Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2016, Article ID 1579468, 5 pages http://dx.doi.org/10.1155/2016/1579468
Research Article Sharp One-Parameter Mean Bounds for Yang Mean Wei-Mao Qian,1 Yu-Ming Chu,2 and Xiao-Hui Zhang2 1
School of Distance Education, Huzhou Broadcast and TV University, Huzhou 313000, China School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China
2
Correspondence should be addressed to Yu-Ming Chu;
[email protected] Received 19 July 2015; Accepted 9 February 2016 Academic Editor: Yann Favennec Copyright Β© 2016 Wei-Mao Qian 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 prove that the double inequality π½πΌ (π, π) < π(π, π) < π½π½ (π, π) holds for all π, π > 0 with π =ΜΈ π if and only if πΌ β€ β2/(π β β2) = 0.8187 β
β
β
and π½ β₯ 3/2, where π(π, π) = (πβπ)/[β2 arctan((πβπ)/β2ππ)], and π½π (π, π) = π(ππ+1 βππ+1 )/[(π+1)(ππ βππ )] (π =ΜΈ 0, β1), π½0 (π, π) = (π β π)/(log π β log π), and π½β1 (π, π) = ππ(log π β log π)/(π β π) are the Yang and πth one-parameter means of π and π, respectively.
1. Introduction Let π β R and π, π > 0 with π =ΜΈ π. Then the πth oneparameter mean π½π (π, π), πth power mean ππ (π, π), harmonic mean π»(π, π), geometric mean πΊ(π, π), logarithmic mean πΏ(π, π), first Seiffert mean π(π, π), identric mean πΌ(π, π), arithmetic mean π΄(π, π), Yang mean π(π, π), second Seiffert mean π(π, π), and quadratic mean π(π, π) are, respectively, defined by π+1
π+1
π (π β π ) { { , π =ΜΈ 0, β1, { { { (π + 1) (ππ β ππ ) { { { { { πβπ π½π (π, π) = { , π = 0, { { log π β log π { { { { { { { ππ (log π β log π) , π = β1, { πβπ ππ (π, π) = [
π
π
π +π ] 2
π0 (π, π) = βππ, 2ππ π» (π, π) = , π+π πΊ (π, π) = βππ,
1/π
(π =ΜΈ 0) ,
πΏ (π, π) =
πβπ , log π β log π
π (π, π) =
πβπ , 2 arcsin ((π β π) / (π + π))
πΌ (π, π) =
1 ππ ( ) π ππ
π΄ (π, π) =
π+π , 2
π (π, π) =
πβπ , β2 arctan ((π β π) /β2ππ)
π (π, π) =
πβπ , 2 arctan ((π β π) / (π + π))
1/(πβπ)
π (π, π) = β
,
π2 + π2 . 2 (1)
It is well known that both the means π½π (π, π) and ππ (π, π) are continuous and strictly increasing with respect to π β R for fixed π, π > 0 with π =ΜΈ π. Recently, the one-parameter mean π½π (π, π) and Yang mean π(π, π) have attracted the attention of many researchers.
2
Mathematical Problems in Engineering Alzer [1] proved that the inequalities
In [12, 13], the authors proved that the double inequalities
πΊ (π, π) < βπ½π (π, π) π½βπ (π, π) < πΏ (π, π)
0 with π =ΜΈ π and π =ΜΈ 0. In [2, 3], the authors discussed the monotonicity and logarithmic convexity properties of the one-parameter mean π½π (π, π). In [4, 5], the authors proved that the double inequalities π½π1 (π, π) < πΌπ΄ (π, π) + (1 β πΌ) πΏ (π, π) < π½π1 (π, π) π½π2 (π, π) < πΌπ΄ (π, π) + (1 β πΌ) π» (π, π) < π½π2 (π, π) ,
(3)
hold for all π, π > 0 with π =ΜΈ π and πΌ β (0, 1) if and only if π1 β€ πΌ/(2 β πΌ), π1 β₯ πΌ, π2 β€ 3πΌ β 2, and π2 β₯ πΌ/(2 β πΌ). Xia et al. [6] proved that the double inequality π½(3πΌβ1)/2 (π, π) < πΌπ΄ (π, π) + (1 β πΌ) πΊ (π, π)
(4)
< π½πΌ/(2βπΌ) (π, π)
holds for all π, π > 0 with π =ΜΈ π if πΌ β (0, 2/3), and inequality (4) is reversed if πΌ β (2/3, 1). Gao and Niu [7] presented the best possible parameters π and π such that the double inequality π½π (π, π) < π΄πΌ (π, π)πΊπ½ (π, π)π»1βπΌβπ½ (π, π) < π½π (π, π) holds for all π, π > 0 with π =ΜΈ π and πΌ + π½ β (0, 1). In [8, 9], the authors proved that the double inequalities π½π 1 (π, π) < π (π, π) < π½π1 (π, π) ,
(5)
π½π 2 (π, π) < πΌ (π, π) < π½π2 (π, π)
hold for all π, π > 0 with π =ΜΈ π if and only if π 1 β€ 2/(2 β π), π1 β₯ 2, π 2 β€ 1/2, and π2 β₯ 1/(π β 1). Xia et al. [10] found that π(1+2π)/3 (π, π) is the best possible lower power mean bound for the one-parameter mean π½π (π, π) if π β (β2, β1/2) βͺ (1, β) and π(1+2π)/3 (π, π) is the best possible upper power mean bound for the oneparameter mean π½π (π, π) if π β (ββ, β2) βͺ (β1/2, 1). For all π, π > 0 with π =ΜΈ π, Yang [11] provided the bounds for the Yang mean π(π, π) in terms of other bivariate means as follows: π (π, π) < π (π, π) < π (π, π) ,
< π2/3 (π, π) [
πΊ (π, π) + π (π, π) ] 2
2 πΊ (π, π) + π (π, π) π 1 π 0 with π =ΜΈ π.
2. Main Result In order to prove our main result we need a lemma, which we present in this section. Lemma 1. Let π β R, and
β (π + 1) π₯4π+1 β π (π + 1) π₯2π+7
2πΊ (π, π) + π (π, π) 1/2 ] < π (π, π) 3 1/3
1/π
π (π₯, π) = ππ₯4π+6 β (π + 1) π₯4π+5 + ππ₯4π+4
π (π, π) π (π, π) πΊ (π, π) π (π, π) < π (π, π) < , π΄ (π, π) π΄ (π, π) π1/2 (π, π) [
2 πΊ (π, π) + π (π, π) π 1 π [ ( ) + π (π, π)] 3 2 3
2
+ 2 (π + 1) π₯2π+5 β 2ππ₯2π+4 (6)
,
β 2π (π + 1) π₯2π+3 β 2ππ₯2π+2 2
+ 2 (π + 1) π₯2π+1 β π (π + 1) π₯2πβ1
πΊ (π, π) + π (π, π) < π (π, π) 2
β (π + 1) π₯5 + ππ₯2 β (π + 1) π₯ + π. 2
2 πΊ (π, π) + π (π, π) 1/2 1 1/2 ) + π (π, π)] . 0 for all π₯ β (1, β);
(9)
Mathematical Problems in Engineering
3
(2) if π = β2/(π β β2) = 0.8187 β
β
β
, then there exists π β (1, β) such that π(π₯, π) < 0 for π₯ β (1, π) and π(π₯, π) > 0 for π₯ β (π, β). Proof. For part (1), if π = 3/2, then (9) becomes π (π₯, π) =
1 (π₯ β 1)6 (π₯2 + 2π₯ + 2) (2π₯2 + 2π₯ + 1) 4
lim π8 (π₯, π) = (π + 1) (2560π5 + 6336π4 + 14176π3
π₯β1
+ 11028π2 β 12680π β 22005) < 0, lim π π₯β+β 8
(π₯, π) = +β,
(21)
(22)
lim π9 (π₯, π) = 6 (512π6 + 1184π5 + 4064π4
π₯β1
(10)
β
(3π₯2 + 4π₯ + 3) .
+ 6372π3 + 4068π2 β 3495π β 5775) = 15.2085
(23)
β
β
β
> 0,
Therefore, part (1) follows from (10). For part (2), let π = β2/(π β β2), π1 (π₯, π) = ππ(π₯, π)/ππ₯, π2 (π₯, π) = (1/2)(ππ1 (π₯, π)/ππ₯), π3 (π₯, π) = (1/(π + 1)π₯2 )(ππ2 (π₯, π)/ππ₯), π4 (π₯, π) = (π₯7β2π /2π)(ππ3 (π₯, π)/ππ₯), π5 (π₯, π) = (1/2π₯)(ππ4 (π₯, π)/ππ₯), π6 (π₯, π) = ππ5 (π₯, π)/ππ₯, π7 (π₯, π) = (1/2)(ππ6 (π₯, π)/ππ₯), π8 (π₯, π) = ππ7 (π₯, π)/ππ₯, π9 (π₯, π) = (1/2(π+1))(ππ8 (π₯, π)/ππ₯), and π10 (π₯, π) = ππ9 (π₯, π)/ππ₯. Then elaborated computations lead to
2
π10 (π₯, π) = (π + 2) (2π + 1) (2π + 3) (2π + 5) (2π + 7) (4π + 1) (4π + 5) π₯2π β 8π (π + 1) (π + 2) (π + 3) (2π + 1) (2π + 3) (4π + 3) (4π + 5) π₯2πβ1 2
+ π (2π β 1) (2π + 1) (2π + 3) (2π + 5) (4π β 1)
(24)
2
β
(4π + 3) π₯2πβ2 β 8π (π β 1) (π β 2) (2π β 1) (2π lim π (π₯, π) = 0,
π₯β1
(11)
lim π (π₯, π) = +β,
+ 7) π₯.
π₯β+β
lim π1 (π₯, π) = 0,
Note that
π₯β1
lim π π₯β+β 1
(12)
(π₯, π) = +β,
π₯β1
(13)
(π₯, π) = +β,
lim π3 (π₯, π) = 0,
(14)
(π₯, π) = +β,
3 lim π4 (π₯, π) = β48 (π + 1) ( β π) < 0, 2
(15)
π₯β+β
3 lim π5 (π₯, π) = β192 (π + 1) ( β π) < 0, π₯β1 2
2
(26)
β
π₯2πβ2 = (1536π7 + 15040π6 + 59440π5
(17)
β 963) < 0,
(18)
+ 122280π4 + 137144π3 + 61850π2 β 49845π β 72450) π₯ + π (2π β 1) (4π β 1) (704π4 + 136π3 + 1120π2 + 248π β 3) π₯2πβ2 > 0,
lim π7 (π₯, π) = (π + 1) (1024π4 + 2096π3 + 1844π2
(19)
β 2876π β 5193) < 0, (π₯, π) = +β,
β
(2π + 1) (2π + 3) (2π + 5) (4π β 1) (4π + 3) β 8π (π β 1) (π β 2) (2π β 1) (2π β 3) (16π2 β 1)]
lim π6 (π₯, π) = 2 (π + 1) (368π3 + 332π2 β 484π
(π₯, π) = +β,
β
(π + 3) (2π + 1) (2π + 3) (4π + 3) (4π + 5)
2
(16)
(π₯, π) = +β,
π₯β1
lim π π₯β+β 7
2
β 720 (π + 3) (2π + 5) (2π + 7)] π₯ + [π (2π β 1)
2
π₯β1
It follows from (24) and (25) that
β
(2π + 7) (4π + 1) (4π + 5) β 8π (π + 1) (π + 2)
lim π4 (π₯, π) = +β,
lim π π₯β+β 6
(25)
π10 (π₯, π) > [(π + 2) (2π + 1) (2π + 3) (2π + 5)
π₯β1
lim π π₯β+β 5
+ 137144π3 + 61850π2 β 49845π β 72450 = 85165.4405 β
β
β
> 0.
π₯β1
lim π π₯β+β 3
2π > 1 > 2π β 1 > 0 > 2π β 2 > 2π β 5, 1536π7 + 15040π6 + 59440π5 + 122280π4
lim π2 (π₯, π) = 0,
lim π π₯β+β 2
β 3) (16π2 β 1) π₯2πβ5 β 720 (π + 3) (2π + 5) (2π
(20)
for π₯ β (1, β). From (23) and (26) we clearly see that π8 (π₯, π) is strictly increasing with respect to π₯ on the interval (1, β). Then (21) and (22) lead to the conclusion that there exists π 1 > 1 such
4
Mathematical Problems in Engineering
that the function π₯ β π7 (π₯, π) is strictly decreasing on (1, π 1 ] and strictly increasing on [π 1 , β). It follows from (19) and (20) together with the piecewise monotonicity of the function π₯ β π7 (π₯, π) that there exists π 2 > 1 such that the function π₯ β π6 (π₯, π) is strictly decreasing on (1, π 2 ] and strictly increasing on [π 2 , β). Making use of (13)β(18) and the same method as the above we know that there exists π π > 1 (π = 3, 4, 5, 6, 7) such that the function π₯ β π8βπ (π₯, π) is strictly decreasing on (1, π π ] and strictly increasing on [π π , β). It follows from (12) and the piecewise monotonicity of the function π₯ β π1 (π₯, π) that there exists πβ > 1 such that the function π₯ β π(π₯, π) is strictly decreasing on (1, πβ ] and strictly increasing on [πβ , β). Therefore, part (2) follows easily from (11) and the piecewise monotonicity of the function π₯ β π(π₯, π).
π½πΌ (π, π) < π (π, π) < π½π½ (π, π)
(27)
holds for all π, π > 0 with π =ΜΈ π if and only if πΌ β€ β2/(πββ2) = 0.8187 β
β
β
and π½ β₯ 3/2. Proof. Since π(π, π) and π½π (π, π) are symmetric and homogeneous of degree one, without loss of generality, we assume that π = π₯2 > 1 and π = 1. Let π β R and π =ΜΈ 0, β1. Then (1) lead to 2
π½π (π, π) β π (π, π) = π½π (π₯ , 1) β π (π₯ , 1) =
Case 2 (π > β2/(π β β2)). Then (1) leads to lim
π½π (π₯, 1)
π₯ββ π (π₯, 1)
=
β2π π > 1. 2 (π + 1)
(34)
Inequality (34) implies that there exists large enough π = π(π) > 1 such that π(π, π) < π½π (π, π) for all π, π > 0 with π/π β (0, 1/π) βͺ (π, β). Case 3 (π = 3/2). Then from Lemma 1(1) and (31) we know that the function π₯ β πΉ(π₯, π) is strictly increasing on the interval (1, β). Therefore, π (π, π) < π½3/2 (π, π)
(35)
follows from (28) and (30) together with the monotonicity of the function π₯ β πΉ(π₯, π).
Theorem 2. The double inequality
2
follows easily from (28), (30), and (32) together with the piecewise monotonicity of the function π₯ β πΉ(π₯, π).
π (π₯2π+2 β 1) (π + 1) (π₯2π β 1) arctan ((π₯2 β 1) /β2π₯)
πΉ (π₯, π) ,
(28)
Case 4 (0 < π < 3/2). Let π₯ > 0 and π₯ β 0; then making use of Taylor expansion we get π (1, 1 + π₯) β π½π (1, 1 + π₯) =
π₯ β2 arctan (π₯/β2 (1 + π₯)) β
π [1 β (1 + π₯)π+1 ] (π + 1) [1 β (1 + π₯)π ]
=
(36) 3 β 2π 2 π₯ + π (π₯2 ) . 24
Equation (36) implies that there exists small enough πΏ > 0 such that π(1, 1 + π₯) > π½π (1, 1 + π₯) for all π₯ β (0, πΏ).
Conflict of Interests
where πΉ (π₯, π) = arctan ( β
The authors declare that there is no conflict of interests regarding the publication of this paper.
π₯2 β 1 ) β2π₯
(π + 1) (π₯2 β 1) (π₯2π β 1) β2π (π₯2π+2 β 1)
(29) ,
lim πΉ (π₯, π) = 0,
(30)
π₯β1
β2 ππΉ (π₯, π) π (π₯, π) , = 2 ππ₯ π (π₯4 + 1) (π₯2π+2 β 1)
(31)
where π(π₯, π) is defined by (9). We divide the proof into four cases. Case 1 (π = β2/(π β β2)). Then it follows from Lemma 1(2), (29), and (31) that there exists π > 1 such that the function π₯ β πΉ(π₯, π) is strictly decreasing on (1, π] and strictly increasing on [π, β), and lim πΉ (π₯, π) = 0.
(32)
π½β2/(πββ2) (π, π) < π (π, π)
(33)
π₯ββ
Therefore,
Acknowledgments The research was supported by the Natural Science Foundation of China under Grants 61374086, 11371125, and 11401191 and the Natural Science Foundation of Zhejiang Broadcast and TV University under Grant XKT-15G17.
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