Bounds for Combinations of Toader Mean and Arithmetic Mean in Terms of Centroidal Mean

Abstract

The authors find the greatest value 𝜆 and the least value 𝜇, such that the double inequality 𝐶(𝜆𝑎+(1−𝜆𝑏),𝜆𝑏+(1−𝜆)𝑎)<𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝑇(𝑎,𝑏)<𝐶(𝜇𝑎+(1−𝜇)𝑏,𝜇𝑏+(1−𝜇)𝑎) holds for all 𝛼∈(0,1) and 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝐶(𝑎,𝑏)=2(𝑎²+𝑎𝑏+𝑏²)/3(𝑎+𝑏), 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, and 𝑇(𝑎,𝑏)=(2/𝜋)∫₀^𝜋/2𝑎²cos²𝜃+𝑏²sin²𝜃𝑑𝜃 denote, respectively, the centroidal, arithmetic, and Toader means of the two positive numbers 𝑎 and 𝑏.

1. Introduction

In [1], Toader introduced a mean 𝑇()=2𝑎,𝑏𝜋₀^𝜋/2𝑎²cos²𝜃+𝑏²sin²=𝜃𝑑𝜃2𝑎ℰ1−(𝑏/𝑎)²𝜋,𝑎>𝑏,2𝑏ℰ1−(𝑎/𝑏)²𝜋,𝑎<𝑏,𝑎,𝑎=𝑏, (1) where ℰ=ℰ(𝑟)=₀^𝜋/21−𝑟²sin²𝜃^1/2𝑑𝜃, (2) for 𝑟∈[0,1] is the complete elliptic integral of the second kind.

In recent years, there have been plenty of literature, such as [2–6], dedicated to the Toader mean.

For 𝑝∈ℝ and 𝑎,𝑏>0, the centroidal mean 𝐶(𝑎,𝑏) and 𝑝th power mean 𝑀_𝑝(𝑎,𝑏) are, respectively, defined by 𝐶()=2𝑎𝑎,𝑏²+𝑎𝑏+𝑏²3(),𝑀𝑎+𝑏_𝑝()=𝑎𝑎,𝑏^𝑝+𝑎^𝑝2^1/𝑝√,𝑝≠0,𝑎𝑏,𝑝=0. (3)

In [7], Vuorinen conjectured that 𝑀_3/2(𝑎,𝑏)<𝑇(𝑎,𝑏), (4) for all 𝑎,𝑏>0 with 𝑎≠𝑏. This conjecture was verified by Qiu and Shen [8] and by Barnard et al. [9], respectively.

In [10], Alzer and Qiu presented a best possible upper power mean bound for the Toader mean as follows: 𝑇(𝑎,𝑏)<𝑀_{log2/log(𝜋/2)}(𝑎,𝑏), (5) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Chu et al. [5] proved that the double inequality 𝐶(()()𝑎)()𝑎𝛼𝑎+1−𝛼𝑏,𝛼𝑏+1−𝛼<𝑇𝑎,𝑏<𝐶𝛽𝑎+1−𝛽𝑏,𝛽𝑏+1−𝛽 (6) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤3/4 and 𝛽≥1/2+√4𝜋−𝜋²/(2𝜋).

Very recently, Hua and Qi [11] proved that the double inequality 𝛼𝐶()+()𝐴()()𝑎,𝑏1−𝛼𝑎,𝑏<𝑇𝑎,𝑏<𝛽𝐶(𝑎,𝑏)+1−𝛽𝐴(𝑎,𝑏) (7) is valid for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤3/4 and 𝛽≥(12/𝜋)−3. Where 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2 denote the arithmetic mean.

For positive numbers 𝑎,𝑏>0 with 𝑎≠𝑏, let 𝐽(𝑥)=𝐶(𝑥𝑎+(1−𝑥)𝑏,𝑥𝑏+(1−𝑥)𝑎) (8) be on [1/2,1]. It is not difficult to directly verify that 𝐽(𝑥) is continuous and strictly increasing on [1/2,1].

The main purpose of the paper is to find the greatest value 𝜆 and the least value 𝜇, such that the double inequality 𝐶(𝜆𝑎+(1−𝜆𝑏),𝜆𝑏+(1−𝜆)𝑎)<𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝑇(𝑎,𝑏)<𝐶(𝜇𝑎+(1−𝜇)𝑏,𝜇𝑏+(1−𝜇)𝑎) holds for all 𝛼∈(0,1) and 𝑎,𝑏>0 with 𝑎≠𝑏. As applications, we also present new bounds for the complete elliptic integral of the second kind.

2. Preliminaries and Lemmas

In order to establish our main result, we need several formulas and Lemmas below.

For 0<𝑟<1 and 𝑟^′=√1−𝑟², Legendre’s complete elliptic integrals of the first and second kinds are defined in [12, 13] by 𝒦=𝒦(𝑟)=₀^𝜋/21−𝑟²sin²𝜃^−1/2𝒦𝑑𝜃,_′=𝒦^′(𝑟)=𝒦𝑟^′,𝒦(0)=𝜋2(1)(𝑟)=,𝒦=∞,ℰ=ℰ₀^𝜋/21−𝑟²sin²𝜃^1/2ℰ𝑑𝜃,_′=ℰ^′(𝑟)=ℰ𝑟^′,ℰ(0)=𝜋2,ℰ(1)=1, (9) respectively.

For 0<𝑟<1, the formulas 𝑑𝒦=𝑑𝑟ℰ−𝑟^′2𝒦𝑟𝑟^′2,𝑑ℰ=𝑑𝑟ℰ−𝒦𝑟,𝑑ℰ−𝑟^′2𝒦𝑑()𝑑𝑟=𝑟𝒦,𝒦−ℰ=𝑑𝑟𝑟ℰ𝑟^′2,ℰ2√𝑟=1+𝑟2ℰ−𝑟^′2𝒦1+𝑟 (10) were presented in [14, Appendix E, pages 474-475].

Lemma 1 (see [14, Theorem 3.21(1), 3.43 exercises 13(a)]). The function  (ℰ−𝑟^′2𝒦)/𝑟² is strictly increasing from (0,1) to (𝜋/4,1), and the function 2ℰ−𝑟^′2𝒦 is increasing from (0,1) to (𝜋/2,2).

Lemma 2. Let 𝑢,𝛼∈(0,1) and 𝑓_𝑢,𝛼(𝑟)=13𝑢𝑟²−()21−𝛼𝜋2ℰ(𝑟)−1−𝑟²𝒦(𝑟)−1. (11) Then, 𝑓_𝑢,𝛼>0, for all 𝑟∈(0,1) if and only if 𝑢≥3(1−𝛼)(4/𝜋−1), and 𝑓_𝑢,𝛼<0, for all 𝑟∈(0,1) if and only if 𝑢≤3(1−𝛼)/4.

Proof. From (11), one has 𝑓_𝑢,𝛼0⁺𝑓=0,_𝑢,𝛼(1⁻)=13()4𝑢−1−𝛼𝜋,𝑓−1_′𝑢,𝛼(𝑟)=23𝑟𝑢−3(1−𝛼)𝑔(𝑟), (12)(13)(14) where 𝑔(𝑟)=(1/𝜋)((ℰ−𝑟^′2𝒦)/𝑟²).
We divide the proof into four cases.
Case 1 (𝑢≥3(1−𝛼)/𝜋).  From (14) and Lemma 1 together with the monotonicity of 𝑔(𝑟), we clearly see that 𝑓_𝑢,𝛼(𝑟) is strictly increasing on (0,1). Therefore, 𝑓_𝑢,𝛼(𝑟)>0, for all 𝑟∈(0,1).
Case 2 (𝑢≤3(1−𝛼)/4). From (14) and Lemma 1 together with the monotonicity of 𝑔(𝑟), we obtain that 𝑓_𝑢,𝛼(𝑟) is strictly decreasing on (0,1). Therefore, 𝑓_𝑢,𝛼(𝑟)<0, for all 𝑟∈(0,1).
Case 3 (3(1−𝛼)/4<𝑢≤3(1−𝛼)(4/𝜋−1)). From (13) and (14) together with the monotonicity of 𝑔(𝑟), we see that there exists 𝜆∈(0,1), such that 𝑓_𝑢,𝛼(𝑟) is strictly increasing in (0,𝜆] and strictly decreasing in [𝜆,1) and 𝑓_𝑢,𝛼(1⁻)≤0. (15) Therefore, making use of (12) and inequality (15) together with the piecewise monotonicity of 𝑓_𝑢,𝛼(𝑟) leads to the conclusion that there exists 0<𝜆<𝜂<1, such that 𝑓_𝑢,𝛼(𝑟)>0 for 𝑟∈(0,𝜂) and 𝑓_𝑢,𝛼(𝑟)<0 for 𝑟∈(𝜂,1).
Case 4 (3(1−𝛼)(4/𝜋−1)≤𝑢<3(1−𝛼)/𝜋). Equation (13) leads to 𝑓_𝑢,𝛼(1⁻)≥0. (16)
From (13) and (14) together with the monotonicity of 𝑔(𝑟), we clearly see that there exists 𝜆∈(0,1), such that 𝑓_𝑢,𝛼(𝑟) is strictly increasing in (0,𝜆] and strictly decreasing in [𝜆,1). Therefore, 𝑓_𝑢,𝛼(𝑟)>0 for 𝑟∈(0,1) follows from (12) and (16) together with the piecewise monotonicity of 𝑓_𝑢,𝛼(𝑟).

3. Main Results

Now, we are in a position to state and prove our main results.

Theorem 3. If 𝛼∈(0,1) and 𝜆,𝜇∈(1/2,1), then the double inequality 𝐶(()()𝑎)()+()𝑇()<𝜆𝑎+1−𝜆𝑏,𝜆𝑏+1−𝜆<𝛼𝐴𝑎,𝑏1−𝛼𝑎,𝑏𝐶𝜇𝑎+1−𝜇𝑏,𝜇𝑏+1−𝜇𝑎 (17) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 1𝜆≤2+3(1−𝛼)4,1𝜇≥21+3()41−𝛼𝜋−1. (18)

Proof. Since 𝐴(𝑎,𝑏), 𝑇(𝑎,𝑏), and 𝐶(𝑎,𝑏) are symmetric and homogeneous of degree one, without loss of generality, we assume that 𝑎>𝑏. Let 𝑝∈(1/2,1), 𝑡=𝑏/𝑎∈(0,1), and 𝑟=(1−𝑡)/(1+𝑡). Then, 𝐶𝑎()−()𝑇()2𝑝𝑎+1−𝑝𝑏,𝑝𝑏+1−𝑝−𝛼𝐴𝑎,𝑏1−𝛼𝑎,𝑏=𝑎3𝑏𝑝+1−𝑝𝑎²+𝑏𝑝+1−𝑝𝑎𝑝𝑏𝑎+𝑝𝑏+1−𝑝𝑎+1−𝑝²𝑏1+𝑎⁻¹()−𝛼𝑎1+𝑏/𝑎2−(1−𝛼)2𝑎𝜋ℰ𝑏1−𝑎²2=𝑎3𝑝+1−𝑝𝑡²+𝑡+𝑝+1−𝑝𝑝𝑡+1−𝑝𝑝𝑡+1−𝑝²(1+𝑡)⁻¹−𝛼1+𝑡2−()21−𝛼𝜋ℰ√1−𝑡²=𝑎1−2𝑝²𝑟²+33()11+𝑟−𝛼−()21+𝑟1−𝛼𝜋2ℰ−𝑟^′2𝒦=𝑎1+𝑟11+𝑟31−2𝑝²𝑟²−()2+1−𝛼1−𝛼𝜋2ℰ−𝑟^′2𝒦. (19) Therefore, Theorem 3 follows easily from Lemma 2 and (19).

Let 𝛼=1/4,  𝜆=7/8,  𝜇=(1/2)(1+(34/𝜋−1/2)). Then, from Theorem 3, we get new bounds for the complete elliptic integral ℰ(𝑟) of the second kind in terms of elementary functions as follows.

Corollary 4. For 𝑟∈(0,1) and 𝑟^′=√1−𝑟², one has 𝜋25+6𝑟^′+5𝑟^′281+𝑟^′(𝑟)𝑟<ℰ<𝜋^′+(2/𝜋)1−𝑟^′21+𝑟^′. (20)

4. Remarks

Remark 5. In the recent past, the complete elliptic integrals have attracted the attention of numerous mathematicians. In [4], it was established that 𝜋2121+𝑟^′22+1+𝑟^′4(𝑟)<𝜋<ℰ24−𝜋√2−1𝜋1+𝑟^′22+√2𝜋−41+𝑟^′2√𝜋,2−1 (21) for all 𝑟∈(0,1).
Guo and Qi [15] proved that 𝜋2−12()log1+𝑟^1−𝑟(1−𝑟)^1+𝑟<ℰ(𝑟)<𝜋−12+1−𝑟²4𝑟log1+𝑟,1−𝑟 (22) for all 𝑟∈(0,1).
Yin and Qi [16] presented that 𝜋26+2√1−𝑟²−3𝑟²2√2(𝑟)≤𝜋≤ℰ210−2√1−𝑟²−5𝑟²2√2, (23) for all 𝑟∈(0,1).
It was pointed out in [4] that the bounds in (21) for ℰ(𝑟) are better than the bounds in (22) for some 𝑟∈(0,1).

Remark 6. The lower bound in (20) for ℰ(𝑟) is better than the lower bound in (21). Indeed, 5+6𝑥+5𝑥²8()−11+𝑥21+𝑥²2+1+𝑥4=3𝑥²+2𝑥+3−221+𝑥²(1+𝑥)8(),1+𝑥3𝑥²+2𝑥+3²−22(1+𝑥²)(1+𝑥)²=(1−𝑥)⁴>0, (24) for all 𝑥∈(0,1).

Remark 7. The following equivalence relations for 𝑥∈(0,1) show that the lower bound in (20) for ℰ(𝑟) is better than the lower bound in (23): 5+6𝑥+5𝑥²8()>1+𝑥6+2𝑥−31−𝑥²2√2⟺5𝑥²+6𝑥+5²>8(𝑥+1)²3𝑥²⟺()+2𝑥+3𝑥−1⁴>0. (25)

Acknowledgments

The author is thankful to the anonymous referees for their valuable and profound comments on and suggestions to the original version of this paper. This work was supported by the project of Shandong Higher Education Science and Technology Program under Grant no. J11LA57.