Optimal Lower Generalized Logarithmic Mean Bound for the Seiffert Mean

Abstract

We present the greatest value 𝑝 such that the inequality 𝑃(𝑎,𝑏)>𝐿_𝑝(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝑃(𝑎,𝑏) and 𝐿_𝑝(𝑎,𝑏) denote the Seiffert and 𝑝th generalized logarithmic means of 𝑎 and 𝑏, respectively.

1. Introduction

For 𝑝∈ℝ, the 𝑝th generalized logarithmic mean 𝐿_𝑝∶(0,∞)²→(0,∞) is defined by 𝐿_𝑝()=𝑏𝑎,𝑏^𝑝+1−𝑎^𝑝+1𝑝+1(𝑏−𝑎)^1/𝑝1,𝑎≠𝑏,𝑝≠0,−1,𝑒𝑏^𝑏𝑎^{𝑎1/(𝑏−𝑎)},𝑎≠𝑏,𝑝=0,𝑏−𝑎log𝑏−log𝑎,𝑎≠𝑏,𝑝=−1,𝑎,𝑎=𝑏, (1) and the Seiffert mean 𝑃∶(0,∞)²→(0,∞) [1] is defined by 𝑃(𝑎,𝑏)=𝑎−𝑏4arctan𝑎/𝑏−𝜋,𝑎≠𝑏,𝑎,𝑎=𝑏. (2)

It is well known that the generalized logarithmic mean 𝐿_𝑝(𝑎,𝑏) is continuous and strictly increasing with respect to 𝑝∈ℝ for fixed 𝑎,𝑏>0 with 𝑎≠𝑏. The special cases of the generalized logarithmic mean are, for example, 𝐺(𝑎,𝑏)=√𝑎𝑏=𝐿₋₂(𝑎,𝑏) is the geometric mean, 𝐿(𝑎,𝑏)=(𝑏−𝑎)/ (log𝑏−log𝑎)=𝐿₋₁(𝑎,𝑏) is the logarithmic mean, 𝐼(𝑎,𝑏)=1/𝑒(𝑏^𝑏/𝑎^𝑎)^{1/(𝑏−𝑎)}=𝐿₀(𝑎,𝑏) is the identric mean, and 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2=𝐿₁(𝑎,𝑏) is the arithmetic mean. The Seiffert mean 𝑃(𝑎,𝑏) can be rewritten as (see [2, equation (2.4)]) 𝑃(𝑎,𝑏)=𝑎−𝑏2arcsin[(𝑎−𝑏)/(𝑎+𝑏)],𝑎≠𝑏,𝑎,𝑎=𝑏. (3)

Recently, the bivariate means have been the subject of intensive research. In particular, many remarkable inequalities and properties for the generalized logarithmic and the Seiffert means can be found in the literature [3–13].

In [1, 11], Seiffert proved that the inequalities 𝐿₋₁(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝐿₀()(),2𝑎,𝑏=𝐼𝑎,𝑏𝜋𝐴()()(),𝑃()>𝐺()𝐴()𝑎,𝑏<𝑃𝑎,𝑏<𝐴𝑎,𝑏𝑎,𝑏𝑎,𝑏𝑎,𝑏𝐿(),𝑃()>()𝐺()𝑎,𝑏𝑎,𝑏3𝐴𝑎,𝑏𝑎,𝑏𝐴(𝑎,𝑏)+2𝐺(𝑎,𝑏) (4) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Sándor [14] presented the bounds for the Seiffert mean 𝑃(𝑎,𝑏) in terms of the arithmetic mean 𝐴(𝑎,𝑏) and geometric mean 𝐺(𝑎,𝑏) as follows: 12𝐴()+1𝑎,𝑏2𝐺()()<2𝑎,𝑏<𝑃𝑎,𝑏3𝐴()+1𝑎,𝑏3𝐺(𝑎,𝑏) (5) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Hästö [15] proved that the double inequality 𝑀_𝛼(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝑀_𝛽(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤log2/log𝜋 and 𝛽≥2/3, where 𝑀_𝑝(𝑎,𝑏)=((𝑎^𝑝+𝑏^𝑝)/2)^1/𝑝 is the 𝑝th power mean of 𝑎 and 𝑏.

In [16], the authors found the greatest value 𝛼 and least value 𝛽 such that the double inequality 𝛼𝐶(𝑎,𝑏)+(1−𝛼)𝐺(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝛽𝐶(𝑎,𝑏)+(1−𝛽)𝐺(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝐶(𝑎,𝑏)=(𝑎²+𝑏²)/(𝑎+𝑏) is the contraharmonic mean of 𝑎 and 𝑏.

Motivated by the first inequality in (4), Gao [17] gave the best possible constants 𝜆 and 𝜇 such that the double inequality 𝜆𝐼(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝜇𝐼(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [18], the author solved the following open problem proposed by Long and Chu [19]: what is the smallest 𝑝 (largest 𝑞) such that the inequality 𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝐺(𝑎,𝑏)<𝐿_𝑝(𝑎,𝑏)(>𝐿_𝑞(𝑎,𝑏)) holds for 𝛼∈(0,1/2)(𝛼∈(1/2,1)) and all 𝑎,𝑏>0 with 𝑎≠𝑏?

Chu et al. [20] proved that the double inequality 𝐿_𝑝(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝐿_𝑞(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝑝≤3 and 𝑞≥𝑞₀=3.152…, where 𝑇(𝑎,𝑏)=(𝑎−𝑏)/[2arctan((𝑎−𝑏)/(𝑎+𝑏))] is the second Seiffert mean of 𝑎 and 𝑏, and 𝑝₀=3.152… is the unique solution of the equation (𝑥+1)^−1/𝑥=2/𝜋.

In [21], the authors answered the question: what are the greatest value 𝑝₁=𝑝₁(𝑞) and the least value 𝑝₂=𝑝₂(𝑞) such that the double inequality 𝐿_𝑝1(𝑎,𝑏)<[𝐿(𝑎^𝑞,𝑏^𝑞)]^1/𝑞<𝐿_𝑝2(𝑎,𝑏) holds for any 𝑞>0 with 𝑞≠1 and all 𝑎,𝑏>0 with 𝑎≠𝑏?

Motivated by the first inequality in (4), it is natural to ask what are the best possible generalized logarithmic mean bounds for the Seiffert mean 𝑃(𝑎,𝑏)? It is the aim of this paper to answer this question.

2. Preliminaries

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

Lemma 1. Let the function 𝐸∶(−1,+∞)→ℝ₊ be defined with (𝑥+1)^−1/𝑥()∪(),1,𝑥∈−1,00,+∞𝑒,𝑥=0. (6) Then, 𝐸 is a continuous and strictly increasing function.

Proof. From (6), we clearly see that lim_𝑥→0𝐸(𝑥)=1𝑒=𝐸(0),lim_𝑥→−1+𝐸(𝑥)=0,lim_𝑥→+∞𝐸(𝑥)=1. (7)(8) If 𝑥∈(−1,0)∪(0,+∞), then simple computation yields 𝐸^′(𝑥)=(1+𝑥)log(1+𝑥)−𝑥𝑥²()𝐸(𝑥).1+𝑥 (9) If we define 𝐹(𝑥)=(1+𝑥)log(1+𝑥)−𝑥, (10) then 𝐹^′(𝑥)=log(1+𝑥). (11) Equation (11) implies that 𝐹^′(𝑥)(),𝐹<0,𝑥∈−1,0_′(𝑥)>0,𝑥∈(0,+∞). (12) Equations (10) and (12) lead to 𝐹(𝑥)>lim_𝑥→0𝐹(𝑥)=0 (13) for 𝑥∈(−1,0)∪(0,+∞).
From (9) and (10) together with (13), we clearly see that 𝐸^′(𝑥)>0 (14) for 𝑥∈(−1,0)∪(0,+∞).
Therefore, the continuity of 𝐸 follows from (6) and (7), and the strict monotonicity of 𝐸 follows from (8), (14) and the continuity of 𝐸.

Remark 2. From Lemma 1, we clearly see that for any fixed 𝜆∈(0,1), there exists a unique 𝑥∈(−1,∞) such 𝐸(𝑥)=𝜆. In particular, for 𝜆=1/𝜋, making use of Mathematica software, we get 𝐸()−1−0.241𝜋𝐸()−1=0.000166⋯>0,−0.242𝜋=−0.000062⋯<0. (15)
Therefore, the unique solution of the equation 𝐸(𝑥)=1/𝜋 belongs to the interval (−0.242,−0.241).

Lemma 3. Let 𝑝∈(−0.242,−0.241) and let 𝑔∶[1,+∞)→ℝ be defined with 𝑔(𝑡)=𝑝𝑡^2𝑝+3+ 2(𝑝−1)𝑡^2𝑝+2−(3𝑝+2)𝑡^2𝑝+1−𝑝(2𝑝+1)𝑡^𝑝+3+(2𝑝²+𝑝+4)𝑡^𝑝+2+(2𝑝²+𝑝+4)𝑡^𝑝+1−𝑝(2𝑝+1)𝑡^𝑝− (3𝑝+2)𝑡²+ 2(𝑝−1)𝑡+𝑝. Then, there exists 𝜆∈(1,+∞) such that 𝑔(𝑡)<0 for 𝑡∈(1,𝜆) and 𝑔(𝑡)>0 for 𝑡∈(𝜆,+∞).

Proof. Simple computations lead to 𝑔(1)=0,lim_𝑡→+∞𝑔(𝑡)𝑔=+∞,_′(𝑡)=𝑝2𝑝+3𝑡^2𝑝+2+4𝑝−1𝑝+1𝑡^2𝑝+1−2𝑝+13𝑝+2𝑡^2𝑝−𝑝𝑝+32𝑝+1𝑡^𝑝+2+𝑝+22𝑝²+𝑝+4𝑡^𝑝+1+×𝑝+12𝑝²+𝑝+4𝑡^𝑝−𝑝²2𝑝+1𝑡^𝑝−1,𝑔−23𝑝+2𝑡+2𝑝−1_′(1)=0,lim_𝑡→+∞𝑔^′(𝑡)𝑔=+∞,_′′(𝑡)=2𝑝𝑝+12𝑝+3𝑡^2𝑝+1+4𝑝−1𝑝+12𝑝+1𝑡^2𝑝×𝑡−2𝑝2𝑝+13𝑝+2^2𝑝−1−𝑝𝑝+2𝑝+32𝑝+1𝑡^𝑝+1+𝑝+1𝑝+22𝑝²+𝑝+4𝑡^𝑝+𝑝𝑝+12𝑝²+𝑝+4𝑡^𝑝−1−𝑝²𝑝−12𝑝+1𝑡^𝑝−2,𝑔−23𝑝+2_′′(1)=0,lim_𝑡→+∞𝑔^′′(𝑡)=+∞. (16)(17)(18)(19)(20)(21)(22)(23)
Let ℎ(𝑡)=𝑡^3−𝑝𝑔^′′′(𝑡). Further computations lead to ℎ(𝑡)=2𝑝𝑝+12𝑝+12𝑝+3𝑡^𝑝+3+8𝑝𝑝−1𝑝+12𝑝+1𝑡^𝑝+2−2𝑝2𝑝−12𝑝+13𝑝+2𝑡^𝑝+1×𝑡−𝑝𝑝+1𝑝+2𝑝+32𝑝+1³×+𝑝𝑝+1𝑝+22𝑝²+𝑝+4𝑡²×+𝑝𝑝−1𝑝+12𝑝²+𝑝+4𝑡−𝑝²,ℎ(1)𝑝−2𝑝−12𝑝+1=0,lim_𝑡→+∞ℎ(𝑡)ℎ=+∞,_′(𝑡)=2𝑝𝑝+1𝑝+32𝑝+12𝑝+3𝑡^𝑝+2+8𝑝𝑝−1𝑝+1𝑝+22𝑝+1𝑡^𝑝+1−2𝑝𝑝+12𝑝−12𝑝+13𝑝+2𝑡^𝑝−3𝑝𝑝+1𝑝+2𝑝+32𝑝+1𝑡²+2𝑝𝑝+1𝑝+22𝑝²×+𝑝+4𝑡+𝑝𝑝−1𝑝+12𝑝²,ℎ+𝑝+4_′(1)=24𝑝³𝑝+1<0,lim_𝑡→+∞ℎ^′(𝑡)ℎ=+∞,_′′(𝑡)=2𝑝𝑝+1𝑝+2𝑝+32𝑝+12𝑝+3𝑡^𝑝+1+8𝑝𝑝−1𝑝+1²𝑝+22𝑝+1𝑡^𝑝−2𝑝²𝑝+12𝑝−12𝑝+13𝑝+2𝑡^𝑝−1−6𝑝𝑝+1𝑝+2𝑝+32𝑝+1𝑡+2𝑝𝑝+1𝑝+22𝑝²,ℎ+𝑝+4_′′(1)=𝑝³72𝑝²+156𝑝+84<0,lim_𝑡→+∞ℎ^′′(𝑡)ℎ=+∞,_′′′(𝑡)=2𝑝𝑝+1²×𝑡𝑝+2𝑝+32𝑝+12𝑝+3^𝑝+8𝑝²𝑝−1𝑝+1²×𝑝+22𝑝+1𝑡^𝑝−1−2𝑝²××𝑡𝑝−1𝑝+12𝑝−12𝑝+13𝑝+2^𝑝−2×,ℎ−6𝑝𝑝+1𝑝+2𝑝+32𝑝+1_′′′(1)=𝑝²112𝑝⁴+312𝑝³+352𝑝²ℎ+192𝑝+40>0,₍₄₎(𝑡)=2𝑝²𝑝+1²×𝑡𝑝+2𝑝+32𝑝+12𝑝+3^𝑝−1+8𝑝²𝑝−1²𝑝+1²𝑝+22𝑝+1𝑡^𝑝−2+2𝑝²×𝑡1−𝑝2−𝑝1−2𝑝𝑝+12𝑝+13𝑝+2^𝑝−3>0. (24)(25)(26)(27)(28)(29)(30)(31)(32)(33)(34)(35)
Inequalities (34) and (35) imply that ℎ^′′′(𝑡)>0 (36) for 𝑡∈(1,∞).
From (31) and (32) together with (36), we clearly see that there exists 𝑡₀∈(1,∞) such that ℎ^′′(𝑡)<0 for 𝑡∈(1,𝑡₀) and ℎ^′′(𝑡)>0 for 𝑡∈(𝑡₀,∞). Hence, ℎ^′ is strictly decreasing on (1,𝑡₀) and strictly increasing on (𝑡₀,∞).
It follows from (28) and (29) together with the monotonicity of ℎ^′ that there exists 𝑡₁∈(1,∞) such that ℎ is strictly decreasing on (1,𝑡₁) and strictly increasing on (𝑡₁,∞).
Equations (25)-(26) and the monotonicity of ℎ lead to the conclusion that there exists 𝑡₂∈(1,∞) such that ℎ(𝑡)<0 for 𝑡∈(1,𝑡₂) and ℎ(𝑡)>0 for 𝑡∈(𝑡₂,∞). Therefore, 𝑔^′′ is strictly decreasing on (1,𝑡₂) and strictly increasing on (𝑡₂,∞).
From (22) and (23) together with the monotonicity of 𝑔^′′, we clearly see that there exists 𝑡₃∈(1,∞) such that 𝑔^′ is strictly decreasing on (1,𝑡₃) and strictly increasing on (𝑡₃,∞).
Equations (19)-(20) and the monotonicity of 𝑔^′ imply that there exists 𝑡₄∈(1,∞) such that 𝑔 is strictly decreasing on (1,𝑡₄) and strictly increasing on (𝑡₄,∞).
Therefore, Lemma 3 follows from (16) and (17) together with the monotonicity of 𝑔.

3. Main Results

Theorem 4. Let 𝐸 be as in Lemma 1 and let 𝑝 be the unique solution of the equation 𝐸(𝑥)=1/𝜋. Then, for all 𝑎,𝑏>0, 𝑎≠𝑏, the inequality 𝐿_𝑝(𝑎,𝑏)<𝑃(𝑎,𝑏) (37) holds, and 𝐿_𝑝(𝑎,𝑏) is the best possible lower generalized logarithmic mean bound for the Seiffert mean 𝑃(𝑎,𝑏).

Proof. From Remark 2, it follows that 𝑝∈(−0.242,−0.241). (38)
We first prove that the inequality (37) holds. Without loss of generality, we assume that 𝑎>𝑏. If 𝑡=𝑎/𝑏>1, then, from (1) and (2), we have log𝐿_𝑝()()=1𝑎,𝑏−log𝑃𝑎,𝑏𝑝𝑡log^𝑝+1−1()√𝑝+1𝑡−1+log4arctan𝑡−𝜋−log(𝑡−1), (39) If 𝑓(𝑡)=1𝑝𝑡log^𝑝+1−1()√𝑝+1𝑡−1+log4arctan𝑡−𝜋−log(𝑡−1), (40) then simple computations lead to lim_𝑡→1𝑓(𝑡)=lim_𝑡→+∞𝑓(𝑡)𝑓=0,_′(𝑡)=𝑝+1𝑡^𝑝−1𝑝𝑡^𝑝+1−1(𝑡−1)4arctan√𝑓𝑡−𝜋¹(𝑡), (41)(42) where 𝑓₁(𝑡)=−4arctan√𝑡𝑡+𝜋+2𝑝^𝑝+1−1(𝑡−1)𝑝+1√𝑡(𝑡+1)(𝑡^𝑝),−1lim_𝑡→1𝑓₁(𝑡)=0,lim_𝑡→+∞𝑓₁(𝑡)𝑓=+∞,_′1(𝑡)=𝑓₂(𝑡)1+𝑝(𝑡+1)²(𝑡^𝑝−1)²𝑡^3/2, (43)(44) where 𝑓₂(𝑡)=𝑝𝑡^2𝑝+3+2𝑝−1𝑡^2𝑝+2−3𝑝+2𝑡^2𝑝+1−𝑝2𝑝+1𝑡^𝑝+3+2𝑝²+𝑝+4𝑡^𝑝+2+2𝑝²+𝑝+4𝑡^𝑝+1−𝑝2𝑝+1𝑡^𝑝−3𝑝+2𝑡²+2𝑝−1𝑡+𝑝. (45)
From (38) and (44)-(45) together with Lemma 3, we clearly see that there exists 𝜆∈(1,+∞) such that 𝑓₁ is strictly decreasing on (1,𝜆) and strictly increasing on (𝜆,∞). Then, (38) and (42)-(43) together with the monotonicity of 𝑓₁ imply that there exists 𝜇∈(1,+∞) such that 𝑓 is strictly decreasing on (1,𝜇) and strictly increasing on (𝜇,+∞).
Therefore, 𝐿_𝑝(𝑎,𝑏)<𝑃(𝑎,𝑏) follows from (39)–(41) and the monotonicity of 𝑓.
Next, we prove that 𝐿_𝑝(𝑎,𝑏) is the best possible lower generalized logarithmic mean bound for the Seiffert mean 𝑃(𝑎,𝑏).
For any 0<𝜀<−𝑝, from (1) and (2), we get lim_𝑡→+∞𝐿_𝑝+𝜀(𝑡,1)𝑃(𝑡,1)=𝜋𝑝+1+𝜀^{−1/(𝑝+𝜀)}. (46) Lemma 1 and (46) lead to lim_𝑡→+∞𝐿_𝑝+𝜀(𝑡,1)𝑃(𝑡,1)>1. (47)
Inequality (47) implies that, for 0<𝜀<−𝑝, there exists 𝑇=𝑇(𝜀)>1 such that 𝐿_𝑝+𝜀(𝑡,1)>𝑃(𝑡,1) for 𝑡∈(𝑇,∞).

Theorem 5. 𝐿₀(𝑎,𝑏) is the best possible upper generalized logarithmic mean bound for the Seiffert mean 𝑃(𝑎,𝑏).

Proof. For any 0<𝜀<1 and 𝑥>0, from (1) and (2), we have [𝑃(1+𝑥,1)]^𝜀−𝐿_−𝜀(1+𝑥,1)^𝜀=𝑓(𝑥)(1+𝑥)^1−𝜀−14arctan√1+𝑥−𝜋^𝜀, (48) where 𝑓(𝑥)=[(1+𝑥)^1−𝜀−1]𝑥^𝜀−(1−𝜀)(4arctan√1+𝑥−𝜋)^𝜀𝑥.
When 𝑥→0, then making use of the Taylor expansion, we get 𝑓(𝑥)=)1(1−𝜀𝑥−2𝜀(1−𝜀)𝑥²+16𝜀(1−𝜀)(1+𝜀)𝑥³+𝑜𝑥³𝑥^𝜀−(11−𝜀)𝑥−2𝑥²+7𝑥24³+𝑜𝑥^3𝜀𝑥=1𝜀24²(1−𝜀)𝑥^3+𝜀+𝑜𝑥^3+𝜀. (49)
Equations (48) and (49) imply that for any 0<𝜀<1, there exists 𝛿=𝛿(𝜀)>0 such that 𝑃(1+𝑥,1)>𝐿_−𝜀(1+𝑥,1) (50) for 𝑥∈(0,𝛿).
Therefore, Theorem 5 follows from inequalities (4) and (50).

Acknowledgments

This paper was supported by the Natural Science Foundation of China under Grant 61173123, the Natural Science Foundation of Zhejiang Province under Grants Z1110551 and LY12F02012, and the Natural Science Foundation of Huzhou City under Grant 2012YZ06.