of the quadratic function π(π) = 990π2 + 1125π + 337 that
+ 9π (π + 1) (3π + 2) π‘3πβ1
π4(4) (π) > 0.
β 3π (3π β 2) (3π + 1) π‘3πβ3
(29)
Therefore, Lemma 1(5) follows from (20)β(27) and (29).
(34)
+ 3π (3π β 1) (3π + 1) π‘3πβ2
+ 6 (3π + 2) ,
(35)
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Chinese Journal of Mathematics β1σΈ (1) = 0,
(36)
Γ (3π + 4) (3π + 5) (6π + 1) π‘3π+2
β2 (π‘) = 36 (π + 1) (2π + 1) (3π + 2) (6π + 1) π‘3π+6
β 36 (π + 1) (3π + 1) (3π + 2)
β 12 (3π + 1) (3π + 2) (6π + 1) π‘3π+4
β 108π (π + 1) (2π + 1) (3π + 1) (3π + 2)
+ 9 (π + 1) (π + 2) (3π + 1) (3π + 4) π‘6
3πβ1
(37)
β 3 (π + 1) (3π + 1) (3π + 2) (3π + 4) π‘4
(38)
β2σΈ (π‘) = 108 (π + 1) (π + 2) (2π + 1) (3π + 2) (6π + 1) π‘3π+5
Γ (3π + 1) (3π + 4) π‘2
Γ (1134π4 + 3429π3 + 3879π2 + 1950π + 448) , (44) β3σΈ (π‘) = 108 (π + 1) (π + 2) (2π + 1) (3π + 2)
β 12 (3π + 1) (3π + 2) (3π + 4) (6π + 1) π‘3π+3
Γ (3π + 4) (3π + 5) (6π + 1) π‘3π+1
β 108π (π + 1) (2π + 1) (3π + 2) (6π + 1) π‘3π+2
β 108π (π + 1) (3π + 1) (3π + 2)
2
2
Γ (3π + 4) (6π + 1) π‘3πβ1
+ 54 (π + 1) (π + 2) (3π + 1) (3π + 4) π‘5
β 108π (π + 1) (2π + 1) (3π β 1) (3π + 1) (3π + 2)
β 45π (π + 1) (3π + 2) (3π + 5) π‘4
2
Γ (6π + 1) π‘3πβ2 + 2160 (π + 1) (π + 2)
β 12 (π + 1) (3π + 1) (3π + 2) (3π + 4) π‘3
Γ (3π + 1) (3π + 4) π‘
+ 18π (π + 1) (3π β 1) (3π + 2) π‘
β 540π (π + 1) (3π + 2) (3π + 5) ,
+ 3π (3π β 1) (3π β 2) (3π + 1) ,
(45) (39)
β2σΈ σΈ
+ 1080 (π + 1) (π + 2)
β3 (1) = 36 (3π + 1) (π + 1)
β 9π (π β 1) (3π β 2) (3π + 1) ,
β2σΈ
(43)
β 72 (π + 1) (3π + 1) (3π + 2) (3π + 4) ,
+ 3π (3π β 2) (3π β 1) (3π + 1) π‘
β2 (1) = 12 (3π + 1) (18π + 21π + 8) ,
Γ (6π + 1) π‘
2
β 540π (π + 1) (3π + 2) (3π + 5) π‘
+ 9π (π + 1) (3π β 1) (3π + 2) π‘2
2
2
Γ (3π + 4) (6π + 1) π‘3π
β 36π (2π + 1) (3π + 2) (6π + 1) π‘3π+3
β 9π (π + 1) (3π + 2) (3π + 5) π‘5
β3 (π‘) = 108 (π + 1) (π + 2) (2π + 1) (3π + 2)
2
(1) = 12 (3π + 1) (9π + 7) (18π + 21π + 8) ,
(40)
β3σΈ (1) = 108 (π + 1) (3π + 1) Γ (1458π5 + 4941π4 + 6903π3
(π‘) = 108 (π + 1) (π + 2) (2π + 1)
(46)
+4809π2 + 1714π + 320) ,
Γ (3π + 2) (3π + 5) (6π + 1) π‘3π+4
β3σΈ σΈ (π‘) = 108 (π + 1) (π + 2) (2π + 1) (3π + 1) 2
β 36 (π + 1) (3π + 1) (3π + 2)
Γ (3π + 2) (3π + 4) (3π + 5) (6π + 1) π‘3π
Γ (3π + 4) (6π + 1) π‘3π+2
β 108π (π + 1) (3π β 1) (3π + 1) (3π + 2) (3π + 4)
2
2
β 108π (π + 1) (2π + 1) (3π + 2) (6π + 1) π‘3π+1
Γ (6π + 1) π‘3πβ2 β 108π (π + 1) (2π + 1) (3π β 2)
+ 270 (π + 1) (π + 2) (3π + 1) (3π + 4) π‘4
Γ (3π β 1) (3π + 1) (3π + 2) (6π + 1) π‘3πβ3
β 180π (π + 1) (3π + 2) (3π + 5) π‘3
+ 2160 (π + 1) (π + 2) (3π + 1) (3π + 4) ,
2
(47)
β 36 (π + 1) (3π + 1) (3π + 2) (3π + 4) π‘2
β3σΈ σΈ (1) = 108 (π + 1) (3π + 1)
+ 18π (π + 1) (3π β 1) (3π + 2) , (41) β2σΈ σΈ (1) = 36 (3π + 1) (π + 1) (270π3 + 567π2 + 384π + 104), (42)
Γ (5346π6 + 18225π5 + 27297π4 + 21996π3 +9672π2 + 2264π + 320) , (48)
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5
β4 (π‘) = 36 (π + 2) (2π + 1) (3π + 4) (3π + 5) π‘3
Then simple computations lead to
β 12 (1 β 3π) (2 β 3π) (3π + 4) π‘
π (1) = 0,
+ 36 (1 β π) (1 β 3π) (2 β 3π) (2π + 1)
πσΈ (π‘) =
> 36 (π + 2) (2π + 1) (3π + 4) (3π + 5) π‘ (49)
β 12 (1 β 3π) (2 β 3π) (3π + 4) π‘ + 36 (1 β π) (1 β 3π) (2 β 3π) (2π + 1)
β (π‘) , π‘ (π‘ + 1) (π‘2 + 1) (π‘3π + 1) (π‘3π+3 + 1)
π (π‘) > 0
It follows from Lemma 1(6) and (49) that (50)
for π‘ > 1. From Lemma 1(1)β(5) and (38)β(48), we clearly see that β2 (1) > 0, β2σΈ (1) > 0, (51)
β3 (1) > 0,
πΌ
πΌ
1βπΌ
Lemma 3. Inequality [π2/3 (π, π)] [πΏ β1/3 (π, π)] > πΏ (πΌβ2)/6 (π, π) holds for any πΌ β (0, 1) and all π, π > 0 with π =ΜΈ π. Proof. Without loss of generality, we assume that π > π. Let π‘ = (π/π)1/3 > 1 and π = (πΌ β 2)/6; then π β (β1/3, β1/6), πΌ = 6π + 2, and from (1) and (4) we have [π2/3 (π, π)] [πΏ β1/3 (π, π)]
(π‘ + 1)6π+1
β29π+3 π‘6π+1 (π‘3π+3 + 1) ])
+ [π2/3 (π, π)] [πΏ β1/3 (π, π)]
1βπΌ
β πΏ (πΌβ2)/6 (π, π) . (57)
Therefore, πΌπΌ (π, π)πΏ1βπΌ (π, π) > πΏ (πΌβ2)/6 (π, π) follows from (7) and (10) together with Lemma 3 and (57). Next, we prove that πΏ (πΌβ2)/6 (π, π) and πΏ 0 (π, π) are the best possible lower and upper Lehmer mean bounds for the product πΌπΌ (π, π)πΏ1βπΌ (π, π). For any 0 < π < 1 and π₯ > 0, from (1) and (2) we have πΌπΌ (1, π₯) πΏ1βπΌ (1, π₯) π₯ β +β πΏ βπ (1, π₯) = lim [
(52)
π₯ β +β
π₯βπ + 1 πΌ((log π₯)/(π₯β1)β1) π π₯πβ1 + 1
Γ(1 β
β1
Γ (29π+3 π‘6π+1 (π‘3π + 1)) .
π₯π 1 1βπΌ ] = +β, ) 1βπΌ π₯ (log π₯)
πΏ (πΌβ2)/6+π (1, 1 + π₯) β πΌπΌ (1, 1 + π₯) πΏ1βπΌ (1, 1 + π₯)
Let π (π‘) = log (π‘3π + 1) + (3π + 2) log (π‘2 + 1) + (6π + 1) log (π‘ + 1) β (6π + 1) log π‘ 3π+3
β log (π‘
+ 1) β 3 (3π + 1) log 2.
1βπΌ
lim
β πΏ (πΌβ2)/6 (π, π)
3π+2
= (π [(π‘3π + 1) (π‘2 + 1)
Proof. Inequality πΌπΌ (π, π)πΏ1βπΌ (π, π) < πΏ 0 (π, π) follows directly from (10) and (11). For the other inequality, we note that
πΌ
(1) > 0.
1βπΌ
Theorem 4. Inequality πΏ (πΌβ2)/6 (π, π) < πΌπΌ (π, π)πΏ1βπΌ (π, π) < πΏ 0 (π, π) holds for any πΌ β (0, 1) and all π, π > 0 with π =ΜΈ π, and πΏ (πΌβ2)/6 (π, π) and πΏ 0 (π, π) are the best possible lower and upper Lehmer mean bounds for the product πΌπΌ (π, π)πΏ1βπΌ (π, π).
= πΌπΌ (π, π) πΏ1βπΌ (π, π) β [π2/3 (π, π)] [πΏ β1/3 (π, π)]
Therefore, Lemma 2 follows from 9π(π + 1)(3π + 1)(3π + 2)2 (6π+1) > 0 and (32)β(36) together with (50) and (51).
πΌ
3. Main Results
πΌπΌ (π, π) πΏ1βπΌ (π, π) β πΏ (πΌβ2)/6 (π, π)
β3σΈ (1) > 0, β3σΈ σΈ
(56)
for π‘ > 1. Therefore, Lemma 3 follows from (52) and (53) together with (56).
+ 36 (1 β π) (1 β 3π) (2 β 3π) (2π + 1) .
β2σΈ σΈ (1) > 0,
(55)
where β(π‘) is defined as in Lemma 2. From (54) and (55) together with Lemma 2, we clearly see that
= 24 (3π + 4) (9π3 + 33π2 + 51π + 14) π‘
β4 (π‘) > 0
(54)
=
(1 + π₯)(πΌ+4)/6+π + 1 β ππΌ[(1+π₯) log(1+π₯)/π₯β1] (1 + π₯)(πΌβ2)/6+π + 1
(53) Γ
π₯1βπΌ πΌ. [log (1 + π₯)]
(58)
6
Chinese Journal of Mathematics
Letting π₯ β 0 and making use of Taylor expansion, one has
=
(1 + π₯)(πΌ+4)/6+π + 1 β ππΌ[(1+π₯) log(1+π₯)/π₯β1] (1 + π₯)(πΌβ2)/6+π + 1 Γ
Letting π₯ β 0 and making use of Taylor expansion, one has (1 β πΌ) π₯ (1 + π₯)(πΌ+4)/6+π + 1 β πΌπ(1+π₯) log(1+π₯)/π₯β1 β (πΌβ2)/6+π log (1 + π₯) +1 (1 + π₯)
πΌ 1 1 π₯ + πΌ ( πΌ β ) π₯2 + π (π₯2 )] 2 8 6
1 πΌ 1 1 1 Γ [1 + ( β ) π₯ + ( πΌ2 β πΌ β ) π₯2 + π (π₯2 )] 2 2 8 24 12 1 πΌ 1 π =1+ π₯+( β + ) π₯2 2 24 12 4 1 πΌ 1 β [1 + π₯ + ( β ) π₯2 ] + π (π₯2 ) 2 24 12 π 2 = π₯ + π (π₯2 ) . 4
(59)
imply that for any 0 < π < 1, > 1 and πΏ1 = πΏ1 (πΌ, π) > 0, > πΏ βπ (1, π₯) for π₯ β (π1 , +β) πΌπΌ (1, 1 + π₯)πΏ1βπΌ (1, 1 + π₯) for
Proof. Inequality πΌπΌ(π, π) + (1 β πΌ)πΏ(π, π) < πΏ 0 (π, π) follows directly from (10) and (11), and inequality πΏ (πΌβ2)/6 (π, π) < πΌπΌ(π, π) + (1 β πΌ)πΏ(π, π) follows from Theorem 4. Next, we prove that πΏ (πΌβ2)/6 (π, π) and πΏ 0 (π, π) are the best possible lower and upper Lehmer mean bounds for the sum πΌπΌ(π, π) + (1 β πΌ)πΏ(π, π). For any 0 < π < 1 and π₯ > 0, from (1) and (2), we have
lim
πΌπΌ (1, π₯) + (1 β πΌ) πΏ (1, π₯) πΏ βπ (1, π₯)
= lim [ π₯ β +β
π₯βπ + 1 π₯πβ1 + 1
Γ (πΌπ(log π₯)/(π₯β1) π₯π +
1 πΌ 1 π =1+ π₯+( β + ) π₯2 + π (π₯2 ) 2 24 12 4 1 1 β πΌ [1 + π₯ β π₯2 + π (π₯2 )] 2 24
(62)
1 1 β (1 β πΌ) [1 + π₯ β π₯2 + π (π₯2 )] 2 12 π = π₯2 + π (π₯2 ) . 4
Theorem 5. Inequality πΏ (πΌβ2)/6 (π, π) < πΌπΌ(π, π) + (1 β πΌ)πΏ(π, π) < πΏ 0 (π, π) holds for any πΌ β (0, 1) and all π, π > 0 with π =ΜΈ π, and πΏ (πΌβ2)/6 (π, π) and πΏ 0 (π, π) are the best possible lower and upper Lehmer mean bounds for the sum πΌπΌ(π, π) + (1 β πΌ)πΏ(π, π).
π₯ β +β
(1 β πΌ) π₯ . log (1 + π₯) (61)
1 πΌ 1 π =1+ π₯+( β + ) π₯2 + π (π₯2 ) 2 24 12 4
Equations (58) and (59) there exist π1 = π1 (πΌ, π) such that πΌπΌ (1, π₯)πΏ1βπΌ (1, π₯) and πΏ (πΌβ2)/6+π (1, 1 + π₯) > π₯ β (0, πΏ1 ).
(1 + π₯)(πΌ+4)/6+π + 1 β πΌπ(1+π₯) log(1+π₯)/π₯β1 (πΌβ2)/6+π + 1 + π₯) (1 β
π₯1βπΌ πΌ [log (1 + π₯)]
β [1 +
πΏ (πΌβ2)/6+π (1, 1 + π₯) β πΌπΌ (1, 1 + π₯) β (1 β πΌ) πΏ (1, 1 + π₯)
(1 β πΌ) (1 β 1/π₯) π π₯ )] log π₯
= +β, (60)
Equations (60)-(62) imply that for any 0 < π < 1, there exist π2 = π2 (πΌ, π) > 1 and πΏ2 = πΏ2 (πΌ, π) > 0, such that πΌπΌ(1, π₯) + (1 β πΌ)πΏ(1, π₯) > πΏ βπ (1, π₯) for π₯ β (π2 , +β) and πΏ (πΌβ2)/6+π (1, 1 + π₯) > πΌπΌ(1, 1 + π₯) + (1 β πΌ)πΏ(1, 1 + π₯) for π₯ β (0, πΏ2 ).
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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