Sharp Power Mean Bounds for the Combination of Seiffert and Geometric Means

Abstract

We answer the question: for 𝛼∈(0,1), what are the greatest value 𝑝 and the least value 𝑞 such that the double inequality 𝑀_𝑝(𝑎,𝑏)<𝑃^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)<𝑀_𝑞(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏. Here, 𝑀_𝑝(𝑎,𝑏), 𝑃(𝑎,𝑏), and 𝐺(𝑎,𝑏) denote the power of order 𝑝, Seiffert, and geometric means of two positive numbers 𝑎 and 𝑏, respectively.

1. Introduction

For 𝑝∈ℝ, the power mean 𝑀_𝑝(𝑎,𝑏) of order 𝑝 and the Seiffert mean 𝑃(𝑎,𝑏) of two positive numbers 𝑎 and 𝑏 are defined by 𝑀_𝑝(𝑎(𝑎,𝑏)=^𝑝+𝑏^𝑝2)^1/𝑝√,𝑝≠0,𝑎𝑏,𝑝=0,𝑃(𝑎,𝑏)=𝑎−𝑏4arctan(𝑎/𝑏)−𝜋,𝑎≠𝑏,𝑎,𝑎=𝑏. (1.1) The main properties of the power mean are given in [1]. It is well known that 𝑀_𝑝(𝑎,𝑏) is strictly increasing with respect to 𝑝 for fixed 𝑎,𝑏>0 with 𝑎≠𝑏. Recently, the power mean has been the subject of intensive research. In particular, many remarkable inequalities for the power mean can be found in the literature [2–16].

The Seiffert mean 𝑃 was introduced by Seiffert in [17], it can be rewritten in the following symmetric form (see [18, (2.4)]): 𝑃(𝑎,𝑏)=𝑎−𝑏2arcsin((𝑎−𝑏)/(𝑎+𝑏)),𝑎≠𝑏,𝑎,𝑎=𝑏. (1.2)

Let 𝐴(𝑎,𝑏)=(1/2)(𝑎+𝑏), 𝐺(𝑎,𝑏)=√𝑎𝑏, 𝐿(𝑎,𝑏)={^{(𝑏−𝑎)/(log𝑏−log𝑎),𝑏≠𝑎,}_{𝑎,𝑏=𝑎,}𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏) and 𝐼(𝑎,𝑏)={^{1/𝑒(𝑏𝑏/𝑎𝑎)1/(𝑏−𝑎),𝑏≠𝑎,}_{𝑎,𝑏=𝑎} be the arithmetic, geometric, logarithmic, harmonic, and identric means of two positive numbers 𝑎 and 𝑏, respectively. Then it is well known that min{𝑎,𝑏}<𝐻(𝑎,𝑏)=𝑀₋₁(𝑎,𝑏)<𝐺(𝑎,𝑏)=𝑀₀(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝐴(𝑎,𝑏)=𝑀₁(𝑎,𝑏)<max{𝑎,𝑏} (1.3) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [9], Alzer and Janous presented the sharp power mean bounds for the sum (2/3)𝐴(𝑎,𝑏)+(1/3)𝐺(𝑎,𝑏) as follows: 𝑀_log2/log32(𝑎,𝑏)<31𝐴(𝑎,𝑏)+3𝐺(𝑎,𝑏)<𝑀_2/3(𝑎,𝑏) (1.4) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [17], Seiffert proved that 𝐿(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝐼(𝑎,𝑏) (1.5) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The following power mean bounds for the Seiffert mean was given by Jagers [19]: 𝑀_1/2(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝑀_2/3(𝑎,𝑏) (1.6) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [20, 21], the authors presented the bounds for the Seiffert mean 𝑃 in terms of 𝐴 and 𝐺 as follows: 𝑃(𝑎,𝑏)>3𝐴(𝑎,𝑏)𝐺(𝑎,𝑏),1𝐴(𝑎,𝑏)+2𝐺(𝑎,𝑏)21𝐴(𝑎,𝑏)+22𝐺(𝑎,𝑏)<𝑃(𝑎,𝑏)<31𝐴(𝑎,𝑏)+3𝐺(𝑎,𝑏),𝑃(𝑎,𝑏)>𝐴^1/3(𝑎,𝑏)𝐺^2/3(𝑎,𝑏) (1.7) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The following sharp lower power mean bounds for (1/3)𝐺(𝑎,𝑏)+(2/3)𝐻(𝑎,𝑏), (2/3)𝐺(𝑎,𝑏)+(1/3)𝐻(𝑎,𝑏), and 𝑃(𝑎,𝑏) can be found in [4, 6]: 132𝐺(𝑎,𝑏)+3𝐻(𝑎,𝑏)>𝑀_−2/32(𝑎,𝑏),31𝐺(𝑎,𝑏)+3𝐻(𝑎,𝑏)>𝑀_−1/3(𝑎,𝑏),𝑃(𝑎,𝑏)>𝑀_log2/log𝜋(𝑎,𝑏) (1.8) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The purpose of this paper is to answer the question: for 𝛼∈(0,1), what are the greatest value 𝑝 and the least value 𝑞 such that the double inequality 𝑀_𝑝(𝑎,𝑏)<𝑃^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)<𝑀_𝑞(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.

2. Lemmas

In order to prove our main result, we need several lemmas which we present in this section.

Lemma 2.1. Let 𝜆∈(0,1/3), 𝑥∈[1,∞) and ℎ(𝑥)=(1−3𝜆)𝑥^2𝜆+1−(1+3𝜆)𝑥^2𝜆−(1+3𝜆)𝑥+(1−3𝜆). Then there exists 𝑥₀>1 such that ℎ(𝑥)<0 for 𝑥∈[1,𝑥₀), ℎ(𝑥)>0 for 𝑥∈(𝑥₀,∞) and ℎ(𝑥₀)=0.

Proof. Simple computations lead to ℎ(1)=−12𝜆<0,lim_𝑥→+∞ℎℎ(𝑥)=+∞,_′(𝑥)=(1−3𝜆)(2𝜆+1)𝑥^2𝜆−2𝜆(1+3𝜆)𝑥^2𝜆−1ℎ−(1+3𝜆),_′(1)=−6𝜆(1+2𝜆)<0,lim_𝑥→+∞ℎ^′ℎ(𝑥)=+∞,_′′(𝑥)=2𝜆𝑥^2𝜆−2[(1+2𝜆)(1−3𝜆)𝑥+(1−2𝜆)(1+3𝜆)]>0. (2.1)(2.2)(2.3)(2.4)(2.5)
Inequality (2.5) implies that ℎ^′(𝑥) is strictly increasing in [1,∞). Then (2.3) and (2.4) lead to that there exists 𝑥₁>1 such that ℎ^′(𝑥)<0 for 𝑥∈[1,𝑥₁) and ℎ^′(𝑥)>0 for 𝑥∈(𝑥₁,∞). Hence, ℎ(𝑥) is strictly decreasing in [1,𝑥₁] and strictly increasing in [𝑥₁,∞).
Therefore, Lemma 2.1 follows from (2.1) and (2.2) together with the monotonicity of ℎ(𝑥).

Lemma 2.2. If 𝜆∈(1/√10,1/3), then the following statements are true:(1)300𝜆⁴+324𝜆³+29𝜆²−45𝜆−8<0;(2)−1176𝜆⁵−24𝜆⁴+114𝜆³+10𝜆²−3𝜆−1<0;(3)−24𝜆⁵+72𝜆⁴+178𝜆³+6𝜆²−25𝜆−3<0.

Proof. Simple computations lead to(1)300𝜆⁴+324𝜆³+29𝜆²−45𝜆−8<300×(1/3)⁴ + 324×(1/3)³+29×(1/3)²−45/√10−8 = (590−243√10)/54<0; (2)−1176𝜆⁵−24𝜆⁴+114𝜆³ + 10𝜆²−3𝜆−1<−1176×(1/√10)⁵−24×(1/√10)⁴ + 114×(1/3)³+10×(1/3)²−3/√10−1 = (3070−1107√10)/750<0; (3)−24𝜆⁵+72𝜆⁴+178𝜆³ + 6𝜆²−25𝜆−3<−24×(1/√10)⁵+72×(1/3)⁴ + 178×(1/3)³+6×(1/3)²−25/√10−3 = (34750−17037√10)/6750<0.

Lemma 2.3. If 𝛼∈(0,3/√10], then 𝐼^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)<𝑀_2𝛼/3(𝑎,𝑏) (2.6) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Proof. Without loss of generality, we assume that 𝑎>𝑏. Let 𝑡=𝑎/𝑏>1 and 𝛽=𝛼/3∈(0,1/√10]. Then log[𝑀_2𝛼/3(𝑎,𝑏)]−log[𝐼^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)]=log[𝑀_2𝛽(𝑎,𝑏)]−log[𝐼^3𝛽(𝑎,𝑏)𝐺^1−3𝛽=1(𝑎,𝑏)]2𝛽log1+𝑡^2𝛽2−3𝛽𝑡𝑡−1log𝑡−1−3𝛽2log𝑡+3𝛽. (2.7) Let 1𝑓(𝑡)=2𝛽log1+𝑡^2𝛽2−3𝛽𝑡𝑡−1log𝑡−1−3𝛽2log𝑡+3𝛽. (2.8) Then simple computations lead to lim_𝑡→1𝑓𝑓(𝑡)=0,_′(𝑡)=𝑔(𝑡)(𝑡−1)², (2.9)(2.10) where 𝑔(𝑡)=(𝑡^2𝛽+1−2𝑡^2𝛽+𝑡^2𝛽−1) / (1+𝑡^2𝛽)−3𝛽(𝑡−1) + 3𝛽log𝑡−(1−3𝛽)(𝑡−2+1/𝑡)/2, 𝑔𝑔(1)=0,_′𝑔(𝑡)=¹(𝑡)2𝑡²(1+𝑡^2𝛽)², (2.11) where 𝑔₁(𝑡)=(1−3𝛽)𝑡^4𝛽+2 + 6𝛽𝑡^4𝛽+1−(1+3𝛽)𝑡^4𝛽−2𝛽𝑡^2𝛽+2 + 4𝛽𝑡^2𝛽+1−2𝛽𝑡^2𝛽−(1+3𝛽)𝑡²+6𝛽𝑡+(1−3𝛽), 𝑔₁𝑔(1)=0,_′1𝑔(1)=0,_1′′(1)=0. (2.12)(2.13)(2.14) Let 𝑔₂(𝑡)=𝑔₁^′′′(𝑡)/(8𝛽𝑡^2𝛽−3) and 𝑔₃(𝑡)=𝑔₂^′′′(𝑡)/(2𝛽𝑡^2𝛽−3). Then 𝑔₂(𝑡)=(1−3𝛽)(2𝛽+1)(4𝛽+1)𝑡^2𝛽+2+3𝛽(4𝛽+1)(4𝛽−1)𝑡^2𝛽+1−(1+3𝛽)(4𝛽−1)(2𝛽−1)𝑡^2𝛽−𝛽(𝛽+1)(2𝛽+1)𝑡²𝑔+𝛽(2𝛽+1)(2𝛽−1)𝑡−𝛽(𝛽−1)(2𝛽−1),₂𝑔(1)=0,_′2(1)=2(1−√10𝛽)(1+√𝑔10𝛽)≥0,_2′′(1)=2(6𝛽+1)(1−√10𝛽)(1+√𝑔10𝛽)≥0,₃(𝑡)=2(1−3𝛽)(2𝛽+1)²(4𝛽+1)(𝛽+1)𝑡²+3𝛽(4𝛽+1)(4𝛽−1)×(2𝛽+1)(2𝛽−1)𝑡−2(1+3𝛽)(4𝛽−1)(2𝛽−1)²𝑔(𝛽−1),₃(1)=9𝛽(3−28𝛽²𝑔)>0,_′3𝑔(𝑡)=(2𝛽+1)(4𝛽+1)[4(1−3𝛽)(2𝛽+1)(𝛽+1)𝑡+3𝛽(4𝛽−1)(2𝛽−1)],_′3(1)=(2𝛽+1)(4𝛽+1)(4+3𝛽−46𝛽²),min_{𝛽∈[0,1/10]}(4+3𝛽−46𝛽²3√)=10−610>0. (2.15)(2.16)(2.17)(2.18)(2.19)(2.20)(2.21)(2.22)(2.23)
From (2.21)–(2.23) we clearly see that 𝑔^′₃(𝑡)>𝑔^′₃(1)≥(3√10−6)/10(2𝛽+1)(4𝛽+1)>0 for 𝑡∈(1,∞), hence 𝑔₃(𝑡) is strictly increasing in [1,∞). Then (2.20) implies that 𝑔₃(𝑡)>0 for 𝑡∈(1,∞), hence 𝑔₂^′′(𝑡) is strictly increasing in [1,∞).
It follows from (2.18) and the monotonicity of 𝑔₂^′′(𝑡) that 𝑔₂^′′(𝑡)>0 for 𝑡∈(1,∞), hence 𝑔^′₂(𝑡) is strictly increasing in [1,∞). Then (2.17) implies that 𝑔^′₂(𝑡)>0 for 𝑡∈(1,∞), therefore 𝑔₂(𝑡) is strictly increasing in [1,∞).
Equation (2.16) and the monotonicity of 𝑔₂(𝑡) lead to that 𝑔₂(𝑡)>0 for 𝑡∈(1,∞), so 𝑔₁^′′(𝑡) is strictly increasing in [1,∞).
From (2.9)–(2.14) and the monotonicity of 𝑔₁^′′(𝑡) we can deduce that 𝑓(𝑡)>0 (2.24) for 𝑡∈(1,∞).
Therefore, Lemma 2.3 follows from (2.7) and (2.8) together with (2.24).

Remark 2.4. In [22, Theorem3.1], the authors proved that𝐼^1/2(𝑎,𝑏)𝐺^1/2(𝑎,𝑏)<𝑀_1/3𝐼(𝑎,𝑏),_2/3(𝑎,𝑏)𝐺^1/3(𝑎,𝑏)<𝑀_4/9𝐼(𝑎,𝑏),_1/3(𝑎,𝑏)𝐺^2/3(𝑎,𝑏)<𝑀_2/9(𝑎,𝑏) (2.25)(2.26)(2.27) for all 𝑎,𝑏>0 with 𝑎≠𝑏.
Obviously, (2.6) is the generalization of (2.25)–(2.27).

Remark 2.5. If 𝛼∈(3/√10,1), then 𝛽=𝛼/3∈(1/√10,1/3) and (2.17) leads to𝑔^′₂(1)=2(1−√10𝛽)(1+√10𝛽)<0. (2.28)
Inequality (2.28) and the continuity of 𝑔^′₂(𝑡) imply that there exists 𝛿>0 such that 𝑔^′₂(𝑡)<0 (2.29) for 𝑡∈[1,1+𝛿).
From (2.29) and (2.9)–(2.16) we can deduce that𝑓(𝑡)<0 (2.30) for 𝑡∈(1,1+𝛿).
Equations (2.7) and (2.8) together with (2.30) lead to𝐼^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)>𝑀_2𝛼/3(𝑎,𝑏) (2.31) for all 𝑎,𝑏>0 with 𝑎/𝑏∈(1,1+𝛿)∪(1/(1+𝛿),1).
Therefore, 𝛼₀=3/√10 is the largest value in (0,1) such that inequality (2.6) holds for 𝛼∈(0,𝛼₀].

3. Main Result

Theorem 3.1. If 𝛼∈(0,1), then 𝑀₀(𝑎,𝑏)<𝑃^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)<𝑀_2𝛼/3(𝑎,𝑏) (3.1) holds for all a,b>0 with a≠b, and M₀(a,b) and M_2𝛼/3(a,b) are the best possible lower and upper power mean bounds for the product P^𝛼(a,b)G^1−𝛼(a,b).

Proof. For all 𝑎,𝑏>0 with 𝑎≠𝑏, from (1.3), (1.5) and Lemma 2.1 we clearly see that𝑃^𝛼(𝑎,𝑏)G^1−𝛼(𝑎,𝑏)>𝑀₀(𝑎,𝑏) (3.2) for all 𝛼∈(0,1), and 𝑃^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)<𝑀_2𝛼/3(𝑎,𝑏) (3.3) for 𝛼∈(0,3/√10].
Next, we prove that (3.3) is also true for 𝛼∈(3/√10,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏.
Without loss of generality, we assume that 𝑎>𝑏. Let 𝑡=𝑎/𝑏>1 and 𝛽=𝛼/3∈(1/√10,1/3). Then (1.1) leads tolog[𝑀_2𝛼/3(𝑎,𝑏)]−log[𝑃^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)]=log[𝑀_2𝛽(𝑎,𝑏)]−log[𝑃^3𝛽(𝑎,𝑏)𝐺^1−3𝛽=1(𝑎,𝑏)]2𝛽log1+𝑡^4𝛽2𝑡−3𝛽log²−14arctan𝑡−𝜋−(1−3𝛽)log𝑡. (3.4) Let 1𝐹(𝑡)=2𝛽log1+𝑡^4𝛽2𝑡−3𝛽log²−14arctan𝑡−𝜋−(1−3𝛽)log𝑡. (3.5) Then simple computations lead to lim_𝑡→1𝐹𝐹(𝑡)=0,_′(𝑡)=(1−3𝛽)𝑡^4𝛽+2−(1+3𝛽)𝑡^4𝛽−(1+3𝛽)𝑡²+(1−3𝛽)𝑡(𝑡²−1)(𝑡^4𝛽++1)12𝛽(𝑡².+1)(4arctan𝑡−𝜋) (3.6)(3.7) From Lemma 2.1 we know that there exists 𝜆₀>1 such that (1−3𝛽)𝜆₀^4𝛽+2−(1+3𝛽)𝜆₀^4𝛽−(1+3𝛽)𝜆₀²+(1−3𝛽)=0,(1−3𝛽)𝑡^4𝛽+2−(1+3𝛽)𝑡^4𝛽−(1+3𝛽)𝑡²+(1−3𝛽)<0 (3.8)(3.9) for 𝑡∈[1,𝜆₀), and (1−3𝛽)𝑡^4𝛽+2−(1+3𝛽)𝑡^4𝛽−(1+3𝛽)𝑡²+(1−3𝛽)>0 (3.10) for 𝑡∈(𝜆₀,∞).
We divide two cases to prove that𝐹^′(𝑡)>0 (3.11) for 𝑡>1.

Case 1. 𝑡∈[𝜆₀,∞). Then (3.11) follows from (3.7) and (3.8) together with (3.10).

Case 2. 𝑡∈(1,𝜆₀). Then (3.7) can be written as𝐹^′(𝑡)=(1−3𝛽)𝑡^4𝛽+2−(1+3𝛽)𝑡^4𝛽−(1+3𝛽)𝑡²+(1−3𝛽)𝑡(𝑡²−1)(𝑡^4𝛽𝐹+1)(4arctan𝑡−𝜋)¹(𝑡), (3.12) where 𝐹₁(𝑡)=4arctan𝑡+12𝛽𝑡(𝑡²−1)(𝑡^4𝛽+1)(𝑡²+1)[(1−3𝛽)𝑡^4𝛽+2−(1+3𝛽)𝑡^4𝛽−(1+3𝛽)𝑡²+(1−3𝛽)]−𝜋. (3.13) Let 𝑥=𝑡²∈(1,𝜆₀²), then (3.13) leads to 𝐹₁𝐹(1)=0,_′1(𝑡)=4𝐹₂(𝑥)(𝑡²+1)²[(1−3𝛽)𝑡^4𝛽+2−(1+3𝛽)𝑡^4𝛽−(1+3𝛽)𝑡²+(1−3𝛽)]², (3.14) where 𝐹₂(𝑥)=(18𝛽²−9𝛽+1)𝑥^4𝛽+3−(18𝛽²−3𝛽+1)𝑥^4𝛽+2−(18𝛽²+3𝛽+1)𝑥^4𝛽+1+(18𝛽²+9𝛽+1)𝑥^4𝛽+2(6𝛽²−1)𝑥^2𝛽+3−2(6𝛽²−1)𝑥^2𝛽+2−2(6𝛽²−1)𝑥^2𝛽+1+2(6𝛽²−1)𝑥^2𝛽+(18𝛽²+9𝛽+1)𝑥³−(18𝛽²+3𝛽+1)𝑥²−(18𝛽²−3𝛽+1)𝑥+(18𝛽²−9𝛽+1). (3.15)(1)(2)(3)(4)(5) Let 𝐹₃(𝑥)=(𝑥^4−2𝛽/8𝛽)𝐹₂⁽⁴⁾(𝑥), then (3.15) leads to 𝐹₂𝐹(1)=0,_′2𝐹(1)=0,_2′′𝐹(1)=0,_2′′′𝐹(1)=0,₃(𝑥)=(18𝛽²−9𝛽+1)(4𝛽+1)(4𝛽+3)(2𝛽+1)𝑥^2𝛽+3−(18𝛽²−3𝛽+1)(2𝛽+1)(4𝛽+1)(4𝛽−1)𝑥^2𝛽+2−(18𝛽²+3𝛽+1)(4𝛽+1)(4𝛽−1)(2𝛽−1)𝑥^2𝛽+1+(18𝛽²+9𝛽+1)(4𝛽−1)(2𝛽−1)(4𝛽−3)𝑥^2𝛽+(6𝛽²−1)(2𝛽+3)(𝛽+1)(2𝛽+1)𝑥³−(6𝛽²−1)(𝛽+1)(2𝛽+1)(2𝛽−1)𝑥²−(6𝛽²−1)(2𝛽+1)(2𝛽−1)(𝛽−1)𝑥+(6𝛽²−1)(2𝛽−1)(𝛽−1)(2𝛽−3). (3.16)(3.17)
From 𝛽∈(1/√10,1/3) and 𝑥>1, we clearly see that (18𝛽²−9𝛽+1)(4𝛽+1)(4𝛽+3)(2𝛽+1)𝑥^2𝛽+3<(18𝛽²−9𝛽+1)(4𝛽+1)(4𝛽+3)(2𝛽+1)𝑥³, −(18𝛽²−3𝛽+1)(2𝛽+1)(4𝛽+1)(4𝛽−1)𝑥^2𝛽+2<−(18𝛽²−3𝛽+1)(2𝛽+1)(4𝛽+1)(4𝛽−1)𝑥², −(18𝛽²+3𝛽+1)(4𝛽+1)(4𝛽−1)(2𝛽−1)𝑥^2𝛽+1<−(18𝛽²+3𝛽+1)(4𝛽+1)(4𝛽−1)(2𝛽−1)𝑥² and (18𝛽²+9𝛽+1)(4𝛽−1)(2𝛽−1)(4𝛽−3)𝑥^2𝛽<(18𝛽²+9𝛽+1)(4𝛽−1)(2𝛽−1)(4𝛽−3)𝑥. These inequalities and Lemma 2.2 lead to𝐹₃(𝑥)<(18𝛽²−9𝛽+1)(4𝛽+1)(4𝛽+3)(2𝛽+1)𝑥³−(18𝛽²−3𝛽+1)(2𝛽+1)(4𝛽+1)(4𝛽−1)𝑥²−(18𝛽²+3𝛽+1)(4𝛽+1)(4𝛽−1)(2𝛽−1)𝑥²+(18𝛽²+9𝛽+1)(4𝛽−1)(2𝛽−1)(4𝛽−3)𝑥+(6𝛽²−1)(2𝛽+3)(𝛽+1)(2𝛽+1)𝑥³−(6𝛽²−1)(𝛽+1)(2𝛽+1)(2𝛽−1)𝑥²−(6𝛽²−1)(2𝛽+1)(2𝛽−1)(𝛽−1)𝑥+(6𝛽²−1)(2𝛽−1)(𝛽−1)(2𝛽−3)=2𝛽(300𝛽⁴+324𝛽³+29𝛽²−45𝛽−8)𝑥³+(−1176𝛽⁵−24𝛽⁴+114𝛽³+10𝛽²−3𝛽−1)𝑥²+2(276𝛽⁵−276𝛽⁴+3𝛽³+43𝛽²−3𝛽−1)𝑥+(24𝛽⁵−72𝛽⁴+62𝛽³−6𝛽²−11𝛽+3)<2𝛽(300𝛽⁴+324𝛽³+29𝛽²−45𝛽−8)𝑥+(−1176𝛽⁵−24𝛽⁴+114𝛽³+10𝛽²−3𝛽−1)𝑥+2(276𝛽⁵−276𝛽⁴+3𝛽³+43𝛽²−3𝛽−1)𝑥+24𝛽⁵−72𝛽⁴+62𝛽³−6𝛽²−11𝛽+3=(−24𝛽⁵+72𝛽⁴+178𝛽³+6𝛽²−25𝛽−3)𝑥+24𝛽⁵−72𝛽⁴+62𝛽³−6𝛽²−11𝛽+3<−24𝛽⁵+72𝛽⁴+178𝛽³+6𝛽²−25𝛽−3+24𝛽⁵−72𝛽⁴+62𝛽³−6𝛽²−11𝛽+3=12𝛽(20𝛽²−3)<0. (3.18)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)(16)(17)(18)(19)(20)(21)(22)(23)(24)(25) From (3.16)–(3.18) we can deduce that 𝐹₂(𝑥)<0 (3.19) for 𝑥∈(1,𝜆₀²).
Equation (3.14) together with (3.19) imply that𝐹₁(𝑡)<0 (3.20) for 𝑡∈(1,𝜆₀).
Therefore, (3.11) follows from (3.9) and (3.12) together with (3.20).
It follows from (3.4)–(3.6) and (3.11) that𝑃^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)<𝑀_2𝛼/3(𝑎,𝑏) (3.21) for 𝛼∈(3/√10,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏.
At last, we prove that 𝑀₀(𝑎,𝑏) and 𝑀_2𝛼/3(𝑎,𝑏) are the best possible lower- and upper-power mean bounds for the product 𝑃^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏), respectively.
For any 0<𝜀<(2/3)𝛼 and 𝑥>0, from (1.1) one haslim_𝑥→+∞𝑀_𝜀(1,𝑥)𝑃^𝛼(1,𝑥)𝐺^1−𝛼(1,𝑥)=lim_𝑥→+∞[((1+𝑥^−𝜀)/2)^1/𝜀((1−𝑥⁻¹)/(4arctan√𝑥−𝜋))^𝛼𝑥^(1−𝛼)/2]=+∞,[𝑃^𝛼(1,1+𝑥)𝐺^1−𝛼(1,1+𝑥)]^{(2𝛼/3)−𝜀}−[𝑀_{(2𝛼/3)−𝜀}(1,1+𝑥)]^{(2𝛼/3)−𝜀}=𝐽(𝑥)(4arctan√1+𝑥−𝜋)^{𝛼(2𝛼−3𝜀)/3}, (3.22)(3.23) where 𝐽(𝑥)=𝑥^{𝛼(2𝛼−3𝜀)/3}(1+𝑥)^{(1−𝛼)(2𝛼−3𝜀)/6}− (1/2) (4arctan√1+𝑥−𝜋)^{𝛼(2𝛼−3𝜀)/3}[1+(1+𝑥)^{(2𝛼−3𝜀)/3}].
Let 𝑥→0, making use of the Taylor expansion we get1𝐽(𝑥)=24𝜀(2𝛼−3𝜀)𝑥^{(1/3)𝛼(2𝛼−3𝜀)+2}+𝑜(𝑥^{(1/3)𝛼(2𝛼−3𝜀)+2}). (3.24)
Equation (3.22) implies that for any 0<𝜀<(2/3)𝛼 there exists 𝑋=𝑋(𝜀,𝛼)>1 such that 𝑀_𝜀(1,𝑥)>𝑃^𝛼(1,𝑥)𝐺^1−𝛼(1,𝑥) for 𝑥∈(𝑋,∞).
Equations (3.23) and (3.24) imply that for any 0<𝜀<(2/3)𝛼 there exists 𝛿=𝛿(𝜀,𝛼)>0 such that 𝑃^𝛼(1,1+𝑥)𝐺^1−𝛼(1,1+𝑥)>𝑀_{2𝛼/3−𝜀}(1,1+𝑥) for 𝑥∈(0,𝛿).

Acknowledgments

The authors wish to thank the anonymous referees for their very careful reading of the paper and fruitful comments and suggestions. This paper is partly supported by N S Foundation of China under Grant 60850005, N S Foundation of Zhejiang Province under Grants Y7080106 and Y607128, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.