On Certain Inequalities for Neuman-Sándor Mean

Abstract

We present several new sharp bounds for Neuman-Sándor mean in terms of arithmetic, centroidal, quadratic, harmonic root square, and contraharmonic means.

1. Introduction

A binary map 𝑚∶ℝ₊×ℝ₊→ℝ (where ℝ₊∶=(0,+∞) is the set of positive numbers) is said to be a bivariate mean if the following statements are satisfied for all 𝑎,𝑏,𝜆>0: (i)𝑚(𝑎,𝑎)=𝑎 (reflexivity property); (ii)𝑚(𝑎,𝑏)=𝑚(𝑏,𝑎) (symmetry property); (iii)𝑚(𝜆𝑎,𝜆𝑏)=𝜆𝑚(𝑎,𝑏) (homogeneous of order one); (iv)𝑚(𝑎,𝑏) is continuous and strictly increasing with respect to 𝑎 and 𝑏.

Let 𝑢,𝑣, and 𝑤 be the bivariate means such that 𝑢(𝑎,𝑏)<𝑤(𝑎,𝑏)<𝑣(𝑎,𝑏) for all 𝑎,𝑏>0 with 𝑎≠𝑏. The problems to find the best possible parameters 𝛼 and 𝛽 such that the inequalities 𝛼𝑢(𝑎,𝑏)+(1−𝛼)𝑣(𝑎,𝑏)<𝑤(𝑎,𝑏)<𝛽𝑢(𝑎,𝑏)+(1−𝛽)𝑣(𝑎,𝑏) and 𝑢^𝛼(𝑎,𝑏)𝑣^1−𝛼(𝑎,𝑏)<𝑤(𝑎,𝑏)<𝑢^𝛽(𝑎,𝑏)𝑣^1−𝛽(𝑎,𝑏) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 have attracted the interest of many mathematicians.

For 𝑎,𝑏>0 with 𝑎≠𝑏, the Neuman-Sándor mean 𝑀(𝑎,𝑏) [1] is defined by 𝑀(𝑎,𝑏)=𝑎−𝑏)/(,2arcsinh[(𝑎−𝑏𝑎+𝑏)] (1) where arcsinh(𝑥)=log(𝑥+√1+𝑥²) is the inverse hyperbolic sine function.

Recently, the bounds for the Neuman-Sándor mean in terms of other bivariate means have been the subject of intensive research.

Let 𝐻(𝑎,𝑏)=√2𝑎𝑏/𝑎²+𝑏², 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐺(𝑎,𝑏)=√𝑎𝑏, 𝐿(𝑎,𝑏)=(𝑎−𝑏)/(log𝑎−log𝑏), 𝑃(𝑎,𝑏)=(𝑎−𝑏)/(4arctan𝑎/𝑏−𝜋), 𝐼(𝑎,𝑏)=1/𝑒(𝑏^𝑏/𝑎^𝑎)^{1/(𝑏−𝑎)}, 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝑇(𝑎,𝑏)=(𝑎−𝑏)/[2arctan((𝑎−𝑏)/(𝑎+𝑏))], 𝐸(𝑎,𝑏)=2(𝑎²+𝑎𝑏+𝑏²)/[3(𝑎+𝑏)], 𝑄(𝑎,𝑏)=(𝑎²+𝑏²)/2, and 𝐶(𝑎,𝑏)=(𝑎²+𝑏²)/(𝑎+𝑏) be the harmonic root square, harmonic, geometric, logarithmic, first Seiffert, identric, arithmetic, second Seiffert, centroidal, quadratic, and contraharmonic means of two distinct positive real numbers 𝑎 and 𝑏, respectively. Then, it is well known that the inequalities 𝐻()()()()()()()()()()()()𝑎,𝑏<𝐻𝑎,𝑏<𝐺𝑎,𝑏<𝐿𝑎,𝑏<𝑃𝑎,𝑏<𝐼𝑎,𝑏<𝐴𝑎,𝑏<𝑀𝑎,𝑏<𝑇𝑎,𝑏<𝐸𝑎,𝑏<𝑄𝑎,𝑏<𝐶𝑎,𝑏 (2) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Neuman and Sándor [1, 2] proved that the inequalities 𝐴()()<𝐴()𝑎,𝑏<𝑀𝑎,𝑏𝑎,𝑏log1+√2,𝜋4𝑇()()(),𝑀()<()()𝑎,𝑏<𝑀𝑎,𝑏<𝑇𝑎,𝑏𝑎,𝑏2𝐴𝑎,𝑏+𝑄𝑎,𝑏3,𝑀()<𝐴𝑎,𝑏²(𝑎,𝑏)𝑃(),𝑎,𝑏𝐴(𝑎,𝑏)𝑇(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐴²(𝑎,𝑏)+𝑇²(𝑎,𝑏)2 (3) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Let 0<𝑎,𝑏<1/2 with 𝑎≠𝑏, 𝑎^′=1−𝑎, and 𝑏^′=1−𝑏. Then, the Ky Fan inequalities 𝐺(𝑎,𝑏)𝐺𝑎^′,𝑏^′<𝐿(𝑎,𝑏)𝐿𝑎^′,𝑏^′<𝑃(𝑎,𝑏)𝑃𝑎^′,𝑏^′<𝐴(𝑎,𝑏)𝐴𝑎^′,𝑏^′<𝑀(𝑎,𝑏)𝑀𝑎^′,𝑏^′<𝑇(𝑎,𝑏)𝑇𝑎^′,𝑏^′ (4) can be found in [1].

Li et al. [3] proved that 𝐿_𝑝0(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐿₂(𝑎,𝑏) for all 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝐿_𝑝(𝑎,𝑏)=[(𝑏^𝑝+1−𝑎^𝑝+1)/((𝑝+1)(𝑏−𝑎))]^1/𝑝(𝑝≠−1,0), 𝐿₀(𝑎,𝑏)=𝐼(𝑎,𝑏) and 𝐿₋₁(𝑎,𝑏)=𝐿(𝑎,𝑏), is the 𝑝th generalized logarithmic mean of 𝑎 and 𝑏, and 𝑝₀=1.843⋯ is the unique solution of the equation (𝑝+1)^1/𝑝=2log(1+√2).

In [4], the author proved that the double inequalities ()+()𝐴()()()+𝐴(),()+()𝐴()()()+𝐴()𝛼𝑄𝑎,𝑏1−𝛼𝑎,𝑏<𝑀𝑎,𝑏<𝛽𝑄𝑎,𝑏1−𝛽𝑎,𝑏𝜆𝐶𝑎,𝑏1−𝜆𝑎,𝑏<𝑀𝑎,𝑏<𝜇𝐶𝑎,𝑏1−𝜇𝑎,𝑏 (5) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤[1−log(1+√2)]/[(√2−1)log(1+√2)]=0.3249⋯, 𝛽≥1/3, 𝜆≤[1−log(1+√2)]/log(1+√2)=0.1345⋯, and 𝜇≥1/6.

In [5, 6], the authors proved that 𝛼=4, 𝛽=log2/[2log(1+√2)], 𝜆=1, and 𝜇=𝑒/[2log(1+√2)] are the best possible constants such that the inequalities 𝑀(𝑎,𝑏)>𝐿_𝛼(𝑎,𝑏), 𝑀(𝑎,𝑏)>𝑀_𝛽(𝑎,𝑏), and 𝜆𝐼(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝜇𝐼(𝑎,𝑏) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏. In here, 𝐿_𝑟(𝑎,𝑏)=[(𝑏^𝑟−𝑎^𝑟)/(𝑟(log𝑏−log𝑎))]^1/𝑟 and 𝑀_𝑝(𝑎,𝑏)=[(𝑎^𝑝+𝑏^𝑝)/2]^1/𝑝 are the 𝑟th generalized logarithmic and 𝑝th power means of 𝑎 and 𝑏, respectively.

Zhao et al. [7, 8] found the least values 𝛼₁, 𝛼₂, 𝛼₃, 𝛼₄, 𝛼₅ and the greatest values 𝛽₁, 𝛽₂, 𝛽₃, 𝛽₄, 𝛽₅ such that the double inequalities 𝛼₁𝐻(𝑎,𝑏)+1−𝛼₁𝑄()()𝑎,𝑏<𝑀𝑎,𝑏<𝛽₁𝐻(𝑎,𝑏)+1−𝛽₁𝑄(),𝛼𝑎,𝑏₂𝐺(𝑎,𝑏)+1−𝛼₂𝑄()()𝑎,𝑏<𝑀𝑎,𝑏<𝛽₂𝐺(𝑎,𝑏)+1−𝛽₂𝑄(),𝛼𝑎,𝑏₃𝐻(𝑎,𝑏)+1−𝛼₃𝐶()()𝑎,𝑏<𝑀𝑎,𝑏<𝛽₃𝐻(𝑎,𝑏)+1−𝛽₃𝐶(),𝐼𝑎,𝑏_𝛼4(𝑎,𝑏)𝑄^1−𝛼4(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐼^𝛽4(𝑎,𝑏)𝑄^1−𝛽4(),𝐼𝑎,𝑏_𝛼5(𝑎,𝑏)𝐶^1−𝛼5(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐼^𝛽5(𝑎,𝑏)𝐶^1−𝛽5(𝑎,𝑏) (6) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [9], the authors proved that if 𝜆,𝜇∈(1/2,1), then the double inequality 𝐶(𝜆𝑎+(1−𝜆)𝑏,𝜆𝑏+(1−𝜆)𝑎)<𝑀(𝑎,𝑏)<𝐶(𝜇𝑎+(1−𝜇)𝑏,𝜇𝑏+(1−𝜇)𝑎) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝜆≤[1+1/log(1+√2)−1]/2 and 𝜇≥(6+√6)/12.

The aim of this paper is to present the sharp bounds for Neuman-Sándor mean in terms of the combinations of either arithmetic and centroidal means, or quadratic and harmonic root square means, contraharmonic and harmonic root square means. Our main results are shown in Theorems 1–4.

Theorem 1. The double inequality 𝛼₁𝐸(𝑎,𝑏)+1−𝛼₁𝐴()()𝑎,𝑏<𝑀𝑎,𝑏<𝛽₁𝐸(𝑎,𝑏)+1−𝛽₁𝐴(𝑎,𝑏) (7) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₁≤(3−3log(1+√2))/log(1+√2)=0.4037 and 𝛽₁≥1/2.

Theorem 2. The double inequality 𝐸^𝜆1(𝑎,𝑏)𝐴^1−𝜆1(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐸^𝜇1(𝑎,𝑏)𝐴^1−𝜇1(𝑎,𝑏) (8) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝜆₁≤log[log(1+√2)]/(log3−2log2)=0.4389⋯ and 𝜇₁≥1/2.

Theorem 3. The double inequality 𝛼₂𝑄(𝑎,𝑏)+1−𝛼₂𝐻()()𝑎,𝑏<𝑀𝑎,𝑏<𝛽₂𝑄(𝑎,𝑏)+1−𝛽₂𝐻(𝑎,𝑏) (9) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₂≤√2/[2log(1+√2)]=0.8022⋯ and 𝛽₂≥5/6.

Theorem 4. The double inequality 𝜆₂𝐶(𝑎,𝑏)+1−𝜆₂𝐻()()𝑎,𝑏<𝑀𝑎,𝑏<𝜇₂𝐶(𝑎,𝑏)+1−𝜇₂𝐻(𝑎,𝑏) (10) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝜆₂≤1/[2log(1+√2)]=0.5672⋯ and 𝜇₂≥2/3.

2. Lemmas

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

Lemma 5 (see [10, Theorem 1.25]). For −∞<𝑎<𝑏<∞, let 𝑓,𝑔∶[𝑎,𝑏]→ℝ be continuous on [𝑎,𝑏] and differentiable on (𝑎,𝑏), and 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 (11) If 𝑓^′(𝑥)/𝑔^′(𝑥) is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 6 (see [11, Lemma 1.1]). Suppose that the power series 𝑓(𝑥)=∑^∞_𝑛=0𝑎_𝑛𝑥^𝑛 and 𝑔(𝑥)=∑^∞_𝑛=0𝑏_𝑛𝑥^𝑛 have the radius of convergence 𝑟>0 and 𝑏_𝑛>0 for all 𝑛∈{0,1,2,…}. Let ℎ(𝑥)=𝑓(𝑥)/𝑔(𝑥). Then,(1)if the sequence {𝑎_𝑛/𝑏_𝑛}^∞_𝑛=0 is (strictly) increasing (decreasing), then ℎ(𝑥) is also (strictly) increasing (decreasing) on (0,𝑟);(2)if the sequence {𝑎_𝑛/𝑏_𝑛} is (strictly) increasing (decreasing) for 0<𝑛≤𝑛₀ and (strictly) decreasing (increasing) for 𝑛>𝑛₀, then there exists 𝑥₀∈(0,𝑟) such that ℎ(𝑥) is (strictly) increasing (decreasing) on (0,𝑥₀) and (strictly) decreasing (increasing) on (𝑥₀,𝑟).

Lemma 7. The function ℎ(𝑡)=𝑡cosh(3𝑡)+11𝑡cosh(𝑡)−sinh(3𝑡)−9sinh(𝑡)2𝑡[cosh(3𝑡)−cosh(𝑡)] (12) is strictly decreasing on (0,log(1+√2)), where sinh(𝑡)=(𝑒^𝑡−𝑒^−𝑡)/2 and cosh(𝑡)=(𝑒^𝑡+𝑒^−𝑡)/2 are the hyperbolic sine and cosine functions, respectively.

Proof. Let ℎ₁(𝑡)()(𝑡)()(𝑡),ℎ=𝑡cosh3𝑡+11𝑡cosh−sinh3𝑡−9sinh₂(𝑡)=2𝑡[cosh(3𝑡)−cosh(𝑡)]. (13) Then, making use of power series formulas, we have ℎ₁(𝑡)=𝑡^∞_𝑛=0(3𝑡)^2𝑛(2𝑛)!+11𝑡^∞_𝑛=0(𝑡)^2𝑛()!−2𝑛_∞𝑛=0(3𝑡)^2𝑛+1(2𝑛+1)!−9^∞_𝑛=0(𝑡)^2𝑛+1()!=2𝑛+1_∞𝑛=0(2𝑛+3)3^2𝑛+2+11−3^2𝑛+3+9()!𝑡2𝑛+3^2𝑛+3,ℎ₂(𝑡)=^∞_𝑛=023^2𝑛+2−1()!𝑡2𝑛+2^2𝑛+3. (14)
It follows from (12)–(14) that ℎ(𝑡)=∑^∞_𝑛=0𝑎_𝑛𝑡^2𝑛∑^∞_𝑛=0𝑏_𝑛𝑡^2𝑛, (15) where 𝑎_𝑛=(2𝑛+3)3^2𝑛+2+11−3^2𝑛+3+9()!,𝑏2𝑛+3_𝑛=23^2𝑛+2−1()!.2𝑛+2 (16)
Equation (16) leads to 𝑎_𝑛+1𝑏_𝑛+1−𝑎_𝑛𝑏_𝑛=18𝑐_𝑛2(2𝑛+3)(2𝑛+5)3^2𝑛+2−13^2𝑛+4,−1 (17) where 𝑐_𝑛=3⋅9^𝑛81⋅9^𝑛−64𝑛²−224𝑛−166−1. (18)
It is not difficult to verify that 𝑥→81⋅9^𝑥−64𝑥²−224𝑥−166 is positive and strictly increasing in [1,∞). Then, from (18), we get that 𝑐_𝑛>0 (19)
for 𝑛≥1. Note that 𝑐₀=−256,𝑐₁=7424. (20)
Equations (17) and (20) together with inequality (19) lead to the conclusion that the sequence {𝑎_𝑛/𝑏_𝑛} is strictly decreasing for 0≤𝑛≤1 and strictly increasing for 𝑛≥2. Then, from Lemma 6(2) and (15), we clearly see that there exists 𝑡₀∈(0,∞) such that ℎ(𝑡) is strictly decreasing on (0,𝑡₀) and strictly increasing on (𝑡₀,∞).
Let 𝑡^∗=log(1+√2). Then, simple computations lead to sinh𝑡^∗=1,cosh𝑡^∗=√2,sinh3𝑡^∗=7,cosh3𝑡^∗=5√2. (21)
Differentiating (12) yields ℎ^′(𝑡)=3𝑡sinh(3𝑡)+11𝑡sinh(𝑡)−2cosh(3𝑡)+2cosh(𝑡)ℎ₂(𝑡)−2𝑡[3sinh(3𝑡)−sinh(𝑡)]+2[cosh(3𝑡)−cosh(𝑡)]ℎ₂(𝑡)²ℎ₁(𝑡). (22)
From (13) together with (21) and (22), we get ℎ^′𝑡^∗=−3√2𝑡^∗2+2𝑡^∗+√2𝑡^∗2=−0.1529⋯<0. (23)
From the piecewise monotonicity of ℎ(𝑡) and inequality (23) we clearly see that 𝑡₀>𝑡^∗=log(1+√2), and the proof of Lemma 7 is completed.

Lemma 8. The function 𝑔(𝑡)=𝑡cosh(3𝑡)+3𝑡cosh(𝑡)−2sinh(2𝑡)𝑡[cosh(3𝑡)+2cosh(2𝑡)+3cosh(𝑡)−6] (24) is strictly increasing from (0,log(1+√2)) onto (1/3,1−1/[2log(1+√2)]).

Proof. Let 𝑔₁(𝑡)()(𝑡)(),𝑔=𝑡cosh3𝑡+3𝑡cosh−2sinh2𝑡₂(𝑡)=𝑡[cosh(3𝑡)+2cosh(2𝑡)+3cosh(𝑡)−6]. (25) Then, making use of power series formulas, we have 𝑔₁(𝑡)=𝑡^∞_𝑛=0(3𝑡)^2𝑛(2𝑛)!+3𝑡^∞_𝑛=0(𝑡)^2𝑛(2𝑛)!−2^∞_𝑛=0(2𝑡)^2𝑛+1()!=2𝑛+1_∞𝑛=0(2𝑛+3)3^2𝑛+2+3−2^2𝑛+4()!𝑡2𝑛+3^2𝑛+3,𝑔₂(𝑡)=^∞_𝑛=03^2𝑛+2+2^2𝑛+3+3()!𝑡2𝑛+2^2𝑛+3. (26)
It follows from (24)–(26) that ℎ(𝑡)=∑^∞_𝑛=0𝑑_𝑛𝑡^2𝑛∑^∞_𝑛=0𝑒_𝑛𝑡^2𝑛, (27) where 𝑑_𝑛=(2𝑛+3)3^2𝑛+2+3−2^2𝑛+4()!,𝑒2𝑛+3_𝑛=3^2𝑛+2+2^2𝑛+3+3()!.2𝑛+2 (28)
Note that 𝑑_𝑛𝑒_𝑛=(2𝑛+3)3^2𝑛+2+3−2^2𝑛+4(2𝑛+3)3^2𝑛+2+2^2𝑛+3=+31−16⋅4^𝑛/(2𝑛+3)9^𝑛+1+31+8⋅4^𝑛/9^𝑛+1.+3 (29)
It is not difficult to verify that the function 𝑥→4^𝑥/(9^𝑥+1+3) is strictly decreasing in (0,∞). Then from (29), we know that the sequence {𝑑_𝑛/𝑒_𝑛} is strictly increasing for 𝑛=0,1,2,…. Hence, from Lemma 6(1), (24), and (27) the monotonicity of 𝑔(𝑡) follows. Moreover, 𝑔(log(1+√2))=1−1/[2log(1+√2)]=0.4327⋯ and lim_𝑡→0𝑔(𝑡)=𝑑₀𝑒₀=13. (30)

3. Proofs of Theorems 1–4

Proof of Theorem 1. Since 𝑀(𝑎,𝑏), 𝐸(𝑎,𝑏), and 𝐴(𝑎,𝑏) are symmetric and homogeneous of degree 1, without loss of generality, we assume that 𝑎>𝑏. Let 𝑥=(𝑎−𝑏)/(𝑎+𝑏) and 𝑡=arcsinh(𝑥). Then, 𝑥∈(0,1), 𝑡∈(0,log(1+√2)), and 𝑀(𝑎,𝑏)−𝐴(𝑎,𝑏)𝐸()()=(𝑥)𝑎,𝑏−𝐴𝑎,𝑏𝑥/arcsinh−13+𝑥²=3[(𝑡)]/3−1sinh−𝑡𝑡sinh²(𝑡)=6[sinh(𝑡)−𝑡]𝑡[()].cosh2𝑡−1 (31)
Let 𝐹(𝑡)=6[sinh(𝑡)−𝑡]𝑡[()].cosh2𝑡−1 (32) Then, simple computations lead to 𝐹(𝑡)=∑^∞_𝑛=0𝑎^∗_𝑛𝑡^2𝑛∑^∞_𝑛=0𝑏^∗_𝑛𝑡^2𝑛, (33) where 𝑎^∗_𝑛=6/(2𝑛+3)! and 𝑏^∗_𝑛=2^2𝑛+2/(2𝑛+2)!. Note that 𝑎^∗_𝑛/𝑏^∗_𝑛=6/[(2𝑛+3)2^2𝑛+2] is strictly decreasing for 𝑛=0,1,2,…. Hence, from Lemma 6(1) and (33), we know that 𝐹(𝑡) is strictly decreasing in (0,log(1+√2)). Moreover, lim_𝑡→0𝐹(𝑡)=𝑎^∗₀𝑏^∗₀=12,lim_{√𝑡→log1+2}𝐹(𝑡)=3−3log1+√2log1+√2=0.4037⋯. (34)
Therefore, Theorem 1 follows from (31), (32), and (34) together with the monotonicity of 𝐹(𝑡).

Proof of Theorem 2. Without loss of generality, we assume that 𝑎>𝑏. Let 𝑥=(𝑎−𝑏)/(𝑎+𝑏) and 𝑡=arcsinh(𝑥). Then, 𝑥∈(0,1), 𝑡∈(0,log(1+√2)), and log[𝑀(𝑎,𝑏)]−log[𝐴(𝑎,𝑏)][𝐸([𝐴(=[(𝑥log𝑎,𝑏)]−log𝑎,𝑏)]log𝑥/arcsinh)]log1+𝑥²=[(𝑡)]/3logsinh/𝑡log1+sinh²(𝑡)./3 (35)
Let 𝐺₁(𝑡)=log[sinh(𝑡)/𝑡], 𝐺₂(𝑡)=log[1+sinh²(𝑡)/3], and 𝐺(𝑡)=log[sinh(𝑡)/𝑡]log1+sinh²(𝑡)./3 (36) Then, 𝐺₁(0⁺)=𝐺₂(0)=0, 𝐺(𝑡)=𝐺₁(𝑡)/𝐺₂(𝑡), and 𝐺^′₁(𝑡)𝐺^′₂(𝑡)=[𝑡cosh(𝑡)−sinh(𝑡)]sinh²(𝑡)+32𝑡sinh²(𝑡)(𝑡)=[(𝑡)(𝑡()]cosh𝑡cosh−sinh)][cosh2𝑡+5()(𝑡)(𝑡),2𝑡sinh2𝑡sinh=ℎ (37) where ℎ(𝑡) is defined as in Lemma 7.
It follows from Lemmas 5 and 7, (36), and (37) that 𝐺(𝑡) is strictly decreasing in (0,log(1+√2)). Moreover, lim_𝑡→0𝐺(𝑡)=lim_𝑡→0𝐺^′₁(𝑡)𝐺^′₂(𝑡)=12,lim_{√𝑡→log(1+2)}𝐺(𝑡)=loglog1+√2log3−2log2=0.4389⋯. (38)
Therefore, Theorem 2 follows easily from (35), (36), and (38) together with the monotonicity of 𝐺(𝑡).

Proof of Theorem 3. Since 𝑀(𝑎,𝑏), 𝑄(𝑎,𝑏), and 𝐻(𝑎,𝑏) are symmetric and homogeneous of degree 1, without loss of generality, we assume that 𝑎>𝑏. Let 𝑥=(𝑎−𝑏)/(𝑎+𝑏) and 𝑡=arcsinh(𝑥). Then, 𝑥∈(0,1), 𝑡∈(0,log(1+√2)), and 𝑀(𝑎,𝑏)−𝑄(𝑎,𝑏)𝐻()()=(𝑥)−√𝑎,𝑏−𝑄𝑎,𝑏𝑥/arcsinh1+𝑥²1−𝑥²/√1+𝑥²−√1+𝑥²=sinh(𝑡)/𝑡−cosh(𝑡)1−sinh²(𝑡(𝑡)(𝑡)=()())/cosh−cosh𝑡cosh2𝑡+𝑡−sinh2𝑡[()].2𝑡cosh2𝑡−1 (39)
Let 𝜑(𝑡)=𝑡cosh(2𝑡)+𝑡−sinh(2𝑡)[()].2𝑡cosh2𝑡−1 (40)
Then, simple computations lead to 𝜑(𝑡)=∑^∞_𝑛=0𝑑^∗_𝑛𝑡^2𝑛∑^∞_𝑛=0𝑒^∗_𝑛𝑡^2𝑛, (41) where 𝑑^∗_𝑛=(2𝑛+1)⋅2^2𝑛+2/(2𝑛+3)!, and 𝑒^∗_𝑛=2^2𝑛+3/(2𝑛+2)!. Note that 𝑑^∗_𝑛/𝑒^∗_𝑛=[1−2/(2𝑛+3)]/2 is strictly increasing for 𝑛=0,1,2,…. Hence, from Lemma 6(1) and (41), we know that 𝜑(𝑡) is strictly increasing in (0,log(1+√2)). Moreover, lim_𝑡→0𝜑(𝑡)=𝑑^∗₀𝑒^∗₀=16,lim_{√𝑡→log1+2}𝜑(𝑡)√=1−22log1+√2=0.1977⋯. (42)
Therefore, Theorem 3 follows from (39), (40), and (42) together with the monotonicity of 𝜑(𝑡).

Proof of Theorem 4. Since 𝑀(𝑎,𝑏), 𝐶(𝑎,𝑏), and 𝐻(𝑎,𝑏) are symmetric and homogeneous of degree 1, without loss of generality, we assume that 𝑎>𝑏. Let 𝑥=(𝑎−𝑏)/(𝑎+𝑏) and 𝑡=arcsinh(𝑥). Then, 𝑥∈(0,1), 𝑡∈(0,log(1+√2)), and 𝑀(𝑎,𝑏)−𝐶(𝑎,𝑏)𝐻()()=(𝑥)−𝑎,𝑏−𝐶𝑎,𝑏𝑥/arcsinh1+𝑥²1−𝑥²/√1+𝑥²−1+𝑥²=sinh(𝑡)/𝑡−cosh²(𝑡)1−sinh²(𝑡)/cosh(𝑡)−cosh²(𝑡)=𝑡cosh³(𝑡)−sinh(𝑡)cosh(𝑡)𝑡cosh³(𝑡)+sinh²(𝑡)(𝑡),−1=𝑔 (43) where 𝑔(𝑡) is defined as in Lemma 8.
Therefore, Theorem 4 follows from (43) and Lemma 8.

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307, the Natural Science Foundation of Huzhou City under Grant 2012YZ06, and the Natural Science Foundation of the Open University of China under Grant Q1601E-Y.