Sharp Power Mean Bounds for the One-Parameter Harmonic Mean

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Apr 27, 2015 - Research Article. Sharp Power Mean Bounds for the One-Parameter. Harmonic Mean. Yu-Ming Chu,1 Li-Min Wu,2 and Ying-Qing Song1.
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 𝛼 = 𝛼(π‘Ÿ) and 𝛽 = 𝛽(π‘Ÿ) such that the double inequality 𝑀𝛼 (π‘Ž, 𝑏) < π»π‘Ÿ (π‘Ž, 𝑏) < 𝑀𝛽 (π‘Ž, 𝑏) holds for all π‘Ÿ ∈ (0, 1/2) and π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏, where 𝑀𝑝 (π‘Ž, 𝑏) = [(π‘Žπ‘ + 𝑏𝑝 )/2]1/𝑝 (𝑝 =ΜΈ 0) and 𝑀0 (π‘Ž, 𝑏) = βˆšπ‘Žπ‘ and π»π‘Ÿ (π‘Ž, 𝑏) = 2[π‘Ÿπ‘Ž + (1 βˆ’ π‘Ÿ)𝑏][π‘Ÿπ‘ + (1 βˆ’ π‘Ÿ)π‘Ž]/(π‘Ž + 𝑏) are the power and one-parameter harmonic means of π‘Ž and 𝑏, respectively.

1. Introduction For 𝑝 ∈ R and π‘Ž, 𝑏 > 0, the 𝑝th power mean 𝑀𝑝 (π‘Ž, 𝑏) of π‘Ž and 𝑏 is defined by 𝑀𝑝 (π‘Ž, 𝑏) = (

𝑝

𝑝

π‘Ž +𝑏 ) 2

positive real numbers π‘Ž and 𝑏, respectively. Then it is well known that the inequalities 𝐻 (π‘Ž, 𝑏) < 𝐺 (π‘Ž, 𝑏) < 𝐿 (π‘Ž, 𝑏) < 𝑃 (π‘Ž, 𝑏) < 𝐼 (π‘Ž, 𝑏) < 𝐴 (π‘Ž, 𝑏) < 𝑇 (π‘Ž, 𝑏) < 𝑄 (π‘Ž, 𝑏) < 𝐢 (π‘Ž, 𝑏)

1/𝑝

(1) (𝑝 =ΜΈ 0) , 𝑀0 (π‘Ž, 𝑏) = βˆšπ‘Žπ‘.

It is well known that 𝑀𝑝 (π‘Ž, 𝑏) is strictly increasing with respect to 𝑝 ∈ R for fixed π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏, symmetric and homogeneous of degree 1. Many classical means are special cases of the power mean: for example, π‘€βˆ’1 (π‘Ž, 𝑏) = 2π‘Žπ‘/(π‘Ž + 𝑏) = 𝐻(π‘Ž, 𝑏) is the harmonic mean, 𝑀0 (π‘Ž, 𝑏) = βˆšπ‘Žπ‘ = 𝐺(π‘Ž, 𝑏) is the geometric mean, 𝑀1 (π‘Ž, 𝑏) = (π‘Ž + 𝑏)/2 = 𝐴(π‘Ž, 𝑏) is the arithmetic mean, and 𝑀2 (π‘Ž, 𝑏) = √(π‘Ž2 + 𝑏2 )/2 = 𝑄(π‘Ž, 𝑏) 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 [2–10]. Let 𝐿(π‘Ž, 𝑏) = (𝑏 βˆ’ π‘Ž)/(log 𝑏 βˆ’ log π‘Ž), 𝑃(π‘Ž, 𝑏) = (π‘Ž βˆ’ 𝑏)/[2 arcsin((π‘Ž βˆ’ 𝑏)/(π‘Ž + 𝑏))], 𝐼(π‘Ž, 𝑏) = 1/𝑒(𝑏𝑏 /π‘Žπ‘Ž )1/(π‘βˆ’π‘Ž) , 𝑇(π‘Ž, 𝑏) = (π‘Ž βˆ’ 𝑏)/[2 arctan((π‘Ž βˆ’ 𝑏)/(π‘Ž + 𝑏))], and 𝐢(π‘Ž, 𝑏) = (π‘Ž2 + 𝑏2 )/(π‘Ž + 𝑏) be the logarithmic, first Seiffert, identric, second Seiffert, and contraharmonic means of two distinct

(2)

hold for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏. Lin [11] proved that the double inequality 𝑀𝑝 (π‘Ž, 𝑏) < 𝐿 (π‘Ž, 𝑏) < π‘€π‘ž (π‘Ž, 𝑏)

(3)

holds for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏 if and only if 𝑝 ≀ 0 and π‘ž β‰₯ 1/3. In [12], Pittenger presented the best possible parameters πœ† = πœ†(𝑝) and πœ‡ = πœ‡(𝑝) such that the double inequality π‘€πœ† (π‘Ž, 𝑏) < 𝐿 𝑝 (π‘Ž, 𝑏) < π‘€πœ‡ (π‘Ž, 𝑏)

(4) 𝑝+1

βˆ’ holds for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏, where 𝐿 𝑝 (π‘Ž, 𝑏) = [(𝑏 1/𝑝 𝑝+1 π‘Ž )/((𝑝 + 1)(𝑏 βˆ’ π‘Ž))] (𝑝 =ΜΈ 0, βˆ’1), 𝐿 0 (π‘Ž, 𝑏) = 𝐼(π‘Ž, 𝑏), and 𝐿 βˆ’1 (π‘Ž, 𝑏) = 𝐿(π‘Ž, 𝑏) is the generalized logarithmic mean of π‘Ž and 𝑏. Jagers [13] and Seiffert [14] proved that the double inequalities 𝑀1/2 (π‘Ž, 𝑏) < 𝑃 (π‘Ž, 𝑏) < 𝑀2/3 (π‘Ž, 𝑏) , 𝑀1 (π‘Ž, 𝑏) < 𝑇 (π‘Ž, 𝑏) < 𝑀2 (π‘Ž, 𝑏) hold for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏.

(5)

2

Journal of Function Spaces In [15, 16], the authors proved that the double inequalities 𝐼 (π‘Ž, 𝑏) + 𝐿 (π‘Ž, 𝑏) < 𝑀1/2 (π‘Ž, 𝑏) , 2 (6) 𝑀0 (π‘Ž, 𝑏) < √𝐼 (π‘Ž, 𝑏) 𝐿 (π‘Ž, 𝑏) < 𝑀1/2 (π‘Ž, 𝑏)

𝑀log 2/(1+log 2) (π‘Ž, 𝑏)
0 with π‘Ž =ΜΈ 𝑏. Costin and Toader [17] proved that the double inequality 𝑀4/3 (π‘Ž, 𝑏) < 𝑇 (π‘Ž, 𝑏) < 𝑀5/3 (π‘Ž, 𝑏)

(7)

holds for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏. In [18–20], the authors proved that the double inequalities 𝑀𝑝 (π‘Ž, 𝑏) < 𝑃 (π‘Ž, 𝑏) < π‘€π‘ž (π‘Ž, 𝑏) , π‘€π‘Ÿ (π‘Ž, 𝑏) < 𝑇 (π‘Ž, 𝑏) < 𝑀𝑠 (π‘Ž, 𝑏)

(8)

hold for all π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏 if and only if 𝑝 ≀ log πœ‹/ log 2, π‘ž β‰₯ 2/3, π‘Ÿ ≀ log 2/(log πœ‹ βˆ’ log 2), and 𝑠 β‰₯ 5/3. Λ‡ Cizmesija [21] proved that 𝑝 = βˆ’π›Ό/2 and π‘ž = log 2/[log 4βˆ’log(1βˆ’π›Ό)] are the best possible parameters such that the double inequality 𝑀𝑝 (π‘Ž, 𝑏) < [𝐻(π‘Ž, 𝑏)+𝛼𝐺(π‘Ž, 𝑏)+(1βˆ’ 𝛼)𝐴(π‘Ž, 𝑏)]/2 < π‘€π‘ž (π‘Ž, 𝑏) holds for all 𝛼 ∈ (0, 1) and π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏. In [22, 23], the authors proved that the inequalities πœ‹ 𝑀𝑝 [K (π‘Ÿ) , E (π‘Ÿ)] > , 2 π‘€πœ† [K (π‘Ÿ) , K (√1 βˆ’ π‘Ÿ2 )]

2. Lemmas In order to prove our main result we need three lemmas, which we present in this section. Lemma 1. The inequality βˆ’8π‘Ÿ2 + 8π‘Ÿ βˆ’ 1 +

Proof. It is not difficult to verify that log[2π‘Ÿ(1 βˆ’ π‘Ÿ)] < 0 for all π‘Ÿ ∈ (0, 1/2). Therefore, we only need to prove that 𝑓(π‘Ÿ) > 0 for π‘Ÿ ∈ (0, 1/2), where 𝑓(π‘Ÿ) = (βˆ’8π‘Ÿ2 +8π‘Ÿβˆ’1) log[2π‘Ÿ(1βˆ’π‘Ÿ)]+log 2. Simple computations lead to 1 𝑓 ( ) = 0, 2

πœ‹/2

(10)

Let 𝑝1 , π‘ž1 , 𝑝2 , π‘ž2 , πœ† 1 , πœ† 2 , πœ† 3 , πœ†, πœ‡ ∈ (0, 1/2) and 𝑝3 , π‘ž3 ∈ (1/2, 1). Then, the authors in [25–28] proved that the inequalities

where 𝑔 (π‘Ÿ) = 8 log [2π‘Ÿ (1 βˆ’ π‘Ÿ)] βˆ’

𝐿 𝑝2 (π‘Ž, 𝑏) < 𝑃 (π‘Ž, 𝑏) < 𝐿 π‘ž2 (π‘Ž, 𝑏) ,

𝑔󸀠 (π‘Ÿ) =

(11)

(14)

(15)

for all π‘Ÿ ∈ (0, 1/2). Inequality (15) implies that 𝑔(π‘Ÿ) < 0 for all π‘Ÿ ∈ (0, 1/2). Then, from (13) we clearly see that 𝑓(π‘Ÿ) > 0 for all π‘Ÿ ∈ (0, 1/2). Lemma 2. The inequality log 2 log [2π‘Ÿ (1 βˆ’ π‘Ÿ)]

(16)

holds for all π‘Ÿ ∈ (0, 1/2). Proof. Let 𝑝 = βˆ’ log 2/ log[2π‘Ÿ(1 βˆ’ π‘Ÿ)] and π‘ž = βˆ’8π‘Ÿ2 + 8π‘Ÿ βˆ’ 1. Then, it is not difficult to verify that

𝐢𝑝3 (π‘Ž, 𝑏) < 𝑇 (π‘Ž, 𝑏) < πΆπ‘ž3 (π‘Ž, 𝑏) , πΊπœ† (π‘Ž, 𝑏) < 𝐼 (π‘Ž, 𝑏) < πΊπœ‡ (π‘Ž, 𝑏)

(1 βˆ’ 2π‘Ÿ) [8π‘Ÿ (1 βˆ’ π‘Ÿ) + 1] >0 π‘Ÿ2 (1 βˆ’ π‘Ÿ)2

2 log 2 ] 𝐺 (π‘Ž, 𝑏) ,

1 + 8, π‘Ÿ (1 βˆ’ π‘Ÿ)

1 𝑔 ( ) = 4 (1 βˆ’ log 4) < 0, 2

βˆ’ 16π‘Ÿ2 + 16π‘Ÿ βˆ’ 2 βˆ’ (8π‘Ÿ2 βˆ’ 8π‘Ÿ βˆ’ 1)

𝐻𝑝1 (π‘Ž, 𝑏) < 𝑃 (π‘Ž, 𝑏) < π»π‘ž1 (π‘Ž, 𝑏) ,

πΊπœ† 3 (π‘Ž, 𝑏) > 𝐿 (π‘Ž, 𝑏) ,

(13)

𝑓 (π‘Ÿ) = (1 βˆ’ 2π‘Ÿ) 𝑔 (π‘Ÿ) ,

∫0 (1 βˆ’ π‘Ÿ2 sin2 πœƒ)βˆ’1/2 π‘‘πœƒ and E(π‘Ÿ) = ∫0 (1 βˆ’ π‘Ÿ2 sin2 πœƒ)1/2 π‘‘πœƒ are, respectively, the complete elliptic integrals of the first and second kinds. Let 𝑝 ∈ [0, 1] and 𝑁 be the bivariate symmetric mean. Then, the one-parameter mean 𝑁𝑝 (π‘Ž, 𝑏) was defined by Neuman [24] as follows:

π»πœ† 2 (π‘Ž, 𝑏) > 𝐿 (π‘Ž, 𝑏) ,

(12)

σΈ€ 

hold for all π‘Ÿ ∈ (0, 1) if and only if 𝑝 β‰₯ βˆ’1/2, πœ† ≀ log 2/(log πœ‹ βˆ’ log 2) βˆ’ log[K(√2/2)] = βˆ’4.18, . . ., and πœ‡ β‰₯ 1 βˆ’ 4K4 (√2/2)/πœ‹2 = βˆ’3.789, . . ., where K(π‘Ÿ) =

𝑁𝑝 (π‘Ž, 𝑏) = 𝑁 [π‘π‘Ž + (1 βˆ’ 𝑝) 𝑏, 𝑝𝑏 + (1 βˆ’ 𝑝) π‘Ž] .

log 2 0 with π‘Ž =ΜΈ 𝑏 if and only if 𝑝1 ≀ 1/2 βˆ’ √1 βˆ’ 2/πœ‹/2, π‘ž1 β‰₯ 1/2 βˆ’ √6/12, 𝑝2 ≀ πœ† 0 , π‘ž2 β‰₯ 1/2 βˆ’ √2/4, πœ† 1 β‰₯ 1/2 βˆ’ √2/4, πœ† 2 β‰₯ 1/2 βˆ’ √3/6, πœ† 3 β‰₯ 1/2 βˆ’ √6/6, 𝑝3 ≀ 1/2 + √4/πœ‹ βˆ’ 1/2, π‘ž3 β‰₯ 1/2 + √3/6, πœ† ≀ 1/2 βˆ’ √1 βˆ’ 4/𝑒2 /2, and πœ‡ β‰₯ 1/2 βˆ’ √3/6, where πœ† 0 is the unique solution of the equation log[(1 βˆ’ πœ†)/πœ†] = πœ‹(1 βˆ’ 2πœ†). The main purpose of this paper is to present the best possible parameters 𝛼 = 𝛼(π‘Ÿ) and 𝛽 = 𝛽(π‘Ÿ) such that the double inequality 𝑀𝛼 (π‘Ž, 𝑏) < π»π‘Ÿ (π‘Ž, 𝑏) < 𝑀𝛽 (π‘Ž, 𝑏) holds for all π‘Ÿ ∈ (0, 1/2) and π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏.

00 log [2π‘Ÿ (1 βˆ’ π‘Ÿ)]

𝑔 (1) = 0, (20)

(27)

𝑔󸀠 (1) = βˆ’8π‘Ÿ2 + 8π‘Ÿ βˆ’ 1 βˆ’ 𝑝,

(28)

𝑔󸀠󸀠 (1) = βˆ’16π‘Ÿ2 + 16π‘Ÿ βˆ’ 2 + 𝑝 (8π‘Ÿ2 βˆ’ 8π‘Ÿ βˆ’ 1) + 𝑝2 ,

(29)

𝑔󸀠󸀠󸀠 (π‘₯) = (1 βˆ’ 𝑝) π‘₯βˆ’π‘βˆ’2 β„Ž (π‘₯) ,

holds for all π‘Ÿ ∈ (0, 1/2). Proof. Let 𝑝 = βˆ’ log 2/ log[2π‘Ÿ(1 βˆ’ π‘Ÿ)] and π‘ž = βˆ’8π‘Ÿ2 + 8π‘Ÿ βˆ’ 1. Then, it follows from (17) and (18) that βˆ’ 6π‘Ÿ2 + 6π‘Ÿ βˆ’ (12π‘Ÿ2 βˆ’ 12π‘Ÿ + 1)

log 2 log [2π‘Ÿ (1 βˆ’ π‘Ÿ)]

3 3 + [1 βˆ’ (π‘ž + 1)] 𝑝 + 𝑝2 = (π‘ž + 1) (1 βˆ’ 2𝑝) 2 4

β„Ž (π‘₯) = (βˆ’π‘Ÿ2 + π‘Ÿ) (3 βˆ’ 𝑝) (2 βˆ’ 𝑝) π‘₯2

β„Ž (1) = 6 (βˆ’π‘Ÿ2 + π‘Ÿ) + 𝑝 (12π‘Ÿ2 βˆ’ 12π‘Ÿ + 1) + 𝑝2 , (21)

β„ŽσΈ€  (π‘₯) = 2 (π‘Ÿ βˆ’ π‘Ÿ2 ) (2 βˆ’ 𝑝) [(3 βˆ’ 𝑝) π‘₯ βˆ’ 𝑝] .

(33)

Case 1 (𝑝 = βˆ’8π‘Ÿ2 + 8π‘Ÿ βˆ’ 1). We divide the discussion into two subcases. Subcase 1.1 (𝑝 = βˆ’8π‘Ÿ2 + 8π‘Ÿ βˆ’ 1 and π‘Ÿ ∈ (0, (2 βˆ’ √2)/4) βˆͺ (2 βˆ’ √2)/4, 1/2)). Then, we clearly see that 𝑝 ∈ (βˆ’1, 0)βˆͺ(0, 1), and (28), (29), (32), and (33) lead to

3. Main Results

𝑔󸀠 (1) = 𝑔󸀠󸀠 (1) = 0,

Theorem 4. The double inequality (22)

β„Ž (1) =

holds for all π‘Ÿ ∈ (0, 1/2) and π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏 if and only if 𝛼 ≀ βˆ’8π‘Ÿ2 + 8π‘Ÿ βˆ’ 1 and 𝛽 β‰₯ βˆ’ log 2/ log[2π‘Ÿ(1 βˆ’ π‘Ÿ)].

β„ŽσΈ€  (π‘₯) =

Proof. Without loss of generality, we assume that π‘Ž = π‘₯ ∈ (1, ∞) and 𝑏 = 1. Let log[π»π‘Ÿ (π‘₯, 1)] βˆ’ log[𝑀𝑝 (π‘₯, 1)] = 𝑓(π‘₯), where π‘Ÿ (1 βˆ’ π‘Ÿ) π‘₯2 + (1 βˆ’ 2π‘Ÿ + 2π‘Ÿ2 ) π‘₯ + π‘Ÿ (1 βˆ’ π‘Ÿ)

(1 + 𝑝) log 2 βˆ’ log (1 + π‘₯𝑝 ) + . 𝑝

(32)

We divide the proof into two cases.

1 = (π‘ž + 1) (3 βˆ’ 2𝑝) = 2π‘Ÿ (1 βˆ’ π‘Ÿ) (3 βˆ’ 2𝑝) > 0 4 for all π‘Ÿ ∈ (0, 1/2).

1+π‘₯

(31)

+ (3π‘Ÿ2 βˆ’ 3π‘Ÿ + 1) 𝑝 (𝑝 + 1) ,

3 (π‘ž + 1) (1 βˆ’ 2𝑝) + 𝑝 (π‘ž + 1) 4

𝑀𝛼 (π‘Ž, 𝑏) < π»π‘Ÿ (π‘Ž, 𝑏) < 𝑀𝛽 (π‘Ž, 𝑏)

(30)

where

βˆ’ (βˆ’2π‘Ÿ2 + 2π‘Ÿ) 𝑝 (2 βˆ’ 𝑝) π‘₯

2 log 2 3 +[ ] = (π‘ž + 1) log [2π‘Ÿ (1 βˆ’ π‘Ÿ)] 4

𝑓 (π‘₯) = log

(26)

+ π‘Ÿ2 βˆ’ π‘Ÿ,

Lemma 3. The inequality

+ 𝑝 (𝑝 + 1) >

(25)

where

= βˆ’ (𝑝 βˆ’ π‘ž) (2 βˆ’ 𝑝) < 0

+[

(24)

𝑓󸀠 (π‘₯)

log 2 βˆ’ 16π‘Ÿ + 16π‘Ÿ βˆ’ 2 βˆ’ (8π‘Ÿ βˆ’ 8π‘Ÿ βˆ’ 1) log [2π‘Ÿ (1 βˆ’ π‘Ÿ)] 2

𝑓 (1) = 0,

(23)

>

1 (𝑝 + 1) (3 βˆ’ 2𝑝) > 0, 4 1 (𝑝 + 1) (2 βˆ’ 𝑝) [(3 βˆ’ 𝑝) π‘₯ βˆ’ 𝑝] 4 1 (𝑝 + 1) (2 βˆ’ 𝑝) (3 βˆ’ 2𝑝) > 0 4

(34) (35)

(36)

for π‘₯ ∈ (1, ∞). It follows easily from (24), (25), (27), (30), and (34)–(36) that π»π‘Ÿ (π‘₯, 1) > π‘€βˆ’8π‘Ÿ2 +8π‘Ÿβˆ’1 (π‘₯, 1) for all π‘Ÿ ∈ (0, 1/2).

(37)

4

Journal of Function Spaces

Subcase 1.2 (𝑝 = βˆ’8π‘Ÿ2 + 8π‘Ÿ βˆ’ 1 and π‘Ÿ = (2 βˆ’ √2)/4). Then, 𝑝 = 0, and (37) follows from π»π‘Ÿ (π‘₯, 1) βˆ’ π‘€βˆ’8π‘Ÿ2 +8π‘Ÿβˆ’1 (π‘₯, 1) 2

=

(√π‘₯ βˆ’ 1) π‘₯2 + 6π‘₯ + 1 βˆ’ √π‘₯ = . 4 (1 + π‘₯) 4 (1 + π‘₯)

(38)

Case 2 (𝑝 = βˆ’ log 2/ log[2π‘Ÿ(1 βˆ’ π‘Ÿ)]). Then, we clearly see that 𝑝 ∈ (0, 1), and Lemmas 1–3 and (23), (26), (28), (29), and (32) lead to

Conflict of Interests

(39)

lim 𝑔 (π‘₯) = +∞,

(40)

The authors declare that there is no conflict of interests regarding the publication of this paper.

lim 𝑔󸀠 (π‘₯) = +∞,

(41)

Acknowledgments

(42)

π‘₯ β†’ +∞ π‘₯ β†’ +∞

𝑔󸀠 (1) < 0,

𝑔󸀠󸀠 (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„Ž (1) > 0

(45)

References

lim 𝑔󸀠󸀠 (π‘₯) = +∞,

π‘₯ β†’ +∞

(43)

and (36) again holds. It follows from (30), (36), and (45) that 𝑔󸀠󸀠 is strictly increasing on (1, ∞). Then, (43) and (44) lead to the conclusion that there exists πœ‡0 > 1 such that 𝑔󸀠 is strictly decreasing on (1, πœ‡0 ] and strictly increasing on [πœ‡0 , ∞). From (41) and (42) together with the piecewise monotonicity of 𝑔󸀠 we clearly see that there exists πœ‡1 > 1 such that 𝑔 is strictly decreasing on (1, πœ‡1 ] and strictly increasing on [πœ‡1 , ∞). Then, (25), (27), and (40) lead to the conclusion that there exists πœ‡2 > 1 such that 𝑓 is strictly decreasing on (1, πœ‡2 ] and strictly increasing on [πœ‡2 , ∞). Therefore, π»π‘Ÿ (π‘₯, 1) < π‘€βˆ’ log 2/ log[2π‘Ÿ(1βˆ’π‘Ÿ)] (π‘₯, 1)

(46)

for all π‘Ÿ ∈ (0, 1/2) follows from (24) and (39) together with the piecewise monotonicity of 𝑓. Next, we prove that 𝛼 = βˆ’8π‘Ÿ2 + 8π‘Ÿ βˆ’ 1 and 𝛽 = βˆ’ log 2/ log[2π‘Ÿ(1 βˆ’ π‘Ÿ)] are the best possible parameters such that the double inequality 𝑀𝛼 (π‘Ž, 𝑏) < π»π‘Ÿ (π‘Ž, 𝑏) < 𝑀𝛽 (π‘Ž, 𝑏)

(47)

holds for all π‘Ÿ ∈ (0, 1/2) and π‘Ž, 𝑏 > 0 with π‘Ž =ΜΈ 𝑏. Let π‘Ÿ ∈ (0, 1/2), 𝛼0 = βˆ’8π‘Ÿ2 +8π‘Ÿβˆ’1, 𝛽0 = βˆ’ log 2/ log[2π‘Ÿ(1βˆ’ π‘Ÿ)], πœ€ ∈ (0, 𝛽0 ), and 𝑑 > 0. Then, we have π»π‘Ÿ (1, 1 + 𝑑) βˆ’ 𝑀𝛼0 +πœ€ (1, 1 + 𝑑) 2 [(1 βˆ’ π‘Ÿ) (1 + 𝑑) + π‘Ÿ] [π‘Ÿ (1 + 𝑑) + 1 βˆ’ π‘Ÿ] 𝑑+2 βˆ’[

Inequality (49) and equation (50) imply that for any π‘Ÿ ∈ (0, 1/2) and πœ€ ∈ (0, 𝛽0 ) there exist 𝑇 = 𝑇(πœ€) > 1 and 𝛿 = 𝛿(πœ€) ∈ (0, 1) such that π»π‘Ÿ (1, 𝑑) > 𝑀𝛽0 βˆ’πœ€ (1, 𝑑) for 𝑑 > 𝑇 and π»π‘Ÿ (1, 1 + 𝑑) < 𝑀𝛼0 +πœ€ (1, 1 + 𝑑) for 𝑑 ∈ (0, 𝛿).

lim 𝑓 (π‘₯) = 0,

π‘₯ β†’ +∞

=

Let 𝑑 β†’ 0+ and make use of the Taylor expansion; then, (48) leads to πœ€ π»π‘Ÿ (1, 1 + 𝑑) βˆ’ 𝑀𝛼0 +πœ€ (1, 1 + 𝑑) = βˆ’ π‘₯2 + π‘œ (π‘₯2 ) . (50) 8

1 + (1 + 𝑑)𝛼0 +πœ€ ] 2

π»π‘Ÿ (1, 𝑑) lim 𝑑 β†’ +∞ 𝑀 𝛽0 βˆ’πœ€ (1, 𝑑)

1/(𝛼0 +πœ€)

(48)

,

= 2πœ€/[𝛽0 (𝛽0 βˆ’πœ€)] > 1.

(49)

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