Optimal Bounds for Neuman Means in Terms of Harmonic and Contraharmonic Means

Abstract

For 𝑎,𝑏>0 with 𝑎≠𝑏, the Schwab-Borchardt mean SB(𝑎,𝑏) is defined as SB(𝑎,𝑏)={𝑏²−𝑎²/cos⁻¹(𝑎/𝑏)if𝑎<𝑏,𝑎²−𝑏²/cosh⁻¹(𝑎/𝑏)if𝑎>𝑏. In this paper, we find the greatest values of 𝛼₁ and 𝛼₂ and the least values of 𝛽₁ and 𝛽₂ in [0,1/2] such that 𝐻(𝛼₁𝑎+(1−𝛼₁)𝑏,𝛼₁𝑏+(1−𝛼₁)𝑎)<𝑆_𝐴𝐻(𝑎,𝑏)<𝐻(𝛽₁𝑎+(1−𝛽₁)𝑏,𝛽₁𝑏+(1−𝛽₁)𝑎) and 𝐻(𝛼₂𝑎+(1−𝛼₂)𝑏,𝛼₂𝑏+(1−𝛼₂)𝑎)<𝑆_𝐻𝐴(𝑎,𝑏)<𝐻(𝛽₂𝑎+(1−𝛽₂)𝑏,𝛽₂𝑏+(1−𝛽₂)𝑎). Similarly, we also find the greatest values of 𝛼₃ and 𝛼₄ and the least values of 𝛽₃ and 𝛽₄ in [1/2,1] such that 𝐶(𝛼₃𝑎+(1−𝛼₃)𝑏,𝛼₃𝑏+(1−𝛼₃)𝑎)<𝑆_𝐶𝐴(𝑎,𝑏)<𝐶(𝛽₃𝑎+(1−𝛽₃)𝑏,𝛽₃𝑏+(1−𝛽₃)𝑎) and 𝐶(𝛼₄𝑎+(1−𝛼₄)𝑏,𝛼₄𝑏+(1−𝛼₄)𝑎)<𝑆_𝐴𝐶(𝑎,𝑏)<𝐶(𝛽₄𝑎+(1−𝛽₄)𝑏,𝛽₄𝑏+(1−𝛽₄)𝑎). Here, 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, and 𝐶(𝑎,𝑏)=(𝑎²+𝑏²)/(𝑎+𝑏) are the harmonic, arithmetic, and contraharmonic means, respectively, and 𝑆_𝐻𝐴(𝑎,𝑏)=SB(𝐻,𝐴), 𝑆_𝐴𝐻(𝑎,𝑏)=SB(𝐴,𝐻), 𝑆_𝐶𝐴(𝑎,𝑏)=SB(𝐶,𝐴), and 𝑆_𝐴𝐶(𝑎,𝑏)=SB(𝐴,𝐶) are four Neuman means derived from the Schwab-Borchardt mean.

1. Introduction

For 𝑎,𝑏>0 with 𝑎≠𝑏, the Schwab-Borchardt mean SB(𝑎,𝑏) is defined as SB(𝑎,𝑏)=𝑏²−𝑎²cos⁻¹(𝑎/𝑏)if𝑎<𝑏,𝑎²−𝑏²cosh⁻¹(𝑎/𝑏)if𝑎>𝑏. (1) It is well known that the mean SB(𝑎,𝑏) is strictly increasing in both 𝑎 and 𝑏, nonsymmetric, and homogeneous of degree 1 in its variables. Several symmetric bivariate means are special cases of the Schwab-Borchardt mean; for example, 𝑃(𝑎,𝑏)=𝑎−𝑏2sin⁻¹)/(()𝑇()=[(𝑎−𝑏𝑎+𝑏)]=SB𝐺,𝐴isthefirstSeiffertmean,𝑎,𝑏𝑎−𝑏2tan⁻¹)/(()𝑀()=[(𝑎−𝑏𝑎+𝑏)]=SB𝐴,𝑄isthesecondSeiffertmean,𝑎,𝑏𝑎−𝑏2sinh⁻¹)/(()́𝐿()=[(𝑎−𝑏𝑎+𝑏)]=SB𝑄,𝐴istheNeuman-Sandormean,𝑎,𝑏𝑎−𝑏2tanh⁻¹)/(()[(𝑎−𝑏𝑎+𝑏)]=SB𝐴,𝐺isthelogarithmicmean, (2) where 𝐺(𝑎,𝑏)=√𝑎𝑏, 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2 and 𝑄(𝑎,𝑏)=(𝑎²+𝑏²)/2 denote the classical geometric mean, arithmetic mean, and quadratic mean, respectively.

The Schwab-Borchardt mean SB(𝑎,𝑏) was firstly investigated in [1–4]. In [3], the authors pointed out that the logarithmic mean, two Seiffert means, and the Neuman-Sándor mean are particular cases of the Schwab-Borchardt mean. Later, SB and its special cases have been the subject of intensive research. In particular, many inequalities for them can be found in the literature [3–13].

Let 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐶(𝑎,𝑏)=(𝑎²+𝑏²)/(𝑎+𝑏) be the harmonic and contraharmonic means of two positive numbers 𝑎 and 𝑏, respectively. Then, it is well known that 𝐻<𝐺<𝐿<𝑃<𝐴<𝑀<𝑇<𝑄<𝐶. (3) for 𝑎,𝑏>0 with 𝑎≠𝑏.

Recently, the second author of this paper reviewed two elegant papers [14, 15] by Neuman and found that the bivariate means 𝑆_𝐴𝐻, 𝑆_𝐻𝐴, 𝑆_𝐶𝐴, and 𝑆_𝐴𝐶, derived from the Schwab-Borchardt mean are very interesting. They are defined as follows: 𝑆_𝐴𝐻=SB(𝐴,𝐻),𝑆_𝐻𝐴(),𝑆=SB𝐻,𝐴_𝐶𝐴=SB(𝐶,𝐴),𝑆_𝐴𝐶=SB(𝐴,𝐶). (4)

We call the means 𝑆_𝐴𝐻, 𝑆_𝐻𝐴, 𝑆_𝐶𝐴, and 𝑆_𝐴𝐶, defined in (4) the Neuman means. Moreover, if we let 𝑣=(𝑎−𝑏)/(𝑎+𝑏)∈(−1,1), then explicit formulas for 𝑆_𝐴𝐻, 𝑆_𝐻𝐴, 𝑆_𝐴𝐶, and 𝑆_𝐶𝐴 are in the following: 𝑆_𝐴𝐻𝑝=𝐴tanh𝑝,𝑆_𝐻𝐴𝑞=𝐴sin𝑞,𝑆_𝐶𝐴(𝑟)=𝐴sinh𝑟,𝑆_𝐴𝐶(𝑠)=𝐴tan𝑠, (5)(6) where 𝑝, 𝑞, 𝑟, and 𝑠 are defined implicitly as sech(𝑝)=1−𝑣², cos(𝑞)=1−𝑣², cosh(𝑟)=1+𝑣² and sec(𝑠)=1+𝑣², respectively. Clearly, 𝑝∈(0,∞), 𝑞∈(0,𝜋/2), 𝑟∈(0,log(2+√3)), and 𝑠∈(0,𝜋/3).

Neuman [14, 15] presented several optimal bounds for 𝑆_𝐻𝐴, 𝑆_𝐴𝐻, 𝑆_𝐶𝐴, and 𝑆_𝐴𝐶. The bounding quantities are arithmetic convex, geometric convex, and harmonic convex combinations of their generating means. Besides, he also proved that 𝐻<𝑆_𝐴𝐻<𝐿<𝑆_𝐻𝐴<𝑃,𝑇<𝑆_𝐶𝐴<𝑄<𝑆_𝐴𝐶<𝐶, (7) for 𝑎,𝑏>0 with 𝑎≠𝑏.

For fixed 𝑎,𝑏>0 with 𝑎≠𝑏, 𝑥∈[0,1/2] and 𝑦∈[1/2,1]. Let 𝑓(𝑥)(()()𝑎),𝑔𝑦𝑎.=𝐻𝑥𝑎+1−𝑥𝑏,𝑥𝑏+1−𝑥=𝐶𝑦𝑎+1−𝑦𝑏,𝑦𝑏+1−𝑦 (8) Then, it is not difficult to verify that 𝑓(𝑥) and 𝑔(𝑦) are continuous and strictly increasing on [0,1/2] and [1/2,1], respectively. Note that 𝑓(0)=𝐻<𝑆_𝐴𝐻<𝑆_𝐻𝐴<𝐴=𝑓(1/2), 𝑔(1/2)=𝐴<𝑆_𝐶𝐴<𝑆_𝐴𝐶<𝐶=𝑔(1). Therefore, it is natural to ask what are the greatest values of 𝛼₁ and 𝛼₂ and the least values of 𝛽₁ and 𝛽₂ in [0,1/2] such that 𝐻(𝛼₁𝑎+(1−𝛼₁)𝑏,𝛼₁𝑏+(1−𝛼₁)𝑎)<𝑆_𝐴𝐻(𝑎,𝑏)<𝐻(𝛽₁𝑎+(1−𝛽₁)𝑏,𝛽₁𝑏+(1−𝛽₁)𝑎) and 𝐻(𝛼₂𝑎+(1−𝛼₂)𝑏,𝛼₂𝑏+(1−𝛼₂)𝑎)<𝑆_𝐻𝐴(𝑎,𝑏)<𝐻(𝛽₂𝑎+(1−𝛽₂)𝑏,𝛽₂𝑏+(1−𝛽₂)𝑎)? And what are the greatest values of 𝛼₃ and 𝛼₄ and the least values of 𝛽₃ and 𝛽₄ in [1/2,1] such that 𝐶(𝛼₃𝑎+(1−𝛼₃)𝑏,𝛼₃𝑏+(1−𝛼₃)𝑎)<𝑆_𝐶𝐴(𝑎,𝑏)<𝐶(𝛽₃𝑎+(1−𝛽₃)𝑏,𝛽₃𝑏+(1−𝛽₃)𝑎) and 𝐶(𝛼₄𝑎+(1−𝛼₄)𝑏,𝛼₄𝑏+(1−𝛼₄)𝑎)<𝑆_𝐴𝐶(𝑎,𝑏)<𝐶(𝛽₄𝑎+(1−𝛽₄)𝑏,𝛽₄𝑏+(1−𝛽₄)𝑎)? The main purpose of this paper is to answer these questions. Our main results are in Theorems 1 and 2.

Theorem 1. Let 𝛼₁,𝛼₂,𝛽₁,𝛽₂∈[0,1/2]. Then, the double inequality 𝐻𝛼₁𝑎+1−𝛼₁𝑏,𝛼₁𝑏+1−𝛼₁𝑎<𝑆_𝐴𝐻<𝐻𝛽₁𝑎+1−𝛽₁𝑏,𝛽₁𝑏+1−𝛽₁𝑎 (9) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₁=0 and 𝛽₁≥[3−√6]/6. Also the double inequality 𝐻𝛼₂𝑎+1−𝛼₂𝑏,𝛼₂𝑏+1−𝛼₂𝑎<𝑆_𝐻𝐴<𝐻𝛽₂𝑎+1−𝛽₂𝑏,𝛽₂𝑏+1−𝛽₂𝑎 (10) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₂≤[1−1−2/𝜋]/2 and 𝛽₂≥[3−√3]/6.

Theorem 2. Let 𝛼₃,𝛼₄,𝛽₃,𝛽₄∈[1/2,1]. Then, the double inequality 𝐶𝛼₃𝑎+1−𝛼₃𝑏,𝛼₃𝑏+1−𝛼₃𝑎<𝑆_𝐶𝐴<𝐶𝛽₃𝑎+1−𝛽₃𝑏,𝛽₃𝑏+1−𝛽₃𝑎 (11) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₃≤[1+√3/log(2+√3)−1]/2 and 𝛽₃≥(3+√3)/6. Also the double inequality 𝐶𝛼₄𝑎+1−𝛼₄𝑏,𝛼₄𝑏+1−𝛼₄𝑎<𝑆_𝐴𝐶<𝐶𝛽₄𝑎+1−𝛽₄𝑏,𝛽₄𝑏+1−𝛽₄𝑎 (12) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₄≤[1+3√3/𝜋−1]/2 and 𝛽₄≥(3+√6)/6.

2. Two Lemmas

In order to prove the desired theorems, we first give two lemmas.

Lemma 1 (see [16, Theorem 1.25]). For −∞<𝑎<𝑏<∞, let 𝑓,𝑔∶[𝑎,𝑏]→ℝ be continuous on [𝑎,𝑏], and be differentiable on (𝑎,𝑏), let 𝑔^′(𝑥)≠0 on (𝑎,𝑏). If 𝑓^′(𝑥)/𝑔^′(𝑥) is increasing (decreasing) on (𝑎,𝑏), then so are 𝑓(𝑥)−𝑓(𝑎)𝑔(𝑥)(𝑎),𝑓(𝑥)(𝑏)−𝑔−𝑓𝑔(𝑥)(𝑏).−𝑔 (13) If 𝑓^′(𝑥)/𝑔^′(𝑥) is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2. (1) The function 𝜑(𝑥)=(𝑥cosh(𝑥)−sinh(𝑥))/[𝑥(cosh(𝑥)−1)] is strictly increasing from (0,∞) onto (2/3,1).
(2) The function 𝜙(𝑥)=(𝑥−sin(𝑥))/[𝑥(1−cos(𝑥))] is strictly increasing from (0,𝜋/2) onto (1/3,(𝜋−2)/𝜋).
(3) The function 𝜉(𝑥)=(sinh(𝑥)−𝑥)/[𝑥(cosh(𝑥)−1)] is strictly decreasing from (0,log(2+√3)) onto ([√3−log(2+√3)]/log(2+√3),1/3).
(4) The function 𝜂(𝑥)=(sin(𝑥)−𝑥cos(𝑥))/[𝑥(1−cos(𝑥))] is strictly decreasing from (0,𝜋/3) onto ((3√3−𝜋)/𝜋,2/3).

Proof. From part (1), let 𝜑₁(𝑥)=𝑥cosh(𝑥)−sinh(𝑥) and 𝜑₂(𝑥)=𝑥(cosh(𝑥)−1). Then, 𝜑(𝑥)=𝜑₁(𝑥)/𝜑₂(𝑥), 𝜑₁(0)=𝜑₂(0)=0, and 𝜑^′₁(𝑥)𝜑^′₂(𝑥)=𝑥sinh(𝑥)(𝑥)(𝑥)=1cosh−1+𝑥sinh((𝑥))/((𝑥=11+cosh−1𝑥sinh))()()/().1+1/2tanh𝑥/2𝑥/2 (14) It is well known that 𝑥→tanh(𝑥)/𝑥 is strictly decreasing on (0,∞). Then, Lemma 1 and (14) lead to the conclusion that 𝜑(𝑥) is strictly increasing on (0,∞). Moreover, by l’Hôptial’s rule we have 𝜑(0⁺)=2/3 and lim_𝑥→+∞𝜑(𝑥)=1.
From part (2), similarly let 𝜙₁(𝑥)=𝑥−sin(𝑥) and 𝜙₂(𝑥)=𝑥(1−cos(𝑥)). Then 𝜙(𝑥)=𝜙₁(𝑥)/𝜙₂(𝑥), 𝜙₁(0)=𝜙₂(0)=0 and 𝜙^′₁(𝑥)𝜙^′₂(𝑥)=1−cos(𝑥)(𝑥)(𝑥)=11−cos+𝑥sin(𝑥)/((𝑥=11+𝑥sin1−cos))()().1+2𝑥/2/tan𝑥/2 (15) It is well known that 𝑥→tan(𝑥)/𝑥 is strictly increasing on (0,𝜋/2). Then, by Lemma 1 and (15) we know that 𝜙(𝑥) is strictly increasing on (0,𝜋/2). Clearly, 𝜙(𝜋/2)=(𝜋−2)/𝜋, while by l’Hôptial’s rule we have 𝜙(0⁺)=1/3.
Parts (3) and (4) have been proven in [14, Theorem 3].

3. Proofs of Theorems 1 and 2

Proof of Theorem 1. Without loss of generality, we assume that 𝑎>𝑏. Let 𝑣=(𝑎−𝑏)/(𝑎+𝑏)∈(0,1) and 𝜆∈[0,1/2]; then, 𝐻(𝜆𝑎+(1−𝜆)𝑏,𝜆𝑏+(1−𝜆)𝑎)−𝑆_𝐴𝐻=𝐴1−(1−2𝜆)²𝑣²𝑝−𝐴tanh𝑝=𝐴1−(1−2𝜆)²𝑝−𝑝1−sechtanh𝑝𝑝𝑝𝑝=𝐴1−sech𝑝cosh−sinh𝑝𝑝−()cosh−11−2𝜆² (16) provided that sech(𝑝)=1−𝑣²(𝑝>0). Thus, inequality (9) follows from (16) and Lemma 2(1). Similarly, 𝐻(𝜆𝑎+(1−𝜆)𝑏,𝜆𝑏+(1−𝜆)𝑎)−𝑆_𝐻𝐴=𝐴1−(1−2𝜆)²𝑣²𝑞−𝐴sin𝑞=𝐴1−(1−2𝜆)²𝑞−𝑞1−cossin𝑞𝑞𝑞=𝐴1−cos𝑞−sin𝑞𝑞−()1−cos1−2𝜆² (17) provided that cos(𝑞)=1−𝑣²(𝑞∈(0,𝜋/2)). Thus, inequality (10) follows from (17) and Lemma 2(2).

Proof of Theorem 2. Without loss of generality, we assume that 𝑎>𝑏. Let 𝑣=(𝑎−𝑏)/(𝑎+𝑏)∈(0,1) and 𝜇∈[1/2,1], then 𝐶𝜇𝑎+1−𝜇𝑏,𝜇𝑏+1−𝜇𝑎−𝑆_𝐶𝐴=𝐴1+1−2𝜇²𝑣²(𝑟)−𝐴sinh𝑟=𝐴1+1−2𝜇²((𝑟))−(𝑟)cosh−1sinh𝑟=𝐴(cosh(𝑟)−1)1−2𝜇²−sinh(𝑟)−𝑟𝑟(cosh(𝑟)−1) (18) provided that cosh(𝑟)=1+𝑣²(𝑟∈(0,cosh⁻¹(2))). Thus, inequality (11) follows from (18) and Lemma 2(3). Similarly, 𝐶𝜇𝑎+1−𝜇𝑏,𝜇𝑏+1−𝜇𝑎−𝑆_𝐴𝐶=𝐴1+1−2𝜇²𝑣²(𝑠)−𝐴tan𝑠=𝐴1+1−2𝜇²((𝑠))−(𝑠)sec−1tan𝑠=𝐴(sec(𝑠)−1)1−2𝜇²−sin(𝑠)−𝑠cos(𝑠)𝑠(1−cos(𝑠)) (19) provided that sec(𝑠)=1+𝑣²(𝑠∈(0,𝜋/3)). Thus, inequality (12) follows from (19) and Lemma 2(4).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 61374086 and 11171307, the Natural Science Foundation of Zhejiang Province under Grant LY13A010004, and the Natural Science Foundation of Huzhou Teachers College under Grant KX21063.