Hindawi Publishing Corporation Journal of Function Spaces Volume 2015, Article ID 517647, 5 pages http://dx.doi.org/10.1155/2015/517647
Research Article Sharp Power Mean Bounds for the One-Parameter Harmonic Mean Yu-Ming Chu,1 Li-Min Wu,2 and Ying-Qing Song1 1
School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China Department of Mathematics, Huzhou University, Huzhou 313000, China
2
Correspondence should be addressed to Yu-Ming Chu;
[email protected] Received 3 November 2014; Accepted 27 April 2015 Academic Editor: David R. Larson Copyright Β© 2015 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. We present the best possible parameters πΌ = πΌ(π) and π½ = π½(π) such that the double inequality ππΌ (π, π) < π»π (π, π) < ππ½ (π, π) holds for all π β (0, 1/2) and π, π > 0 with π =ΜΈ π, where ππ (π, π) = [(ππ + ππ )/2]1/π (π =ΜΈ 0) and π0 (π, π) = βππ and π»π (π, π) = 2[ππ + (1 β π)π][ππ + (1 β π)π]/(π + π) are the power and one-parameter harmonic means of π and π, respectively.
1. Introduction For π β R and π, π > 0, the πth power mean ππ (π, π) of π and π is defined by ππ (π, π) = (
π
π
π +π ) 2
positive real numbers π and π, respectively. Then it is well known that the inequalities π» (π, π) < πΊ (π, π) < πΏ (π, π) < π (π, π) < πΌ (π, π) < π΄ (π, π) < π (π, π) < π (π, π) < πΆ (π, π)
1/π
(1) (π =ΜΈ 0) , π0 (π, π) = βππ.
It is well known that ππ (π, π) is strictly increasing with respect to π β R for fixed π, π > 0 with π =ΜΈ π, symmetric and homogeneous of degree 1. Many classical means are special cases of the power mean: for example, πβ1 (π, π) = 2ππ/(π + π) = π»(π, π) is the harmonic mean, π0 (π, π) = βππ = πΊ(π, π) is the geometric mean, π1 (π, π) = (π + π)/2 = π΄(π, π) is the arithmetic mean, and π2 (π, π) = β(π2 + π2 )/2 = π(π, π) is the quadratic mean. The main properties of the power mean are given in [1]. Recently, the power mean has attracted the attention of many researchers. In particular, many remarkable inequalities for the power mean can be found in the literature [2β10]. Let πΏ(π, π) = (π β π)/(log π β log π), π(π, π) = (π β π)/[2 arcsin((π β π)/(π + π))], πΌ(π, π) = 1/π(ππ /ππ )1/(πβπ) , π(π, π) = (π β π)/[2 arctan((π β π)/(π + π))], and πΆ(π, π) = (π2 + π2 )/(π + π) be the logarithmic, first Seiffert, identric, second Seiffert, and contraharmonic means of two distinct
(2)
hold for all π, π > 0 with π =ΜΈ π. Lin [11] proved that the double inequality ππ (π, π) < πΏ (π, π) < ππ (π, π)
(3)
holds for all π, π > 0 with π =ΜΈ π if and only if π β€ 0 and π β₯ 1/3. In [12], Pittenger presented the best possible parameters π = π(π) and π = π(π) such that the double inequality ππ (π, π) < πΏ π (π, π) < ππ (π, π)
(4) π+1
β holds for all π, π > 0 with π =ΜΈ π, where πΏ π (π, π) = [(π 1/π π+1 π )/((π + 1)(π β π))] (π =ΜΈ 0, β1), πΏ 0 (π, π) = πΌ(π, π), and πΏ β1 (π, π) = πΏ(π, π) is the generalized logarithmic mean of π and π. Jagers [13] and Seiffert [14] proved that the double inequalities π1/2 (π, π) < π (π, π) < π2/3 (π, π) , π1 (π, π) < π (π, π) < π2 (π, π) hold for all π, π > 0 with π =ΜΈ π.
(5)
2
Journal of Function Spaces In [15, 16], the authors proved that the double inequalities πΌ (π, π) + πΏ (π, π) < π1/2 (π, π) , 2 (6) π0 (π, π) < βπΌ (π, π) πΏ (π, π) < π1/2 (π, π)
πlog 2/(1+log 2) (π, π)
0 with π =ΜΈ π. Costin and Toader [17] proved that the double inequality π4/3 (π, π) < π (π, π) < π5/3 (π, π)
(7)
holds for all π, π > 0 with π =ΜΈ π. In [18β20], the authors proved that the double inequalities ππ (π, π) < π (π, π) < ππ (π, π) , ππ (π, π) < π (π, π) < ππ (π, π)
(8)
hold for all π, π > 0 with π =ΜΈ π if and only if π β€ log π/ log 2, π β₯ 2/3, π β€ log 2/(log π β log 2), and π β₯ 5/3. Λ Cizmesija [21] proved that π = βπΌ/2 and π = log 2/[log 4βlog(1βπΌ)] are the best possible parameters such that the double inequality ππ (π, π) < [π»(π, π)+πΌπΊ(π, π)+(1β πΌ)π΄(π, π)]/2 < ππ (π, π) holds for all πΌ β (0, 1) and π, π > 0 with π =ΜΈ π. In [22, 23], the authors proved that the inequalities π ππ [K (π) , E (π)] > , 2 ππ [K (π) , K (β1 β π2 )]
2. Lemmas In order to prove our main result we need three lemmas, which we present in this section. Lemma 1. The inequality β8π2 + 8π β 1 +
Proof. It is not difficult to verify that log[2π(1 β π)] < 0 for all π β (0, 1/2). Therefore, we only need to prove that π(π) > 0 for π β (0, 1/2), where π(π) = (β8π2 +8πβ1) log[2π(1βπ)]+log 2. Simple computations lead to 1 π ( ) = 0, 2
π/2
(10)
Let π1 , π1 , π2 , π2 , π 1 , π 2 , π 3 , π, π β (0, 1/2) and π3 , π3 β (1/2, 1). Then, the authors in [25β28] proved that the inequalities
where π (π) = 8 log [2π (1 β π)] β
πΏ π2 (π, π) < π (π, π) < πΏ π2 (π, π) ,
πσΈ (π) =
(11)
(14)
(15)
for all π β (0, 1/2). Inequality (15) implies that π(π) < 0 for all π β (0, 1/2). Then, from (13) we clearly see that π(π) > 0 for all π β (0, 1/2). Lemma 2. The inequality log 2 log [2π (1 β π)]
(16)
holds for all π β (0, 1/2). Proof. Let π = β log 2/ log[2π(1 β π)] and π = β8π2 + 8π β 1. Then, it is not difficult to verify that
πΆπ3 (π, π) < π (π, π) < πΆπ3 (π, π) , πΊπ (π, π) < πΌ (π, π) < πΊπ (π, π)
(1 β 2π) [8π (1 β π) + 1] >0 π2 (1 β π)2
2 log 2 ] πΊ (π, π) ,
1 + 8, π (1 β π)
1 π ( ) = 4 (1 β log 4) < 0, 2
β 16π2 + 16π β 2 β (8π2 β 8π β 1)
π»π1 (π, π) < π (π, π) < π»π1 (π, π) ,
πΊπ 3 (π, π) > πΏ (π, π) ,
(13)
π (π) = (1 β 2π) π (π) ,
β«0 (1 β π2 sin2 π)β1/2 ππ and E(π) = β«0 (1 β π2 sin2 π)1/2 ππ are, respectively, the complete elliptic integrals of the first and second kinds. Let π β [0, 1] and π be the bivariate symmetric mean. Then, the one-parameter mean ππ (π, π) was defined by Neuman [24] as follows:
π»π 2 (π, π) > πΏ (π, π) ,
(12)
σΈ
hold for all π β (0, 1) if and only if π β₯ β1/2, π β€ log 2/(log π β log 2) β log[K(β2/2)] = β4.18, . . ., and π β₯ 1 β 4K4 (β2/2)/π2 = β3.789, . . ., where K(π) =
ππ (π, π) = π [ππ + (1 β π) π, ππ + (1 β π) π] .
log 2 0 with π =ΜΈ π if and only if π1 β€ 1/2 β β1 β 2/π/2, π1 β₯ 1/2 β β6/12, π2 β€ π 0 , π2 β₯ 1/2 β β2/4, π 1 β₯ 1/2 β β2/4, π 2 β₯ 1/2 β β3/6, π 3 β₯ 1/2 β β6/6, π3 β€ 1/2 + β4/π β 1/2, π3 β₯ 1/2 + β3/6, π β€ 1/2 β β1 β 4/π2 /2, and π β₯ 1/2 β β3/6, where π 0 is the unique solution of the equation log[(1 β π)/π] = π(1 β 2π). The main purpose of this paper is to present the best possible parameters πΌ = πΌ(π) and π½ = π½(π) such that the double inequality ππΌ (π, π) < π»π (π, π) < ππ½ (π, π) holds for all π β (0, 1/2) and π, π > 0 with π =ΜΈ π.
00 log [2π (1 β π)]
π (1) = 0, (20)
(27)
πσΈ (1) = β8π2 + 8π β 1 β π,
(28)
πσΈ σΈ (1) = β16π2 + 16π β 2 + π (8π2 β 8π β 1) + π2 ,
(29)
πσΈ σΈ σΈ (π₯) = (1 β π) π₯βπβ2 β (π₯) ,
holds for all π β (0, 1/2). Proof. Let π = β log 2/ log[2π(1 β π)] and π = β8π2 + 8π β 1. Then, it follows from (17) and (18) that β 6π2 + 6π β (12π2 β 12π + 1)
log 2 log [2π (1 β π)]
3 3 + [1 β (π + 1)] π + π2 = (π + 1) (1 β 2π) 2 4
β (π₯) = (βπ2 + π) (3 β π) (2 β π) π₯2
β (1) = 6 (βπ2 + π) + π (12π2 β 12π + 1) + π2 , (21)
βσΈ (π₯) = 2 (π β π2 ) (2 β π) [(3 β π) π₯ β π] .
(33)
Case 1 (π = β8π2 + 8π β 1). We divide the discussion into two subcases. Subcase 1.1 (π = β8π2 + 8π β 1 and π β (0, (2 β β2)/4) βͺ (2 β β2)/4, 1/2)). Then, we clearly see that π β (β1, 0)βͺ(0, 1), and (28), (29), (32), and (33) lead to
3. Main Results
πσΈ (1) = πσΈ σΈ (1) = 0,
Theorem 4. The double inequality (22)
β (1) =
holds for all π β (0, 1/2) and π, π > 0 with π =ΜΈ π if and only if πΌ β€ β8π2 + 8π β 1 and π½ β₯ β log 2/ log[2π(1 β π)].
βσΈ (π₯) =
Proof. Without loss of generality, we assume that π = π₯ β (1, β) and π = 1. Let log[π»π (π₯, 1)] β log[ππ (π₯, 1)] = π(π₯), where π (1 β π) π₯2 + (1 β 2π + 2π2 ) π₯ + π (1 β π)
(1 + π) log 2 β log (1 + π₯π ) + . π
(32)
We divide the proof into two cases.
1 = (π + 1) (3 β 2π) = 2π (1 β π) (3 β 2π) > 0 4 for all π β (0, 1/2).
1+π₯
(31)
+ (3π2 β 3π + 1) π (π + 1) ,
3 (π + 1) (1 β 2π) + π (π + 1) 4
ππΌ (π, π) < π»π (π, π) < ππ½ (π, π)
(30)
where
β (β2π2 + 2π) π (2 β π) π₯
2 log 2 3 +[ ] = (π + 1) log [2π (1 β π)] 4
π (π₯) = log
(26)
+ π2 β π,
Lemma 3. The inequality
+ π (π + 1) >
(25)
where
= β (π β π) (2 β π) < 0
+[
(24)
πσΈ (π₯)
log 2 β 16π + 16π β 2 β (8π β 8π β 1) log [2π (1 β π)] 2
π (1) = 0,
(23)
>
1 (π + 1) (3 β 2π) > 0, 4 1 (π + 1) (2 β π) [(3 β π) π₯ β π] 4 1 (π + 1) (2 β π) (3 β 2π) > 0 4
(34) (35)
(36)
for π₯ β (1, β). It follows easily from (24), (25), (27), (30), and (34)β(36) that π»π (π₯, 1) > πβ8π2 +8πβ1 (π₯, 1) for all π β (0, 1/2).
(37)
4
Journal of Function Spaces
Subcase 1.2 (π = β8π2 + 8π β 1 and π = (2 β β2)/4). Then, π = 0, and (37) follows from π»π (π₯, 1) β πβ8π2 +8πβ1 (π₯, 1) 2
=
(βπ₯ β 1) π₯2 + 6π₯ + 1 β βπ₯ = . 4 (1 + π₯) 4 (1 + π₯)
(38)
Case 2 (π = β log 2/ log[2π(1 β π)]). Then, we clearly see that π β (0, 1), and Lemmas 1β3 and (23), (26), (28), (29), and (32) lead to
Conflict of Interests
(39)
lim π (π₯) = +β,
(40)
The authors declare that there is no conflict of interests regarding the publication of this paper.
lim πσΈ (π₯) = +β,
(41)
Acknowledgments
(42)
π₯ β +β π₯ β +β
πσΈ (1) < 0,
πσΈ σΈ (1) < 0,
(44)
The research was supported by the Natural Science Foundation of China under Grants 61374086 and 11371125, the Natural Science Foundation of Zhejiang Province under Grant LY13A010004, and the Natural Science Foundation of Hunan Province under Grant 14JJ2127.
β (1) > 0
(45)
References
lim πσΈ σΈ (π₯) = +β,
π₯ β +β
(43)
and (36) again holds. It follows from (30), (36), and (45) that πσΈ σΈ is strictly increasing on (1, β). Then, (43) and (44) lead to the conclusion that there exists π0 > 1 such that πσΈ is strictly decreasing on (1, π0 ] and strictly increasing on [π0 , β). From (41) and (42) together with the piecewise monotonicity of πσΈ we clearly see that there exists π1 > 1 such that π is strictly decreasing on (1, π1 ] and strictly increasing on [π1 , β). Then, (25), (27), and (40) lead to the conclusion that there exists π2 > 1 such that π is strictly decreasing on (1, π2 ] and strictly increasing on [π2 , β). Therefore, π»π (π₯, 1) < πβ log 2/ log[2π(1βπ)] (π₯, 1)
(46)
for all π β (0, 1/2) follows from (24) and (39) together with the piecewise monotonicity of π. Next, we prove that πΌ = β8π2 + 8π β 1 and π½ = β log 2/ log[2π(1 β π)] are the best possible parameters such that the double inequality ππΌ (π, π) < π»π (π, π) < ππ½ (π, π)
(47)
holds for all π β (0, 1/2) and π, π > 0 with π =ΜΈ π. Let π β (0, 1/2), πΌ0 = β8π2 +8πβ1, π½0 = β log 2/ log[2π(1β π)], π β (0, π½0 ), and π‘ > 0. Then, we have π»π (1, 1 + π‘) β ππΌ0 +π (1, 1 + π‘) 2 [(1 β π) (1 + π‘) + π] [π (1 + π‘) + 1 β π] π‘+2 β[
Inequality (49) and equation (50) imply that for any π β (0, 1/2) and π β (0, π½0 ) there exist π = π(π) > 1 and πΏ = πΏ(π) β (0, 1) such that π»π (1, π‘) > ππ½0 βπ (1, π‘) for π‘ > π and π»π (1, 1 + π‘) < ππΌ0 +π (1, 1 + π‘) for π‘ β (0, πΏ).
lim π (π₯) = 0,
π₯ β +β
=
Let π‘ β 0+ and make use of the Taylor expansion; then, (48) leads to π π»π (1, 1 + π‘) β ππΌ0 +π (1, 1 + π‘) = β π₯2 + π (π₯2 ) . (50) 8
1 + (1 + π‘)πΌ0 +π ] 2
π»π (1, π‘) lim π‘ β +β π π½0 βπ (1, π‘)
1/(πΌ0 +π)
(48)
,
= 2π/[π½0 (π½0 βπ)] > 1.
(49)
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