Best Possible Bounds for Neuman-Sándor Mean by the Identric, Quadratic and Contraharmonic Means

Abstract

We prove that the double inequalities 𝐼^𝛼1(𝑎,𝑏)𝑄^1−𝛼1(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐼^𝛽1(𝑎,𝑏)𝑄^1−𝛽1(𝑎,𝑏),𝐼^𝛼2(𝑎,𝑏)𝐶^1−𝛼2(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐼^𝛽2(𝑎,𝑏)𝐶^1−𝛽2(𝑎,𝑏) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼₁≥1/2, 𝛽₁≤log[√2log(1+√2)]/(1−log√2), 𝛼₂≥5/7, and 𝛽₂≤log[2log(1+√2)], where 𝐼(𝑎,𝑏), 𝑀(𝑎,𝑏), 𝑄(𝑎,𝑏), and 𝐶(𝑎,𝑏) are the identric, Neuman-Sándor, quadratic, and contraharmonic means of 𝑎 and 𝑏, respectively.

1. Introduction

For 𝑝∈ℝ and 𝑎,𝑏>0 with 𝑎≠𝑏, the identric mean 𝐼(𝑎,𝑏), Neuman-Sándor mean 𝑀(𝑎,𝑏) [1], quadratic mean 𝑄(𝑎,𝑏), contraharmonic mean 𝐶(𝑎,𝑏), and 𝑝th power mean 𝑀_𝑝(𝑎,𝑏) are defined by 𝐼()=1𝑎,𝑏𝑒𝑏^𝑏𝑎^{𝑎1/(𝑏−𝑎)},𝑀(𝑎,𝑏)=𝑎−𝑏2sinh⁻¹)/(,𝑄()=[(𝑎−𝑏𝑎+𝑏)]𝑎,𝑏𝑎²+𝑏²2()=𝑎,𝐶𝑎,𝑏²+𝑏²,𝑀𝑎+𝑏_𝑝()=𝑎𝑎,𝑏^𝑝+𝑏^𝑝2^1/𝑝√,𝑝≠0,𝑎𝑏,𝑝=0, (1) respectively, where sinh⁻¹(𝑥)=log(𝑥+√1+𝑥²) is the inverse hyperbolic sine function.

Recently, the identric, Neuman-Sándor, quadratic, and contraharmonic means have attracted the interest of numerous eminent mathematicians. In particular, many remarkable inequalities for these means can be found in the literature [1–18].

Let 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐺(𝑎,𝑏)=√𝑎𝑏, 𝐿(𝑎,𝑏)=(𝑏−𝑎)/(log𝑏−log𝑎), 𝑃(𝑎,𝑏)=(𝑎−𝑏)/(4arctan𝑎/𝑏−𝜋), 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, and 𝑇(𝑎,𝑏)=(𝑎−𝑏)/[2arctan((𝑎−𝑏)/(𝑎+𝑏))] be the harmonic, geometric, logarithmic, first Seiffert, arithmetic, and second Seiffert means of two distinct positive numbers 𝑎 and 𝑏, respectively. Then it is well known that the inequalities 𝐻(𝑎,𝑏)=𝑀₋₁(𝑎,𝑏)<𝐺(𝑎,𝑏)=𝑀₀()()()()()𝑎,𝑏<𝐿𝑎,𝑏<𝑃𝑎,𝑏<𝐼𝑎,𝑏<𝐴𝑎,𝑏<𝑀₁()()()()𝑎,𝑏<𝑀𝑎,𝑏<𝑇𝑎,𝑏<𝑄𝑎,𝑏=𝑀₂()()𝑎,𝑏<𝐶𝑎,𝑏 (2) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Neuman and Sándor [1, 8] established that 𝐴()()<𝐴()𝑎,𝑏<𝑀𝑎,𝑏𝑎,𝑏log1+√2,𝜋4𝑇()()(),𝑀()<𝐴𝑎,𝑏<𝑀𝑎,𝑏<𝑇𝑎,𝑏𝑎,𝑏²(𝑎,𝑏)𝑃(),𝑎,𝑏𝐴()𝑇()()<𝑎,𝑏𝑎,𝑏<𝑀𝑎,𝑏𝐴²(𝑎,𝑏)+𝑇²(𝑎,𝑏)2,𝑀()<()()𝑎,𝑏2𝐴𝑎,𝑏+𝑄𝑎,𝑏3 (3) 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) were presented in [1].

Li et al. [19] found the best possible bounds for the Neuman-Sándor mean in terms of the generalized logarithmic mean 𝐿_𝑟(𝑎,𝑏). Neuman [20] and Zhao et al. [21] proved that the inequalities ()+()𝐴()()()+𝐴(),()+()𝐴()()()+𝐴(),𝛼𝛼𝑄𝑎,𝑏1−𝛼𝑎,𝑏<𝑀𝑎,𝑏<𝛽𝑄𝑎,𝑏1−𝛽𝑎,𝑏𝜆𝐶𝑎,𝑏1−𝜆𝑎,𝑏<𝑀𝑎,𝑏<𝜇𝐶𝑎,𝑏1−𝜇𝑎,𝑏₁𝐻(𝑎,𝑏)+1−𝛼₁𝑄()()𝑎,𝑏<𝑀𝑎,𝑏<𝛽₁𝐻(𝑎,𝑏)+1−𝛽₁𝑄(),𝛼𝑎,𝑏₂𝐶(𝑎,𝑏)+1−𝛼₂𝑄()()𝑎,𝑏<𝑀𝑎,𝑏<𝛽₂𝐶(𝑎,𝑏)+1−𝛽₂𝑄(𝑎,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)], 𝛽≥1/3, 𝜆≤[1−log(1+√2)]/log(1+√2), 𝜇≥1/6, 𝛼₁≥2/9, 𝛽₁≤1−1/[√2log(1+√2)], 𝛼₂≥1/3, and 𝛽₂≤1−1/[√2log(1+√2)].

In [22], Chu and Long gave the best possible constants 𝑝,𝑞,𝛼, and 𝛽 such that the double inequalities 𝑀_𝑝(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝑀_𝑞(𝑎,𝑏) and 𝛼𝐼(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝛽𝐼(𝑎,𝑏) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The ratio of identric means leads to the weighted geometric mean 𝐼𝑎²,𝑏²𝐼()=𝑎𝑎,𝑏^𝑎𝑏^{𝑏1/(𝑎+𝑏)}, (6) which has been investigated in [23–25]. Alzer [26] proved that the inequalities 𝐴(𝑎,𝑏)𝐺(𝑎,𝑏)<𝐼()𝐿()<𝐼()()𝑎,𝑏𝑎,𝑏𝑎,𝑏+𝐿𝑎,𝑏2<𝐴(𝑎,𝑏)+𝐺(𝑎,𝑏)2 (7) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The following sharp bounds for 𝐼, (𝐼𝐿)^1/2, and (𝐼+𝐿)/2 in terms of the power mean and the convex combination of arithmetic and geometric means are given in [27] as 𝑀_2/3(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝑀_log2(),𝑀𝑎,𝑏₀(𝑎,𝑏)<𝐼(𝑎,𝑏)𝐿(𝑎,𝑏)<𝑀_1/2(),𝑀𝑎,𝑏_{log2/(1+log2)}()<𝐼()()𝑎,𝑏𝑎,𝑏+𝐿𝑎,𝑏2<𝑀_1/2(),2𝑎,𝑏3𝐴()+1𝑎,𝑏3𝐺()()<2𝑎,𝑏<𝐼𝑎,𝑏𝑒𝐴()+2𝑎,𝑏1−𝑒𝐺(𝑎,𝑏) (8) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Chu et al. [28] presented the optimal constants 𝛼₁,𝛽₁,𝛼₂, and 𝛽₂ such that the double inequalities 𝛼₁𝑄(𝑎,𝑏)+1−𝛼₁𝐴()<2𝑎,𝑏𝜋₀^𝜋/2𝑎²cos²𝜃+𝑏²sin²𝜃𝑑𝜃<𝛽₁𝑄(𝑎,𝑏)+1−𝛽₁𝐴(),𝑄𝑎,𝑏_𝛼2(𝑎,𝑏)𝐴^1−𝛼2()<2𝑎,𝑏𝜋₀^𝜋/2𝑎²cos²𝜃+𝑏²sin²𝜃𝑑𝜃<𝑄^𝛽2(𝑎,𝑏)𝐴^1−𝛽2(𝑎,𝑏) (9) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The aim of this paper is to find the best possible constants 𝛼₁,𝛽₁,𝛼₂ and 𝛽₂ such that the double inequalities 𝐼^𝛼1(𝑎,𝑏)𝑄^1−𝛼1(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐼^𝛽1(𝑎,𝑏)𝑄^1−𝛽1(),𝐼𝑎,𝑏_𝛼2(𝑎,𝑏)𝐶^1−𝛼2(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐼^𝛽2(𝑎,𝑏)𝐶^1−𝛽2(𝑎,𝑏) (10) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏. All numerical computations are carried out using MATHEMATICA software.

2. Lemmas

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

Lemma 1. The double inequality 𝑥𝑥+³3−2𝑥⁵<√151+𝑥²sinh⁻¹(𝑥)𝑥<𝑥+³3−2𝑥⁵+158𝑥⁷105 (11) holds for 𝑥∈(0,1).

Proof. To prove Lemma 1, it suffices to prove that 𝜂₁(𝑥)=√1+𝑥²sinh⁻¹(𝑥)−𝑥𝑥+³3−2𝑥⁵𝜂15>0,₂(𝑥)=√1+𝑥²sinh⁻¹(𝑥)−𝑥𝑥+³3−2𝑥⁵+158𝑥⁷105<0 (12)(13) for 𝑥∈(0,1).
From the expressions of 𝜂₁(𝑥) and 𝜂₂(𝑥), we get 𝜂₁(0)=𝜂₂(0)𝜂=0,_′1(𝑥)=𝑥𝜂^∗₁(𝑥)√1+𝑥²,𝜂^′₂(𝑥)=𝑥𝜂^∗₂(𝑥)√1+𝑥², (14) where 𝜂^∗₁(𝑥)=sinh⁻¹(𝑥)−𝑥−2𝑥³3√1+𝑥²,𝜂^∗₂(𝑥)=sinh⁻¹(𝑥)−𝑥−2𝑥³3+8𝑥⁵√151+𝑥²,𝜂^∗₁(0)=𝜂^∗₂(0)𝜂=0,_∗1′(𝑥)=8𝑥⁴3√1+𝑥²𝜂>0,_∗2′(𝑥)=−16𝑥⁶5√1+𝑥²<0, (15)(16)(17) for 𝑥∈(0,1).
Therefore, inequality (12) follows from (14)–(16), and inequality (13) follows from (14)–(17).

Lemma 2. Let 𝐿(𝑥)()=log1+𝑥^1+𝑥(1−𝑥)^1−𝑥−2𝑥−𝑥log1−𝑥². (18) Then 𝐿(𝑥)>2𝑥³3+2𝑥⁵5+2𝑥⁷7 (19) for 𝑥∈(0,1), and 𝐿(𝑥)<2𝑥³3+2𝑥⁵5+2𝑥⁷7+𝑥⁹ (20) for 𝑥∈(0,3/4).

Proof. To prove inequalities (19) and (20), it suffices to show that 𝜄₁(𝑥)()∶=log1+𝑥^1+𝑥(1−𝑥)^1−𝑥−2𝑥−𝑥log1−𝑥²−2𝑥³3+2𝑥⁵5+2𝑥⁷7>0 (21) for 𝑥∈(0,1), and 𝜄₂(𝑥)()∶=log1+𝑥^1+𝑥(1−𝑥)^1−𝑥−2𝑥−𝑥log1−𝑥²−2𝑥³3+2𝑥⁵5+2𝑥⁷7+𝑥⁹<0 (22) for 𝑥∈(0,3/4).
From (21) and (22), one has 𝜄₁0⁺=𝜄₂0⁺𝜄=0,_′1(𝑥)=2𝑥⁸1−𝑥²>0 (23)(24) for 𝑥∈(0,1), and 𝜄^′₂(𝑥)=−9𝑥⁸1−𝑥²79−𝑥²<0 (25) for 𝑥∈(0,3/4).
Therefore, inequality (21) follows from (23) and (24), and inequality (22) follows from (23) and (25).

Lemma 3. Let Φ₁(𝑥)=1√1+𝑥²sinh⁻¹(𝑥)−1𝑥1+𝑥². (26)
Then the double inequality 2𝑥3−34𝑥³+𝑥452⁵<Φ₁(𝑥)<2𝑥3−34𝑥³+454𝑥⁵5 (27) holds for 𝑥∈(0,0.7).

Proof. To prove inequality (27), it suffices to show that 𝜙₁(𝑥)=𝑥√1+𝑥²−sinh⁻¹(𝑥)−𝑥1+𝑥²sinh⁻¹(𝑥)2𝑥3−34𝑥³+𝑥452⁵𝜙>0,₂(𝑥)=𝑥√1+𝑥²−sinh⁻¹(𝑥)−𝑥1+𝑥²sinh⁻¹(𝑥)2𝑥3−34𝑥³+454𝑥⁵5<0 (28)(29) for 𝑥∈(0,0.7).
First, we prove inequality (28). From the expression of 𝜙₁(𝑥), we have 𝜙₁(0)=0,𝜙₁()𝜙0.7=0.0033⋯,_′1(𝑥)=𝑥𝜙^∗₁(𝑥)90√1+𝑥², (30)(31) where 𝜙^∗₁(𝑥)=120𝑥+8𝑥³+23𝑥⁵−45𝑥⁷−260−16𝑥²−69𝑥⁴+180𝑥⁶×√1+𝑥²sinh⁻¹(𝑥). (32)
Equation (32) leads to 𝜙^∗₁(0.6)=3.017⋯,𝜙^∗₁()𝜙0.7=−3.551⋯,_∗1′(𝑥)=−𝑥𝜙₁^∗∗(𝑥)1+𝑥², (33) where 𝜙₁^∗∗(𝑥)=−56𝑥−309𝑥³+422𝑥⁵+675𝑥⁷+228−324𝑥²+735𝑥⁴+1260𝑥⁶×√1+𝑥²sinh⁻¹(𝑥). (34)
Note that 60−16𝑥²−69𝑥⁴+180𝑥⁶>0 (35) for 𝑥∈(0,0.6], and 28−324𝑥²+735𝑥⁴+1260𝑥⁶>0 (36) for 𝑥∈[0.6,0.7).
It follows from (32) and (34)–(36) together with Lemma 1 that 𝜙^∗₁(𝑥)>120𝑥+8𝑥³+23𝑥⁵−45𝑥⁷−260−16𝑥²−69𝑥⁴+180𝑥⁶𝑥𝑥+³3=𝑥⁵3515−1077𝑥²−360𝑥⁴≥𝑥⁵39515−1077×25−360×81=62510078𝑥⁵375>0 (37) for 𝑥∈(0,0.6], and 𝜙₁^∗∗(𝑥)>−56𝑥−309𝑥³+422𝑥⁵+675𝑥⁷+228−324𝑥²+735𝑥⁴+1260𝑥⁶×𝑥𝑥+³3−2𝑥⁵=𝑥15³×15−14075+25028𝑥²+56571𝑥⁴+9660𝑥⁶−5040𝑥⁸>𝑥³()15−14075+25028×0.6²+56571×(0.6)⁴=1416676𝑥³9375>0 (38) for 𝑥∈[0.6,0.7).
From (33), (37), and (38), we clearly see that there exists 𝑥₁∈(0.6,0.7) such that 𝜙^∗₁(𝑥)>0 for 𝑥∈(0,𝑥₁) and 𝜙^∗₁(𝑥)<0 for 𝑥∈(𝑥₁,0.7). Then (31) leads to the conclusion that 𝜙₁(𝑥) is strictly increasing on (0,𝑥₁] and strictly decreasing on [𝑥₁,0.7).
Therefore, inequality (28) follows from (30) and the piecewise monotonicity of 𝜙₁(𝑥).
Next, we prove inequality (29). From the expression of 𝜙₂(𝑥), we get 𝜙₂(0)𝜙=0,_′2(𝑥)=−2𝑥𝜙^∗₂(𝑥)45√1+𝑥², (39) where 𝜙^∗₂(𝑥)=𝑥18𝑥⁶+𝑥⁴−2𝑥²−30+215−4𝑥²+3𝑥⁴+72𝑥⁶×√1+𝑥²sinh⁻¹(𝑥). (40)
It follows from Lemma 1 and (40) that 𝜙^∗₂(𝑥)>𝑥18𝑥⁶+𝑥⁴−2𝑥²−30+215−4𝑥²+3𝑥⁴+72𝑥⁶𝑥𝑥+³3−2𝑥⁵=𝑥15₅155+2476𝑥²+708𝑥⁴−288𝑥⁶>0 (41) for 𝑥∈(0,0.7).
Therefore, inequality (29) follows from (39) together with (41).

Lemma 4. Let Φ₂(𝑥)=1√1+𝑥²sinh⁻¹(𝑥)−1−𝑥²𝑥1+𝑥². (42) Then the double inequality 5𝑥3−79𝑥³+4511𝑥⁵10<Φ₂(𝑥)<5𝑥3−79𝑥³+459𝑥⁵5 (43) holds for 𝑥∈(0,3/4).

Proof. To prove Lemma 4, it suffices to prove that 𝜑₁(𝑥)∶=𝑥√1+𝑥²−1−𝑥²sinh⁻¹(𝑥)−𝑥1+𝑥²sinh⁻¹(𝑥)5𝑥3−79𝑥³+4511𝑥⁵𝜑10>0,₂(𝑥)∶=𝑥√1+𝑥²−1−𝑥²sinh⁻¹(𝑥)−𝑥1+𝑥²sinh⁻¹(𝑥)5𝑥3−79𝑥³+459𝑥⁵5<0 (44)(45) for 𝑥∈(0,3/4).
We first prove inequality (44). From the expression of 𝜑₁(𝑥), we obtain 𝜑₁(0)=0,𝜑₁34𝜑=0.008457⋯>0,_′1(𝑥)=𝑥𝜑^∗₁(𝑥)90√1+𝑥², (46)(47) where 𝜑^∗₁(𝑥)=120𝑥+8𝑥³+59𝑥⁵−99𝑥⁷×−260−16𝑥²−177𝑥⁴+396𝑥⁶√1+𝑥²sinh⁻¹(𝑥). (48)
Equation (48) leads to 𝜑^∗₁(0.66)=6.02⋯>0,𝜑^∗₁34𝜑=−19.299⋯<0,_∗1′(𝑥)=−𝑥𝜙₁^∗∗(𝑥)1+𝑥², (49)(50) where 𝜑₁^∗∗(𝑥)=−56𝑥−705𝑥³+836𝑥⁵+1485𝑥⁷+144−108𝑥²+213𝑥⁴+396𝑥⁶×√1+𝑥²sinh⁻¹(𝑥). (51)
Note that 60−16𝑥²−177𝑥⁴+396𝑥⁶>60−16×(0.66)²−177×(0.66)⁴=19.4451>0 (52) for 𝑥∈(0,0.66), and 4−108𝑥²+213𝑥⁴+396𝑥⁶3>4−108×4²+213×(0.66)⁴+396×(0.66)⁶=16.3972>0 (53) for 𝑥∈[0.66,3/4).
It follows from Lemma 1, (48), and (51)–(53) that 𝜑^∗₁(𝑥)>120𝑥+8𝑥³+59𝑥⁵−99𝑥⁷−260−16𝑥²−177𝑥⁴+396𝑥⁶×𝑥𝑥+³3−2𝑥⁵+158𝑥⁷=𝑥105₅10546165−82573𝑥²−32420𝑥⁴+7584𝑥⁶+6336𝑥⁶1−𝑥²>𝑥⁵()10546165−82573×0.66²−32420×(0.66)⁴=𝑥⁵105×4044.5917⋯>0 (54) for 𝑥∈(0,0.66), and 𝜑₁^∗∗(𝑥)>−56𝑥−705𝑥³+836𝑥⁵+1485𝑥⁷+144−108𝑥²+213𝑥⁴+396𝑥⁶𝑥𝑥+³3−2𝑥⁵=𝑥15₃15−32975+49598𝑥²+123369𝑥⁴+10668𝑥⁶+11088𝑥⁶1−𝑥²>𝑥³()15−32975+49598×0.66²+123369×(0.66)⁴=𝑥³15×12038.83⋯>0 (55) for 𝑥∈[0.66,3/4).
From (50) and (55), we know that 𝜑^∗₁(𝑥) is strictly decreasing on [0.66,3/4), and this in conjunction with (49) and (54) leads to the conclusion that there exists 𝑥₁∈(0.66,3/4) such that 𝜑^∗₁(𝑥)>0 for 𝑥∈(0,𝑥₁) and 𝜑^∗₁(𝑥)<0 for 𝑥∈(𝑥₁,3/4). Then (47) implies that 𝜑₁(𝑥) is strictly increasing on (0,𝑥₁] and strictly decreasing on [𝑥₁,3/4). Therefore, inequality (44) follows from (46) and the piecewise monotonicity of 𝜑₁(𝑥).
Next, we prove inequality (45). From the expression of 𝜑₂(𝑥) one has 𝜑₂(0)𝜙=0,_′2(𝑥)=−𝑥𝜙^∗₂(𝑥)45√1+𝑥², (56) where 𝜙^∗₂(𝑥)=−60𝑥−4𝑥³+2𝑥⁵+81𝑥⁷+415−4𝑥²+3𝑥⁴+162𝑥⁶×√1+𝑥²sinh⁻¹(𝑥). (57)
It follows from Lemma 1 and (52) that 𝜙^∗₂(𝑥)>−60𝑥−4𝑥³+2𝑥⁵+81𝑥⁷+415−4𝑥²+3𝑥⁴+162𝑥⁶𝑥𝑥+³3−2𝑥⁵=𝑥15₅1510+11027𝑥²+3216𝑥⁴−1296𝑥⁶>0 (58) for 𝑥∈(0,3/4).
Therefore, inequality (45) follows from (56) together with (58).

Lemma 5. Let 𝐿(𝑥) be defined as in Lemma 2 and Υ₁(𝑥)=𝐿(𝑥)2𝑥²+𝑥1+𝑥². (59)
Then the double inequality 4𝑥3−4𝑥³5+4𝑥⁵5<Υ₁(𝑥)<4𝑥3−4𝑥³5+8𝑥⁵7 (60) holds for 𝑥∈(0,0.7).

Proof. From Lemma 2, one has Υ₁(𝑥)−4𝑥3−4𝑥³5+4𝑥⁵5>12𝑥²2𝑥³3+2𝑥⁵5+2𝑥⁷7+𝑥1+𝑥²−4𝑥3−4𝑥³5+4𝑥⁵5=23𝑥⁵351+𝑥²1223−𝑥²Υ>0,₁(𝑥)−4𝑥3−4𝑥³5+8𝑥⁵7<12𝑥²2𝑥³3+2𝑥⁵5+2𝑥⁷7+𝑥⁹+𝑥1+𝑥²−4𝑥3−4𝑥³5+8𝑥⁵7𝑥=−⁷1−𝑥²21+𝑥²<0 (61) for 𝑥∈(0,0.7).
Therefore, Lemma 5 follows easily from (61).

Lemma 6. Let 𝐿(𝑥) be defined as in Lemma 2 and Υ₂(𝑥)=𝐿(𝑥)2𝑥²+2𝑥1+𝑥². (62)
Then the double inequality 7𝑥3−9𝑥³5+7𝑥⁵5<Υ₂(𝑥)<7𝑥3−9𝑥³5+15𝑥⁵7 (63) holds for 𝑥∈(0,3/4).

Proof. It follows from Lemma 2 that Υ₂(𝑥)−7𝑥3−9𝑥³5+7𝑥⁵5>12𝑥²2𝑥³3+2𝑥⁵5+2𝑥⁷7+2𝑥1+𝑥²−7𝑥3−9𝑥³5+7𝑥⁵5=44𝑥⁵351+𝑥²1322−𝑥²Υ>0,₂(𝑥)−7𝑥3−9𝑥³5+15𝑥⁵7<12𝑥²2𝑥³3+2𝑥⁵5+2𝑥⁷7+𝑥⁹+2𝑥1+𝑥²−7𝑥3−9𝑥³5+15𝑥⁵7𝑥=−⁷3−𝑥²21+𝑥²<0 (64) for 𝑥∈(0,3/4).
Therefore, Lemma 6 follows from (64).

Lemma 7. The inequality 𝑥³√1+𝑥²>sinh⁻¹(𝑥)³ (65) holds for 𝑥∈(0,1).

Proof. Let 𝜁(𝑥)=𝑥³√1+𝑥²−sinh⁻¹(𝑥)³. (66) Then 𝜁(0)𝜁=0,_′(𝑥)=𝜁₁(𝑥)1+𝑥^23/2, (67) where 𝜁₁(𝑥)=𝑥²3+2𝑥²−3√1+𝑥²sinh⁻¹(𝑥)². (68)
It follows from Lemma 1 and (68) that 𝜁₁(𝑥)>𝑥²3+2𝑥²𝑥−3𝑥+³3−2𝑥⁵+158𝑥⁷105²=𝑥⁶37+525208+52536𝑥²+17564𝑥⁶×36751−𝑥²+32𝑥⁶735>0 (69) for 𝑥∈(0,1).
Therefore, Lemma 7 follows from (67) together with (69).

Lemma 8. Let 𝜇₁(𝑥)=1+3𝑥²𝑥+𝑥³²−11+𝑥²sinh⁻¹(𝑥)²−𝑥1+𝑥^23/2sinh⁻¹(𝑥). (70)
Then 𝜇₁(𝑥)<0.2 for 𝑥∈[0.7,1).

Proof. Let 𝜔₁(𝑥)=1𝑥²−1sinh⁻¹(𝑥)²,𝜔₂(𝑥)=2√1+𝑥²−𝑥sinh⁻¹(𝑥). (71) Then 𝜇₁(𝑥)=𝜔₁(𝑥)1+𝑥²+𝜔₂(𝑥)1+𝑥^23/2. (72)
Lemma 7 and 𝑥>sinh⁻¹(𝑥) give 𝜔₁(𝑥)<0 and 𝜔^′₁(𝑥)=2𝑥³sinh⁻¹(𝑥)³𝑥³√1+𝑥²−sinh⁻¹(𝑥)³>0 (73) for 𝑥∈(0,1). This in turn implies that 𝜔₁(𝑥)1+𝑥^2′=𝜔^′₁(𝑥)1+𝑥²−2𝑥𝜔₁(𝑥)1+𝑥²²>0 (74) for 𝑥∈(0,1).
On the other hand, from the expression of 𝜔₂(𝑥), we get 𝜔₂(1)𝜔=0.2796⋯>0,_′2(𝑥)=−2𝑥1+𝑥^23/2+𝜔^∗₂(𝑥)sinh⁻¹(𝑥)², (75) where 𝜔^∗₂(𝑥)=𝑥√1+𝑥²−sinh⁻¹(𝑥),𝜔^∗₂(0)𝜔=0,_2∗′(𝑥)𝑥=−²1+𝑥^23/2<0 (76) for 𝑥∈(0,1).
From (75)–(76), we clearly see that 𝜔^′₂(𝑥)<0 and 𝜔₂(𝑥)>0 for 𝑥∈(0,1). This in turn implies that 𝜔₂(𝑥)1+𝑥^23/2′=𝜔^′₂(𝑥)1+𝑥^23/2−3𝑥√1+𝑥²𝜔₂(𝑥)1+𝑥²³<0 (77) for 𝑥∈(0,1).
Equation (72) together with inequalities (74) and (77) lead to the conclusion that 𝜇₁(𝑥)≤𝜔₁(1)2+𝜔₂(0.7)1+(0.7)^23/2=0.167⋯<0.2 (78) for 𝑥∈[0.7,1).

Lemma 9. Let 𝜇₂(𝑥)=1+4𝑥²−𝑥⁴𝑥+𝑥³²−11+𝑥²sinh⁻¹(𝑥)²−𝑥1+𝑥^23/2sinh⁻¹(𝑥). (79) Then 𝜇₂(𝑥)<0.51 for 𝑥∈[0.65,1).

Proof. Let 𝜏₁(𝑥)=1𝑥²−1sinh⁻¹(𝑥)²=𝜇₁(𝑥),𝜏₂(𝑥)=3−𝑥²√1+𝑥²−𝑥sinh⁻¹(𝑥), (80) then 𝜇₂(𝑥)=𝜏₁(𝑥)1+𝑥²+𝜏₂(𝑥)1+𝑥^23/2. (81)
From (74), we clearly see that 𝜏₁(𝑥)1+𝑥^2′=𝜔₁(𝑥)1+𝑥^2′>0 (82) for 𝑥∈(0,1).
On the other hand, from the expression of 𝜏₂(𝑥) together with Lemma 1, we get 𝜏₂(1)𝜏=0.2796⋯>0,_′2(𝑥)1=−sinh⁻¹(𝑥)−𝑥𝜏^∗₂(𝑥)1+𝑥^23/2sinh⁻¹(𝑥)²,𝜏^∗₂(𝑥)=5+𝑥²sinh⁻¹(𝑥)²−1+𝑥²,𝜏^∗₂()𝜏0.65=0.6033⋯,_2∗′(𝑥)=2𝑥sinh⁻¹(𝑥)²+25+𝑥²1+𝑥²√1+𝑥²sinh⁻¹(𝑥)−𝑥>0 (83) for 𝑥∈(0,1).
From (83), we clearly see that 𝜏^′₂(𝑥)<0 and 𝜏₂(𝑥)>0 for 𝑥∈[0.65,1). This in turn implies that 𝜏₂(𝑥)1+𝑥^23/2′=𝜏^′₂(𝑥)1+𝑥^23/2−3𝑥√1+𝑥²𝜏₂(𝑥)1+𝑥²³<0 (84) for 𝑥∈[0.65,1).
Equation (81) together with inequalities (82) and (84) lead to the conclusion that 𝜇₂(𝑥)≤𝜏₁(1)2+𝜏₂(0.65)1+(0.65)^23/2=0.503⋯<0.51 (85) for 𝑥∈[0.65,1).

Lemma 10. Let 𝐿(𝑥) be defined as in Lemma 2 and 𝜈₁(𝑥)=21+𝑥⁴1−𝑥²1+𝑥²²−𝐿(𝑥)𝑥³. (86) Then 𝜈₁(𝑥)>1.2 for x∈[0.7,1).

Proof. Differentiating 𝜈₁(𝑥) yields 𝜈^′₁(𝑥)=3𝐿(𝑥)𝑥⁴−2+8𝑥²−20𝑥⁴−6𝑥⁸𝑥1−𝑥²²1+𝑥²³. (87)
It follows from (19) and (87) that 𝜈^′₁(𝑥)>1𝑥323+2𝑥²5+2𝑥⁴7−2+8𝑥²−20𝑥⁴−6𝑥⁸1−𝑥²²1+𝑥²³=2𝑥−84+316𝑥²−97𝑥⁴+68𝑥⁶+26𝑥⁸+36𝑥¹⁰+15𝑥¹²351−𝑥²²1+𝑥²³>2𝑥351−𝑥²²1+𝑥²³−84+316×(0.7)²−3495+68𝑥⁴𝑥²−25>2𝑥351−𝑥²²1+𝑥²³>0 (88)for 𝑥∈[0.7,1).
Therefore, 𝜈₁(𝑥)≥𝜈₁(0.7)=1.214⋯>1.2 for 𝑥∈