Bounds for the Arithmetic Mean in Terms of the Neuman-Sándor and Other Bivariate Means

Abstract

We present the largest values 𝛼₁, 𝛼₂, and 𝛼₃ and the smallest values 𝛽₁, 𝛽₂, and 𝛽₃ such that the double inequalities 𝛼₁𝑀(𝑎,𝑏)+(1−𝛼₁)𝐻(𝑎,𝑏)<𝐴(𝑎,𝑏)<𝛽₁𝑀(𝑎,𝑏)+ (1−𝛽₁)𝐻(𝑎,𝑏), 𝛼₂𝑀(𝑎,𝑏)+(1−𝛼₂) 𝐻(𝑎,𝑏) < 𝐴(𝑎,𝑏)<𝛽₂𝑀(𝑎,𝑏)+(1−𝛽₂)𝐻(𝑎,𝑏), and 𝛼₃𝑀(𝑎,𝑏)+(1−𝛼₃)𝐻𝑒(𝑎,𝑏)< 𝐴(𝑎,𝑏)<𝛽₃𝑀 (𝑎,𝑏)+(1−𝛽₃)𝐻𝑒(𝑎,𝑏) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝑀(𝑎,𝑏), 𝐴(𝑎,𝑏), 𝐻𝑒(𝑎,𝑏), 𝐻(𝑎,𝑏) and 𝐻(𝑎,𝑏) denote the Neuman-Sándor, arithmetic, Heronian, harmonic, and harmonic root-square means of 𝑎 and 𝑏, respectively.

1. Introduction

For 𝑎,𝑏>0 with 𝑎≠𝑏 the Neuman-Sándor mean 𝑀(𝑎,𝑏) [1] is defined by 𝑀(𝑎,𝑏)=𝑎−𝑏2sinh⁻¹)/(,((𝑎−𝑏𝑎+𝑏)) (1) where sinh⁻¹(𝑥)=log(𝑥+√𝑥²+1) is the inverse hyperbolic sine function.

Recently, the Neuman-Sándor mean has been the object intensive research. In particular, many remarkable inequalities for the Neuman-Sándor mean 𝑀(𝑎,𝑏) can be found in the literature [1–10].

Let 𝐻(𝑎,𝑏)=√2𝑎𝑏/𝑎²+𝑏², 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐺(𝑎,𝑏)=√𝑎𝑏, 𝐻𝑒(𝑎,𝑏)=(𝑎+√𝑎𝑏+𝑏)/3, 𝐿(𝑎,𝑏)=(𝑏−𝑎)/(log𝑏−log𝑎), 𝑃(𝑎,𝑏)=(𝑎−𝑏)/[4arctan𝑎/𝑏−𝜋], 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝑇(𝑎,𝑏)=(𝑎−𝑏)/ [2arctan((𝑎−𝑏)/(𝑎+𝑏))], 𝑄(𝑎,𝑏)=(𝑎²+𝑏²)/2, and 𝐶(𝑎,𝑏)=(𝑎²+𝑏²)/(𝑎+𝑏) be the harmonic root-square, harmonic, geometric, Heronian, logarithmic, first Seiffert, arithmetic, second Seiffert, quadratic, and contraharmonic means of 𝑎 and 𝑏, respectively. Then it is known that the inequalities 𝐻()()()()()()()()()()()𝑎,𝑏<𝐻𝑎,𝑏<𝐺𝑎,𝑏<𝐿𝑎,𝑏<𝐻𝑒𝑎,𝑏<𝑃𝑎,𝑏<𝐴𝑎,𝑏<𝑀𝑎,𝑏<𝑇𝑎,𝑏<𝑄𝑎,𝑏<𝐶𝑎,𝑏 (2) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Neuman and Sándor [1, 2] proved that the inequalities 𝜋4log1+√2𝑇()()<𝐴()𝑎,𝑏<𝑀𝑎,𝑏𝑎,𝑏log1+√2,2𝑇²(𝑎,𝑏)−𝑄²()()<𝑇𝑎,𝑏<𝑀𝑎,𝑏²(𝑎,𝑏)𝑄(),𝐻(𝑇()(()(𝐴()(,𝐻(𝑀()(()()<𝐴𝑎,𝑏𝑎,𝑏,𝐴𝑎,𝑏))<𝑀𝑎,𝑏<𝐿𝑎,𝑏,𝑄𝑎,𝑏))𝑎,𝑏,𝑄𝑎,𝑏))<𝑇𝑎,𝑏,𝑀𝑎,𝑏²(𝑎,𝑏)𝑃(),𝐴𝑎,𝑏_2/3(𝑎,𝑏)𝑄^1/3()()<()()𝑎,𝑏<𝑀𝑎,𝑏2𝐴𝑎,𝑏+𝑄𝑎,𝑏3,𝐴(𝑎,𝑏)𝑇(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐴²(𝑎,𝑏)+𝑇²(),𝐺𝑎,𝑏𝑥,𝑦𝐺<𝐿1−𝑥,1−𝑦𝑥,𝑦𝐿<𝑃1−𝑥,1−𝑦𝑥,𝑦𝑃<𝐴1−𝑥,1−𝑦𝑥,𝑦𝐴<𝑀1−𝑥,1−𝑦𝑥,𝑦𝑀<𝑇1−𝑥,1−𝑦𝑥,𝑦𝑇,11−𝑥,1−𝑦𝐴−11−𝑥,1−𝑦𝐴<1𝑥,𝑦𝑀−11−𝑥,1−𝑦𝑀<1𝑥,𝑦𝑇−11−𝑥,1−𝑦𝑇,𝐴𝐴𝑀𝑇𝑥,𝑦𝑥,𝑦1−𝑥,1−𝑦<𝑀𝑥,𝑦1−𝑥,1−𝑦<𝑇𝑥,𝑦1−𝑥,1−𝑦 (3) hold for all 𝑎,𝑏>0 and 𝑥,𝑦∈(0,1/2] with 𝑎≠𝑏 and 𝑥≠𝑦. All the results stated above are in fact particular cases of more general and stronger results for the Schwab-Borchardt means [1, 2]. Some of them are based on the sequential method of Sándor [11]. In particular, Neuman and Sándor [1] also found that the inequality 𝑀(𝑎,𝑏)<𝐿𝑎_𝑛,𝑏_𝑛 (4) holds for all 𝑛≥0 and 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝑎₀=𝑄(𝑎,𝑏), 𝑏₀=𝐴(𝑎,𝑏), 𝑎_𝑛+1=(𝑎_𝑛+𝑏_𝑛)/2, and 𝑏_𝑛+1=𝑎_𝑛𝑏_𝑛.

Li et al. [3] proved that the double inequality 𝐿_𝑝0(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐿₂(𝑎,𝑏) holds 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)=1/𝑒(𝑏^𝑏/𝑎^𝑎)^{1/(𝑏−𝑎)}, and 𝐿₋₁(𝑎,𝑏)=(𝑏−𝑎)/(log𝑏−log𝑎) is the 𝑝th generalized logarithmic mean of 𝑎 and 𝑏; 𝑝₀=1.843⋯ is the unique solution of the equation (𝑝+1)^1/𝑝=2log(1+√2).

In [4], Neuman 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.

The main purpose of this paper is to find the largest values 𝛼₁, 𝛼₂, and 𝛼₃ and the smallest values 𝛽₁, 𝛽₂, and 𝛽₃ such that the double inequalities 𝛼₁𝑀(𝑎,𝑏)+1−𝛼₁𝐻()()𝑎,𝑏<𝐴𝑎,𝑏<𝛽₁𝑀(𝑎,𝑏)+1−𝛽₁𝐻(),𝛼𝑎,𝑏₂𝑀(𝑎,𝑏)+1−𝛼₂𝐻()()𝑎,𝑏<𝐴𝑎,𝑏<𝛽₂𝑀(𝑎,𝑏)+1−𝛽₂𝐻(),𝛼𝑎,𝑏₃𝑀(𝑎,𝑏)+1−𝛼₃()()𝐻𝑒𝑎,𝑏<𝐴𝑎,𝑏<𝛽₃𝑀(𝑎,𝑏)+1−𝛽₃𝐻𝑒(𝑎,𝑏) (6) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏. All numerical computations are carried out using Mathematical software.

2. Lemmas

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

Lemma 1 (see [12, 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 𝑎_𝑛,𝑏_𝑛>0 for all 𝑛∈{0,1,2,…}. Let ℎ(𝑥)=𝑓(𝑥)/𝑔(𝑥); then the following hold.(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 2. The function𝑔(𝑡)=𝑡[cosh(2𝑡)+2cosh(𝑡)−3]sinh(2𝑡)+𝑡cosh(2𝑡)−3𝑡 (7) is strictly decreasing on (0,log(1+√2)), where sinh(𝑡)=(𝑒^𝑡−𝑒^−𝑡)/2 and cosh(𝑡)=(𝑒^𝑡+𝑒^−𝑡)/2 denote the hyperbolic sine and hyperbolic cosine functions, respectively.

Proof. Making use of power series sinh(𝑡)=∑^∞_𝑛=0𝑡^2𝑛+1/(2𝑛+1)! and cosh(𝑡)=∑^∞_𝑛=0𝑡^2𝑛/(2𝑛)!, the function 𝑔(𝑡) can be written as 𝑔(𝑡)=∑^∞_𝑛=02^2𝑛+2+2/(2𝑛+2)!𝑡^2𝑛∑^∞_𝑛=0(2𝑛+5)2^2𝑛+2/(2𝑛+3)!𝑡^2𝑛. (8)
Let 𝑎_𝑛=2^2𝑛+2+2(2𝑛+2)!,𝑏_𝑛=(2𝑛+5)2^2𝑛+2()!.2𝑛+3 (9)
Then simple computation leads to 𝑎_𝑛+1𝑏_𝑛+1−𝑎_𝑛𝑏_𝑛=𝑐_𝑛(2𝑛+5)(2𝑛+7)2^2𝑛+3, (10) where 𝑐_𝑛=2^2𝑛+5−12𝑛²+60𝑛+59. (11)
It follows from (11) that 𝑐₀=−27,𝑐₁𝑐=−3,_𝑛≥128𝑛²−12𝑛²+60𝑛+59=116𝑛²−60𝑛−59>0 (12)(13) for all 𝑛≥2.
Equations (10) and (12) together with inequality (13) lead to the conclusion that the sequence {𝑎_𝑛/𝑏_𝑛}^∞_𝑛=0 is strictly decreasing for 0≤𝑛≤1 and strictly increasing for 𝑛≥2. Then from Lemma 1(2) and (8) together with (9) we clearly see that there exists 𝑡₀∈(0,∞) such that 𝑔(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,sinh2𝑡^∗=2√2,cosh2𝑡^∗=3. (14)
It is not difficult to verify that 𝑔^′𝑡^∗=−2𝑡^∗2+2−√2𝑡^∗+1=−0.03734⋯<0. (15)
From the piecewise monotonicity of 𝑔(𝑡) and inequality (15) we clearly see that 𝑡^∗=log(1+√2)<𝑡₀, which implies that 𝑔(𝑡) is strictly decreasing on (0,log(1+√2)).

Lemma 3. The inequality √1−𝑡²>1−𝑡²2−𝑡⁴8−𝑡⁶−165𝑡⁸−12835𝑡¹⁰128 (16) holds for all 𝑡∈(0,1).

Proof. Simple computations lead to 𝑡1−²2−𝑡⁴8−𝑡⁶−165𝑡⁸−12835𝑡¹⁰128²=1−𝑡²−𝑡¹⁰×163848064−4704𝑡²−1200𝑡⁴−585𝑡⁶−350𝑡⁸−1225𝑡¹⁰<1−𝑡²−𝑡¹⁰()163848064−4704−1200−585−350−1225<1−𝑡² (17) for all 𝑡∈(0,1).

Lemma 4. The function 𝑓(𝑡)=[𝑡−sinh⁻¹(𝑡)]/[(1−√1−𝑡²)sinh⁻¹(𝑡)] is strictly decreasing in (0,1).

Proof. Differentiating 𝑓(𝑡) gives 𝑓^′(𝑡)=𝑡1−𝑡²−√1−𝑡²+√1+𝑡²𝑡sinh⁻¹(𝑡)+√1−𝑡²−1sinh⁻¹(𝑡)×1−√1−𝑡²²√1−𝑡⁴sinh⁻¹(𝑡)²⁻¹. (18) Making use of the power series √1+𝑡²=^∞_𝑛=0(−1)^𝑛−1(2𝑛)!(2𝑛−1)4^𝑛(𝑛!)²𝑡^2𝑛𝑡=1+²2−𝑡⁴8+𝑡⁶−165𝑡⁸+1287𝑡¹⁰√256−⋯,1−𝑡²=^∞_𝑛=0(2𝑛)!(1−2𝑛)4^𝑛(𝑛!)²𝑡^2𝑛𝑡=1−²2−𝑡⁴8−𝑡⁶−165𝑡⁸−1287𝑡¹⁰256−⋯,sinh⁻¹(𝑡)=^∞_𝑛=0(−1)^𝑛(2𝑛)!(2𝑛+1)4^𝑛(𝑛!)²𝑡^2𝑛+1𝑡=𝑡−³6+3𝑡⁵−405𝑡⁷112+⋯, (19)(20)(21) we get √1+𝑡²sinh⁻¹(𝑡)=𝑡+𝑡³3−2𝑡⁵+158𝑡⁷−1051091𝑡⁹𝑡13440+⋯<𝑡+³3−2𝑡⁵+158𝑡⁷,105𝑡sinh⁻¹(𝑡)+√1−𝑡²=𝑡−1₂2−7𝑡⁴+𝑡24⁶−8075𝑡⁸𝑡896+⋯<²2−7𝑡⁴+𝑡34⁶,80 (22) for 𝑡∈(0,1).
Let 𝑔(𝑡)=𝑡1−𝑡²−√1−𝑡²+√1+𝑡²𝑡sinh⁻¹(𝑡)+√1−𝑡²−1sinh⁻¹(𝑡)=−1−√1−𝑡²𝑡√1−𝑡²+√1+𝑡²sinh⁻¹(𝑡√)+𝑡1+𝑡²sinh⁻¹(𝑡)². (23)
We divide the proof into two cases.
Case 1 (𝑡∈(0,√10/5)). Then from Lemma 3, (22), and (23) we have 𝑔(𝑡)<𝑡1−𝑡²−𝑡1−²2−𝑡⁴8−𝑡⁶−165𝑡⁸−12835𝑡¹⁰+𝑡128𝑡+³3−2𝑡⁵+158𝑡⁷×𝑡1052²−7𝑡⁴+𝑡24⁶=𝑡80₂20160024235𝑡²+50309𝑡⁴+192𝑡⁶<𝑡−17920₂√20160024235105²√[phone]√+192105⁶−17920<0. (24)
Case 2 (𝑡∈[√10/5,1)). Then from Lemma 3, (19), (20), and (23) together with sinh⁻¹(√10/5)=0.5964⋯ we get 𝑔(𝑡)𝑡<−1−1−²2−𝑡⁴8−𝑡⁶×𝑡𝑡161−²2−𝑡⁴8−𝑡⁶−165𝑡⁸−12835𝑡¹⁰𝑡128+0.5961+²2−𝑡⁴8+(0.596)²𝑡1+²2−𝑡⁴8+𝑡⁶𝑔16∶=₁(𝑡)=0.356409𝑡−0.298𝑡²−0.3217955𝑡³−0.2235𝑡⁴+0.08448875𝑡⁵−0.03725𝑡⁶+0.0847755625𝑡⁷−0.0093125𝑡⁸+0.078125𝑡⁹+0.00465625𝑡¹⁰+0.03515625𝑡¹¹+0.1455078125𝑡¹³+0.03662109375𝑡¹⁵+0.01708984375𝑡¹⁷. (25)
Numerical computations show that𝑔₁√105=−0.0000435,𝑔₁(1)𝑔=−0.0510683⋯,_′1√105=−0.52459⋯,𝑔^′₁(1)𝑔=2.4665⋯,_1′′√105=−1.86198⋯,𝑔₁^′′(1)𝑔=43.2711⋯,_1′′′√105𝑔=4.63578⋯,₁₍₄₎(𝑡)>(−5.364+9.653865𝑡)+𝑡²−13.41+71.2114725𝑡−15.645𝑡²>√−5.364+9.653865×105+𝑡²√−13.41+71.2114725×105−15.645=0.7416⋯+𝑡²×15.9830⋯>0. (26)(27)(28)(29)(30)
It follows from (29) and (30) that 𝑔₁^′′(𝑡) is strictly increasing in [√10/5,1). Then (28) leads to the conclusion that there exists 𝑡₀∈(√10/5,1) such that 𝑔^′₁(𝑡) is strictly decreasing in [√10/5,𝑡₀] and strictly increasing in [𝑡₀,1).
From (27) and the piecewise monotonicity of 𝑔^′₁(𝑡) we clearly see that there exists 𝑡₁∈(𝑡₀,1) such that 𝑔₁(𝑡) is strictly decreasing in [√10/5,𝑡₁] and strictly increasing in [𝑡₁,1). Therefore, 𝑔(𝑡)<𝑔₁(𝑡)<0 (31) for 𝑡∈[√10/5,1) follows from (25) and (26) together with the piecewise monotonicity of 𝑔₁(𝑡).

Lemma 5. Let 𝑝∈(0,1), 𝜆₀=log(1+√2)/[3−2log(1+√2)]=0.7123⋯, and 𝜑_𝑝(𝑡)=3𝑝√1−𝑡²𝑝−1𝑡+2𝑝−1−sinh⁻¹√1−𝑡². (32) Then 𝜑_1/2(𝑡)<0 and 𝜑_𝜆0(𝑡)>0 for all 𝑡∈(0,1).

Proof. We first prove that 𝜑_1/2(𝑡)<0 for 𝑡∈(0,1). From (32) one has 𝜑_1/2(1)𝜑=0,_′1/2(𝑡)=4𝑡−2𝑡²+(1/4)𝑡³−(3𝑡−3/4)√2−𝑡²√1−𝑡²√2−𝑡²(2−(1/2)𝑡)². (33)(34) Let 𝑓(𝑡)=4𝑡−2𝑡²+14𝑡³−33𝑡−4√2−𝑡². (35) Then 𝑓(1)𝑓=0,_′(𝑡)=24𝑡²−3𝑡−24+16−16𝑡+3𝑡²√2−𝑡²4√2−𝑡². (36)
We divide the proof into two cases.
Case 1 (𝑡∈(0,0.7)). Then we clearly see that 24𝑡²−3𝑡−24+16−16𝑡+3𝑡²√2−𝑡²√2−𝑡²<24𝑡²−3𝑡+24√2+16−16𝑡+3𝑡²=27𝑡²−32√2+16𝑡+16−12√23√=27𝑡−2+32+5376√2−5870×3√108𝑡−2+32−5376√2−5870()108=27𝑡−0.72105⋯)(𝑡+0.04985⋯<0. (37)
Case 2 (𝑡∈[0.7,1)). Then we get 24𝑡²−3𝑡−24+16−16𝑡+3𝑡²√2−𝑡²√2−𝑡²=24𝑡²−3𝑡−24√2−𝑡²+32−32𝑡−10𝑡²+16𝑡³−3𝑡⁴<24𝑡²−3𝑡−24+32−32𝑡−10𝑡²+16𝑡³−3𝑡⁴=−(1−𝑡)27𝑡+13𝑡²−3𝑡³−8<0. (38)
It follows from (32) and (33) together with Cases 1 and 2 that 𝑓(𝑡)<0 for 𝑡∈(0,1). Then from (34) and (35) we know that 𝜑_1/2(𝑡) is strictly decreasing in (0,1).
Therefore, 𝜑_1/2(𝑡)<0 for 𝑡∈(0,1) follows from (33) and the monotonicity of 𝜑_1/2(𝑡).
Next, we prove that 𝜑_𝜆0(𝑡)>0 for 𝑡∈(0,1). From (32) we clearly see that we only have to prove that𝜑_𝜆0√1−𝑡²=3𝜆₀𝑡𝜆₀−1√1−𝑡²+2𝜆₀+1−sinh⁻¹(𝑡)>0 (39) for all 𝑡∈(0,1).
Inequality (39) can be rewritten as 𝜆₀>1−√1−𝑡²sinh⁻¹(𝑡)3𝑡−2+√1−𝑡²sinh⁻¹(𝑡). (40) Let ℎ(𝑡)=1−√1−𝑡²sinh⁻¹(𝑡)3𝑡−2+√1−𝑡²sinh⁻¹(𝑡)=11+3𝑡−sinh⁻¹(𝑡)/1−√1−𝑡²sinh⁻¹(𝑡.) (41) Then inequality (40) follows from Lemma 4 and (41) together with ℎ(1)=𝜆₀.

3. Main Results

Theorem 6. The double inequality 𝛼₁𝑀(𝑎,𝑏)+1−𝛼₁𝐻()()𝑎,𝑏<𝐴𝑎,𝑏<𝛽₁𝑀(𝑎,𝑏)+1−𝛽₁𝐻(𝑎,𝑏) (42) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₁≤6/7=0.8571⋯ and 𝛽₁≥log(1+√2)=0.8813⋯.

Proof. Since 𝐻(𝑎,𝑏), 𝑀(𝑎,𝑏), and 𝐴(𝑎,𝑏) are symmetric and homogeneous of degree one, without loss of generality, we assume that 𝑎>𝑏. Let 𝑥=(𝑎−𝑏)/(𝑎+𝑏) and 𝑡=sinh⁻¹(𝑥). Then 𝑥∈(0,1), 𝑡∈(0,log(1+√2)), and 𝐴(𝑎,𝑏)−𝐻(𝑎,𝑏)𝑀()()=𝑥𝑎,𝑏−𝐻𝑎,𝑏𝑥/sinh⁻¹(𝑥)²−1−𝑥²=𝑡cosh(2𝑡)−𝑡(𝑡)().2sinh+𝑡cosh2𝑡−3𝑡 (43) Let 𝑓(𝑡)=𝑡cosh(2𝑡)−𝑡(𝑡)().2sinh+𝑡cosh2𝑡−3𝑡 (44) Then 𝑓(𝑡) can be rewritten as 𝑓(𝑡)=𝑡∑^∞_𝑛=02^2𝑛/(2𝑛)!𝑡^2𝑛−𝑡2∑^∞_𝑛=0𝑡^2𝑛+1/(2𝑛+1)!+∑^∞_𝑛=02^2𝑛/(2𝑛)!𝑡^2𝑛+1=∑−3𝑡_∞𝑛=12^2𝑛/(2𝑛)!𝑡^2𝑛∑^∞_𝑛=12+(2𝑛+1)2^2𝑛/(2𝑛+1)!𝑡^2𝑛+1=∑^∞_𝑛=02^2𝑛+2/(2𝑛+2)!𝑡^2𝑛∑^∞_𝑛=02+(2𝑛+3)2^2𝑛+2/(2𝑛+3)!𝑡^2𝑛∑∶=^∞𝑛=0𝑎_𝑛𝑡^2𝑛∑^∞_𝑛=0𝑏_𝑛𝑡^2𝑛,𝑎_𝑛+1𝑏_𝑛+1−𝑎_𝑛𝑏_𝑛=(12𝑛+34)2^2𝑛+2(2𝑛+3)2^2𝑛+2+2(2𝑛+5)2^2𝑛+4+2>0 (45) for all 𝑛≥0.
It follows from (45) together with Lemma 1(1) that 𝑓(𝑡) is strictly increasing in (0,log(1+√2)). Note that lim_𝑡→0𝑓(𝑡)=𝑎₀𝑏₀=67,𝑓log1+√2=log1+√2. (46)
Therefore, Theorem 6 follows from (43) and (44) together with (46) and the monotonicity of 𝑓(𝑡).

Theorem 7. The double inequality 𝛼₂𝑀(𝑎,𝑏)+1−𝛼₂𝐻()()𝑎,𝑏<𝐴𝑎,𝑏<𝛽₂𝑀(𝑎,𝑏)+1−𝛽₂𝐻(𝑎,𝑏) (47) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₂≤log(1+√2)=0.8813⋯ and 𝛽₂≥9/10.

Proof. Since 𝐻(𝑎,𝑏), 𝑀(𝑎,𝑏), and 𝐴(𝑎,𝑏) are symmetric and homogeneous of degree one, without loss of generality, we assume that 𝑎>𝑏. Let 𝑥=(𝑎−𝑏)/(𝑎+𝑏) and 𝑡=sinh⁻¹(𝑥). Then 𝑥∈(0,1), 𝑡∈(0,log(1+√2)), and 𝐴(𝑎,𝑏)−𝐻(𝑎,𝑏)𝑀(𝑎,𝑏)−𝐻()=√𝑎,𝑏1+𝑥²+𝑥²−1sinh⁻¹(𝑥)𝑥√1+𝑥²−1−𝑥²sinh⁻¹(𝑥)=𝑡[cosh(2𝑡)+2cosh(𝑡)−3]()().sinh2𝑡+𝑡cosh2𝑡−3𝑡 (48)
Simple computations lead to lim_𝑡→0𝑡[cosh(2𝑡)+2cosh(𝑡)−3]()()=9sinh2𝑡+𝑡cosh2𝑡−3𝑡,10lim_{√𝑡→log(1+2)}𝑡[cosh(2𝑡)+2cosh(𝑡)−3]()()√sinh2𝑡+𝑡cosh2𝑡−3𝑡=log1+2. (49)
Therefore, Theorem 7 follows from Lemma 2, (48), and (49).

Theorem 8. The double inequality 𝛼₃𝑀(𝑎,𝑏)+1−𝛼₃()()𝐻𝑒𝑎,𝑏<𝐴𝑎,𝑏<𝛽₃𝑀(𝑎,𝑏)+1−𝛽₃𝐻𝑒(𝑎,𝑏) (50) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₃≤1/2 and 𝛽₃≥log(1+√2)/ [3−2log(1+√2)]=0.7123⋯.

Proof. Since 𝐻(𝑎,𝑏), 𝑀(𝑎,𝑏), and 𝐴(𝑎,𝑏) are symmetric and homogeneous of degree one, without loss of generality, we assume that 𝑎>𝑏. Let 𝑥=(𝑎−𝑏)/(𝑎+𝑏)∈(0,1), 0<𝑝<1, and 𝜆₀=log(1+√2)/[3−2log(1+√2)]; then 𝐴(𝑎,𝑏)−𝐻𝑒(𝑎,𝑏)𝑀()()=√𝑎,𝑏−𝐻𝑒𝑎,𝑏1−1−𝑥²sinh⁻¹(𝑥)3𝑥−2+√1−𝑥²sinh⁻¹(𝑥)()+()()=𝐴(),𝑝𝑀𝑎,𝑏1−𝑝𝐻𝑒𝑎,𝑏−𝐴𝑎,𝑏𝑎,𝑏33𝑝𝑥sinh⁻¹(𝑥)−𝑝−1√1−𝑥²−=𝐴(−√1+2𝑝𝑎,𝑏)1+2𝑝1−𝑝1−𝑥²3sinh⁻¹(𝑥)𝜑_𝑝√1−𝑥², (51)(52) where 𝜑_𝑝(𝑡) is defined as in Lemma 5.
Note that lim_𝑥→01−√1−𝑥²sinh⁻¹(𝑥)3𝑥−2+√1−𝑥²sinh⁻¹(𝑥)=12,lim_𝑥→11−√1−𝑥²sinh⁻¹(𝑥)3𝑥−2+√1−𝑥²sinh⁻¹(𝑥)=𝜆₀. (53)
Therefore, Theorem 8 follows from (51)–(53) together with Lemma 5.

Acknowledgments

This work was supported by the Natural Science Foundation of China under Grant 61374086, the Natural Science Foundation of the Department of Education of Zhejiang Province under Grant Y201223519, and the Natural Science Foundation of Zhejiang Radio and Television University under Grant XKT-13Z04.