Sharp bounds for Neuman means in terms of one-parameter family of bivariate means

Abstract

We present the best possible parameters $p_1,p_2,p_3,p_4,q_1,q_2,q_3,q_4\in [0,1]$ such that the double inequalities $G_{p_1}(a,b)<S_{HA}(a,b)<G_{q_1}(a,b)$, $Q_{p_2}(a,b)<S_{CA}(a,b)<Q_{q_2}(a,b)$, $H_{p_3}(a,b)<S_{AH}(a,b)<H_{q_3}(a,b)$, $C_{p_4}(a,b)<S_{AC}(a,b)<C_{q_4}(a,b)$ hold for all $a,b>0$ with $a\ne b$, where $S_{HA}$, $S_{CA}$, $S_{AH}$, $S_{AC}$ are the Neuman means, and $G_p$, $Q_p$, $H_p$, $C_p$ are the one-parameter means.

MSC:26E60.

1 Introduction

Let $a,b>0$ with $a\ne b$. Then the Schwab-Borchardt mean $SB(a,b)$ [1–3], and the Neuman means $S_{HA}(a,b)$, $S_{AH}(a,b)$, $S_{CA}(a,b)$, and $S_{AC}(a,b)$ [4, 5] of a and b are given by

$$\begin{array}{c}SB(a,b)=\frac{\sqrt{b^2-a^2}}{{cos}^{-1}(a/b)}(a<b),SB(a,b)=\frac{\sqrt{a^2-b^2}}{{cosh}^{-1}(a/b)}(a>b),\hfill \\ S_{HA}(a,b)=SB[H(a,b),A(a,b)],S_{AH}(a,b)=SB[A(a,b),H(a,b)],\hfill \\ S_{CA}(a,b)=SB[C(a,b),A(a,b)],S_{AC}(a,b)=SB[A(a,b),C(a,b)],\hfill \end{array}$$

respectively. Here, ${cos}^{-1}(x)$ and ${cosh}^{-1}(x)=log(x+\sqrt{x^2-1})$ are, respectively, the inverse cosine and inverse hyperbolic cosine functions, and $H(a,b)=2ab/(a+b)$, $A(a,b)=(a+b)/2$, and $C(a,b)=(a^2+b^2)/(a+b)$ are, respectively, the classical harmonic, arithmetic, and contraharmonic means of a and b.

Let $v=(a-b)/(a+b)\in (-1,1)$, and $p\in (0,\mathrm{\infty })$, $q\in (0,\pi /2)$, $r\in (0,log(2+\sqrt{3}))$, and $s\in (0,\pi /3)$ be the parameters such that $1/cosh(p)=cos(q)=1-v^2$ and $cosh(r)=1/cosh(s)=1+v^2$. Then the following explicit formulas were found by Neuman [4]:

$$S_{AH}(a,b)=A(a,b)\frac{tanh(p)}{p},S_{HA}(a,b)=A(a,b)\frac{sin(q)}{q},$$

(1.1)

$$S_{CA}(a,b)=A(a,b)\frac{sinh(r)}{r},S_{AC}(a,b)=A(a,b)\frac{tan(s)}{s}.$$

(1.2)

Let $p\in [0,1]$ and N be a bivariate symmetric mean. Then the one-parameter bivariate mean $N_p(a,b)$ was defined by Neuman [6] as follows:

$$N_p(a,b)=N[\frac{(1+p)}{2}a+\frac{(1-p)}{2}b,\frac{(1+p)}{2}b+\frac{(1-p)}{2}a].$$

(1.3)

Recently, the Neuman means $S_{AH}$, $S_{HA}$, $S_{CA}$, and $S_{AC}$, and the one-parameter bivariate mean $N_p$ have been the subject of intensive research. He et al. [7] found the greatest values ${\alpha }_1,{\alpha }_2\in [0,1/2]$, and ${\alpha }_3,{\alpha }_4\in [1/2,1]$, and the least values ${\beta }_1,{\beta }_2\in [0,1/2]$, and ${\beta }_3,{\beta }_4\in [1/2,1]$ such that the double inequalities

$$\begin{array}{c}H[{\alpha }_{1}a+(1-{\alpha }_1)b,{\alpha }_{1}b+(1-{\alpha }_1)a]<S_{AH}(a,b)<H[{\beta }_{1}a+(1-{\beta }_1)b,{\beta }_{1}b+(1-{\beta }_1)a],\hfill \\ H[{\alpha }_{2}a+(1-{\alpha }_2)b,{\alpha }_{2}b+(1-{\alpha }_2)a]<S_{HA}(a,b)<H[{\beta }_{2}a+(1-{\beta }_2)b,{\beta }_{2}b+(1-{\beta }_2)a],\hfill \\ C[{\alpha }_{3}a+(1-{\alpha }_3)b,{\alpha }_{3}b+(1-{\alpha }_3)a]<S_{CA}(a,b)<C[{\beta }_{3}a+(1-{\beta }_3)b,{\beta }_{3}b+(1-{\beta }_3)a],\hfill \\ C[{\alpha }_{4}a+(1-{\alpha }_4)b,{\alpha }_{4}b+(1-{\alpha }_4)a]<S_{AC}(a,b)<C[{\beta }_{4}a+(1-{\beta }_4)b,{\beta }_{4}b+(1-{\beta }_4)a]\hfill \end{array}$$

hold for all $a,b>0$ with $a\ne b$.

In [4, 5], Neuman proved that the inequalities

$$\begin{array}{r}H(a,b)<S_{AH}(a,b)<L(a,b)<S_{HA}(a,b)<P(a,b),\\ T(a,b)<S_{CA}(a,b)<Q(a,b)<S_{AC}(a,b)<C(a,b),\\ H^{1/3}(a,b)A^{2/3}(a,b)<S_{HA}(a,b)<\frac{1}{3}H(a,b)+\frac{2}{3}A(a,b),\\ C^{1/3}(a,b)A^{2/3}(a,b)<S_{CA}(a,b)<\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b),\\ A^{1/3}(a,b)H^{2/3}(a,b)<S_{AH}(a,b)<\frac{1}{3}A(a,b)+\frac{2}{3}H(a,b),\\ A^{1/3}(a,b)C^{2/3}(a,b)<S_{AC}(a,b)<\frac{1}{3}A(a,b)+\frac{2}{3}C(a,b)\end{array}$$

(1.4)

hold for all $a,b>0$ with $a\ne b$, where $L(a,b)=(a-b)/(loga-logb)$, $P(a,b)=(a-b)/[2arcsin((a-b)/(a+b))]$, $Q(a,b)=\sqrt{(a^2+b^2)/2}$, and $T(a,b)=(a-b)/[2arctan((a-b)/(a+b))]$ are, respectively, the logarithmic, first Seiffert, quadratic, and second Seiffert means of a and b.

Qian and Chu [8] proved that the double inequalities

$$\begin{array}{c}{\alpha }_{1}A(a,b)+(1-{\alpha }_1)G(a,b)<S_{HA}(a,b)<{\beta }_{1}A(a,b)+(1-{\beta }_1)G(a,b),\hfill \\ {\alpha }_{2}A(a,b)+(1-{\alpha }_2)Q(a,b)<S_{CA}(a,b)<{\beta }_{2}A(a,b)+(1-{\beta }_2)Q(a,b)\hfill \end{array}$$

hold for all $a,b>0$ with $a\ne b$ if and only if ${\alpha }_1\le 1/3$, ${\beta }_1\ge \pi /2$, ${\alpha }_2\ge 1/3$, and ${\beta }_2\le [\sqrt{2}log(2+\sqrt{3})-\sqrt{3}]/[(\sqrt{2}-1)log(2+\sqrt{3})]=0.2394\cdots$, where $G(a,b)=\sqrt{ab}$ is the geometric mean of a and b.

In [9], the authors proved that the double inequalities

$$\begin{array}{c}\begin{array}{r}{\alpha }_1[\frac{H(a,b)}{3}+\frac{2A(a,b)}{3}]+(1-{\alpha }_1)H^{1/3}(a,b)A^{2/3}(a,b)<S_{HA}(a,b)\\ <{\beta }_1[\frac{H(a,b)}{3}+\frac{2A(a,b)}{3}]+(1-{\beta }_1)H^{1/3}(a,b)A^{2/3}(a,b),\end{array}\hfill \\ {\alpha }_2[\frac{C(a,b)}{3}+\frac{2A(a,b)}{3}]+(1-{\alpha }_2)C^{1/3}(a,b)A^{2/3}(a,b)<S_{CA}(a,b)\hfill \\ <{\beta }_2[\frac{C(a,b)}{3}+\frac{2A(a,b)}{3}]+(1-{\beta }_2)C^{1/3}(a,b)A^{2/3}(a,b),\hfill \\ {\alpha }_3[\frac{A(a,b)}{3}+\frac{2H(a,b)}{3}]+(1-{\alpha }_3)A^{1/3}(a,b)H^{2/3}(a,b)<S_{AH}(a,b)\hfill \\ <{\beta }_3[\frac{A(a,b)}{3}+\frac{2H(a,b)}{3}]+(1-{\beta }_3)A^{1/3}(a,b)H^{2/3}(a,b),\hfill \\ {\alpha }_4[\frac{A(a,b)}{3}+\frac{2C(a,b)}{3}]+(1-{\alpha }_4)A^{1/3}(a,b)C^{2/3}(a,b)<S_{AC}(a,b)\hfill \\ <{\beta }_4[\frac{A(a,b)}{3}+\frac{2C(a,b)}{3}]+(1-{\beta }_4)A^{1/3}(a,b)C^{2/3}(a,b)\hfill \end{array}$$

hold for all $a,b>0$ with $a\ne b$ if and only if ${\alpha }_1\le 4/5$, ${\beta }_1\ge 3/\pi$, ${\alpha }_2\le 3[\sqrt[3]{2}log(2+\sqrt{3})-\sqrt{3}]/[(3\sqrt[3]{2}-4)log(2+\sqrt{3})]=0.7528\cdots$, ${\beta }_2\ge 4/5$, ${\alpha }_3\le 0$, ${\beta }_3\ge 4/5$, ${\alpha }_4\le 4/5$, and ${\beta }_4\ge 3(3\sqrt{3}-\sqrt[3]{4}\pi )/[(5-3\sqrt[3]{4})\pi ]=0.8400\cdots$.

Let $p,p_i,q_i,{\alpha }_j,{\beta }_j\in [0,1]$ ($i,j=1,2,\dots ,8$). Then Neuman [6, 10] proved that the inequalities

$$\begin{array}{c}{H}_{p_1}(a,b)<P(a,b)<H_{q_1}(a,b),G_{p_2}(a,b)<P(a,b)<G_{q_2}(a,b),\hfill \\ Q_{p_3}(a,b)<T(a,b)<Q_{q_3}(a,b),C_{p_4}(a,b)<T(a,b)<C_{q_4}(a,b),\hfill \\ Q_{p_5}(a,b)<M(a,b)<Q_{q_5}(a,b),C_{p_6}(a,b)<M(a,b)<C_{q_6}(a,b),\hfill \\ H_{p_7}(a,b)<L(a,b)<H_{q_7}(a,b),G_{p_8}(a,b)<L(a,b)<G_{q_8}(a,b),\hfill \\ {\alpha }_{1}A(a,b)+(1-{\alpha }_1)G_p(a,b)<P_p(a,b)<{\beta }_{1}A(a,b)+(1-{\beta }_1)G_p(a,b),\hfill \\ {\alpha }_2{Q}_p(a,b)+(1-{\alpha }_2)A(a,b)<T_p(a,b)<{\beta }_2{Q}_p(a,b)+(1-{\beta }_2)A(a,b),\hfill \\ {\alpha }_3{Q}_p(a,b)+(1-{\alpha }_3)A(a,b)<M_p(a,b)<{\beta }_3{Q}_p(a,b)+(1-{\beta }_3)A(a,b),\hfill \\ {\alpha }_{4}A(a,b)+(1-{\alpha }_4)G_p(a,b)<L_p(a,b)<{\beta }_{4}A(a,b)+(1-{\beta }_4)G_p(a,b),\hfill \\ A^{{\alpha }_5}(a,b)G_p^{1-{\alpha }_5}(a,b)<P_p(a,b)<A^{{\beta }_5}(a,b)G_p^{1-{\beta }_5}(a,b),\hfill \\ Q_p^{{\alpha }_6}(a,b)A^{1-{\alpha }_6}(a,b)<T_p(a,b)<Q_p^{{\beta }_6}(a,b)A^{1-{\beta }_6}(a,b),\hfill \\ Q_p^{{\alpha }_7}(a,b)A^{1-{\alpha }_7}(a,b)<M_p(a,b)<Q_p^{{\beta }_7}(a,b)A^{1-{\beta }_7}(a,b),\hfill \\ A^{{\alpha }_8}(a,b)G_p^{1-{\alpha }_8}(a,b)<L_p(a,b)<A^{{\beta }_8}(a,b)G_p^{1-{\beta }_8}(a,b),\hfill \end{array}$$

hold for all $a,b>0$ with $a\ne b$ if and only if $p_1\ge \sqrt{1-2/\pi }$, $q_1\le \sqrt{6}/6$, $p_2\ge \sqrt{1-4/{\pi }^2}$, $q_2\le \sqrt{3}/3$, $p_3\le \sqrt{16/{\pi }^2-1}$, $q_3\ge \sqrt{6}/3$, $p_4\le \sqrt{4/\pi -1}$, $q_4\ge \sqrt{3}/3$, $p_5\le \sqrt{1/{log}^2(1+\sqrt{2})-1}$, $q_5\ge \sqrt{3}/3$, $p_6\le \sqrt{1/log(1+\sqrt{2})-1}$, $q_6\ge \sqrt{6}/6$, $p_7=1$, $q_7\le \sqrt{3}/3$, $p_8=1$, $q_8\le \sqrt{6}/3$, ${\alpha }_1\le 2/\pi$, ${\beta }_1\ge 2/3$, ${\alpha }_2\le (4-\pi )/[(\sqrt{2}-1)\pi ]$, ${\beta }_2\ge 2/3$, ${\alpha }_3\le [1-log(1+\sqrt{2})]/[(\sqrt{2}-1)log(1+\sqrt{2})]$, ${\beta }_3\ge 1/3$, ${\alpha }_4=0$, ${\beta }_4\ge 1/3$, ${\alpha }_5\le 2/3$, ${\beta }_5=1$, ${\alpha }_6\le 2/3$, ${\beta }_6\ge (4log2-2log\pi )/log2$, ${\alpha }_7\le 1/3$, ${\beta }_7\ge -log[log(1+\sqrt{2})]/log[cosh(log(1+\sqrt{2}))]$, ${\alpha }_8\le 1/3$, ${\beta }_8=1$, where $M(a,b)=(a-b)/[2{sinh}^{-1}((a-b)/(a+b))]$ is the Neuman-Sándor mean of a and b.

The main purpose of this paper is to present the best possible parameters $p_1$, $p_2$, $p_3$, $p_4$, $q_1$, $q_2$, $q_3$, $q_4$ on the interval $[0,1]$ such that the double inequalities

$$\begin{array}{c}{G}_{p_1}(a,b)<S_{HA}(a,b)<G_{q_1}(a,b),Q_{p_2}(a,b)<S_{CA}(a,b)<Q_{q_2}(a,b),\hfill \\ H_{p_3}(a,b)<S_{AH}(a,b)<H_{q_3}(a,b),C_{p_4}(a,b)<S_{AC}(a,b)<C_{q_4}(a,b)\hfill \end{array}$$

hold for all $a,b>0$ with $a\ne b$.

2 Main results

Theorem 2.1 Let $p_1,q_1\in [0,1]$. Then the double inequality

$$G_{p_1}(a,b)<S_{HA}(a,b)<G_{q_1}(a,b)$$

(2.1)

holds for all $a,b>0$ with $a\ne b$ if and only if $p_1\ge \sqrt{6}/3$ and $q_1\le \sqrt{1-4/{\pi }^2}$.

Proof Without loss of generality, we assume that $a>b$. Let $v=(a-b)/(a+b)$, $\lambda =v\sqrt{2-v^2}$, $x=\sqrt{1-{\lambda }^2}$ and $p\in [0,1]$. Then $v,\lambda ,x\in (0,1)$, and (1.1) and (1.3) lead to

$$\begin{array}{rl}{S}_{HA}(a,b)-G_p(a,b)& =A(a,b)[\frac{\lambda }{arcsin(\lambda )}-\sqrt{1-p^2(1-\sqrt{1-{\lambda }^2})}]\\ =\frac{A(a,b)\sqrt{1-p^2(1-\sqrt{1-{\lambda }^2})}}{arcsin(\lambda )}F(x),\end{array}$$

(2.2)

where

$$F(x)=\frac{\sqrt{1-x^2}}{\sqrt{1-p^2(1-x)}}-arcsin\left(\sqrt{1-x^2}\right),$$

(2.3)

$$F(0)=\frac{1}{\sqrt{1-p^2}}-\frac{\pi }{2},F(1)=0,$$

(2.4)

$$F^{\prime }(x)=-\frac{(1-x)f(x)}{2\sqrt{1-x^2}{(p^{2}x+1-p^2)}^{3/2}[2{(p^{2}x+1-p^2)}^{3/2}+p^{2}x+2(1-p^2)x+p^2]},$$

(2.5)

where

$$\begin{array}{rl}f(x)=& -p^4{x}^3+(4p^6+3p^4-4p^2)x^2\\ +(-8p^6+9p^4+4p^2-4)x+(4p^6-11p^4+12p^2-4),\end{array}$$

(2.6)

$$f^{\prime }(x)=-3p^4{x}^2+2(4p^6+3p^4-4p^2)x+(-8p^6+9p^4+4p^2-4).$$

(2.7)

We divide the discussion into two cases.

Case 1 $p=\sqrt{6}/3$. Then (2.6) becomes

$$f(x)=\frac{4}{27}(1-x)(3x^2+4x+2).$$

(2.8)

From (2.5) and (2.8) we clearly see that $F(x)$ is strictly decreasing on $[0,1]$, then (2.4) leads to the conclusion that

$$F(x)>0$$

(2.9)

for all $x\in (0,1)$.

Therefore,

$$S_{HA}(a,b)>G_{\sqrt{6}/3}(a,b)$$

(2.10)

for all $a,b>0$ with $a\ne b$ follows from (2.2) and (2.9).

Case 2 $p=\sqrt{1-4/{\pi }^2}$. Then numerical computations lead to

$$4p^6+3p^4-4p^2=\frac{3{\pi }^6-56{\pi }^4+240{\pi }^2-256}{{\pi }^6}<0,$$

(2.11)

$$-8p^6+9p^4+4p^2-4=\frac{{\pi }^6+8{\pi }^4-240{\pi }^2+512}{{\pi }^6}<0,$$

(2.12)

$$f(0)=4p^6-11p^4+12p^2-4=\frac{{\pi }^6-8{\pi }^4+16{\pi }^2-256}{{\pi }^6}>0,$$

(2.13)

$$f(1)=4(3p^2-2)=-\frac{4(12-{\pi }^2)}{{\pi }^2}<0.$$

(2.14)

It follows from (2.7) and (2.11) together with (2.12) that $f(x)$ is strictly decreasing on $[0,1]$. Then inequalities (2.13) and (2.14) together with (2.5) lead to the conclusion that there exists ${\lambda }_1\in (0,1)$ such that $F(x)$ is strictly decreasing on $[0,{\lambda }_1]$ and strictly increasing on $[{\lambda }_1,1]$.

Note that inequality (2.4) becomes

$$F(0)=F(1)=0.$$

(2.15)

From (2.2), (2.15), and the piecewise monotonicity of $F(x)$ we clearly see that the inequality

$$S_{HA}(a,b)<G_{\sqrt{1-4/{\pi }^2}}(a,b)$$

(2.16)

holds for all $a,b>0$ with $a\ne b$.

Note that

$$\underset{\lambda \to 0^{+}}{lim}\frac{\sqrt{{arcsin}^2(\lambda )-{\lambda }^2}}{arcsin(\lambda )\sqrt{1-\sqrt{1-{\lambda }^2}}}=\frac{\sqrt{6}}{3},$$

(2.17)

$$\underset{\lambda \to 1}{lim}\frac{\sqrt{{arcsin}^2(\lambda )-{\lambda }^2}}{arcsin(\lambda )\sqrt{1-\sqrt{1-{\lambda }^2}}}=\sqrt{1-\frac{4}{{\pi }^2}}.$$

(2.18)

Therefore, Theorem 2.1 follows from (2.10) and (2.16)-(2.18) together with the fact that inequality (2.1) is equivalent to the inequality (2.19) as follows:

$$q_1<\frac{\sqrt{{arcsin}^2(\lambda )-{\lambda }^2}}{arcsin(\lambda )\sqrt{1-\sqrt{1-{\lambda }^2}}}<p_1.$$

(2.19)

 □

Theorem 2.2 Let $p_2,q_2\in [0,1]$. Then the double inequality

$$Q_{p_2}(a,b)<S_{CA}(a,b)<Q_{q_2}(a,b)$$

(2.20)

holds for all $a,b>0$ with $a\ne b$ if and only if $p_2\le \sqrt{6}/3$ and $q_2\ge \sqrt{3/{log}^2(2+\sqrt{3})-1}=0.8542\cdots$.

Proof Without loss of generality, we assume that $a>b$. Let $v=(a-b)/(a+b)$, $\mu =v\sqrt{2+v^2}$, $x=\sqrt{1+{\mu }^2}$, and $p\in [0,1]$. Then $v\in (0,1)$, $\mu \in (0,\sqrt{3})$, $x\in (1,2)$, and (1.2) and (1.3) lead to

$$\begin{array}{rl}{S}_{CA}(a,b)-Q_p(a,b)& =A(a,b)[\frac{\mu }{{sinh}^{-1}(\mu )}-\sqrt{1+p^2(\sqrt{1+{\mu }^2}-1)}]\\ =\frac{A(a,b)\sqrt{1+p^2(\sqrt{1+{\mu }^2}-1)}}{{sinh}^{-1}(\mu )}G(x),\end{array}$$

(2.21)

where

$$\begin{array}{r}G(x)=\frac{\sqrt{x^2-1}}{\sqrt{1+p^2(x-1)}}-{sinh}^{-1}\left(\sqrt{x^2-1}\right),\\ G(1)=0,G(2)=\frac{\sqrt{3}}{\sqrt{1+p^2}}-log(2+\sqrt{3}),\end{array}$$

(2.22)

$$G^{\prime }(x)=-\frac{(x-1)f(x)}{2\sqrt{x^2-1}{(p^{2}x+1-p^2)}^{3/2}[p^2{x}^2+2{(p^{2}x+1-p^2)}^{3/2}+2(1-p^2)x+p^2]},$$

(2.23)

where $f(x)$ is defined by (2.6).

We divide the discussion into two cases.

Case 1 $p=\sqrt{6}/3$. Then it follows from (2.6) that

$$f(x)=-\frac{4}{27}(x-1)(3x^2+4x+2)<0$$

(2.24)

for all $x\in (1,2)$.

Therefore,

$$S_{CA}(a,b)>Q_{\sqrt{6}/3}(a,b)$$

(2.25)

for all $a,b>0$ with $a\ne b$ follows easily from (2.21)-(2.24).

Case 2 $p=\sqrt{3/{log}^2(2+\sqrt{3})-1}$. Then numerical computations lead to

$$4p^6+3p^4-4p^2=0.2329\cdots >0,$$

(2.26)

$$-8p^6+9p^4+4p^2-4=0.6027\cdots >0,$$

(2.27)

$$3p^4-p^2-1=-0.1322\cdots <0,$$

(2.28)

$$f(1)=4(3p^2-2)=0.7567\cdots >0,$$

(2.29)

$$f(2)=4p^6+11p^4+4p^2-12=-1.669\cdots <0.$$

(2.30)

It follows from (2.7) and (2.26)-(2.28) that

$$\begin{array}{rl}{f}^{\prime }(x)& <-3p^4{x}^2+2(4p^6+3p^4-4p^2)x^2+(-8p^6+9p^4+4p^2-4)x^2\\ =4(3p^4-p^2-1)x^2<0\end{array}$$

(2.31)

for $x\in (1,2)$.

Equation (2.23) and inequalities (2.29)-(2.31) lead to the conclusion that there exists ${\lambda }_2\in (1,2)$ such that $G(x)$ is strictly decreasing on $[0,{\lambda }_2]$ and strictly increasing on $[{\lambda }_2,1]$.

Note that (2.22) becomes

$$G(1)=G(2)=0.$$

(2.32)

Therefore,

$$S_{CA}(a,b)<Q_{\sqrt{3/{log}^2(2+\sqrt{3})-1}}(a,b)$$

(2.33)

for all $a,b>0$ with $a\ne b$ follows from (2.21) and (2.32) together with the piecewise monotonicity of $G(x)$.

Note that

$$\underset{\mu \to 0^{+}}{lim}\frac{\sqrt{{\mu }^2-{[{sinh}^{-1}(\mu )]}^2}}{{sinh}^{-1}(\mu )\sqrt{\sqrt{1+{\mu }^2}-1}}=\frac{\sqrt{6}}{3},$$

(2.34)

$$\underset{\mu \to \sqrt{3}}{lim}\frac{\sqrt{{\mu }^2-{[{sinh}^{-1}(\mu )]}^2}}{{sinh}^{-1}(\mu )\sqrt{\sqrt{1+{\mu }^2}-1}}=\sqrt{\frac{3}{{log}^2(2+\sqrt{3})}-1}.$$

(2.35)

Therefore, Theorem 2.2 follows from (2.25) and (2.33)-(2.35) together with the fact that inequality (2.20) is equivalent to the inequality (2.36) as follows:

$$p_2<\frac{\sqrt{{\mu }^2-{[{sinh}^{-1}(\mu )]}^2}}{{sinh}^{-1}(\mu )\sqrt{\sqrt{1+{\mu }^2}-1}}<q_2.$$

(2.36)

 □

Theorem 2.3 Let $p_3,q_3\in [0,1]$. Then the double inequality

$$H_{p_3}(a,b)<S_{AH}(a,b)<H_{q_3}(a,b)$$

(2.37)

holds for all $a,b>0$ with $a\ne b$ if and only if $p_3=1$ and $q_3\le \sqrt{6}/3$.

Proof Without loss of generality, we assume that $a>b$. Let $v=(a-b)/(a+b)$, $\lambda =v\sqrt{2-v^2}$, $x=\sqrt{1-{\lambda }^2}$ and $p\in [0,1]$. Then $v,\lambda ,x\in (0,1)$, and (1.1) and (1.3) lead to

$$\begin{array}{rl}{S}_{AH}(a,b)-H_p(a,b)& =A(a,b)[\frac{\lambda }{{tanh}^{-1}(\lambda )}+p^2(1-\sqrt{1-{\lambda }^2})-1]\\ =\frac{A(a,b)[1-p^2(1-\sqrt{1-{\lambda }^2})]}{{tanh}^{-1}(\lambda )}H(x),\end{array}$$

(2.38)

where

$$\begin{array}{r}H(x)=\frac{\sqrt{1-x^2}}{p^{2}x+(1-p^2)}-{tanh}^{-1}\left(\sqrt{1-x^2}\right),\\ H(1)=0,\end{array}$$

(2.39)

$$H^{\prime }(x)=-\frac{1-x}{x\sqrt{1-x^2}{[p^{2}x+(1-p^2)]}^2}g(x),$$

(2.40)

where

$$g(x)=(p^4+p^2-1)x-p^4+2p^2-1.$$

(2.41)

We divide the discussion into two cases.

Case 1 $p=\sqrt{6}/3$. Then (2.41) leads to

$$g(x)=-\frac{1}{9}(1-x)<0$$

(2.42)

for $x\in (0,1)$.

Therefore,

$$S_{AH}(a,b)<H_{\sqrt{6}/3}(a,b)$$

(2.43)

for all $a,b>0$ with $a\ne b$ follows easily from (2.38)-(2.40) and (2.42).

Case 2 $p=1$. Then it follows from (1.3) and (1.4) that

$$S_{AH}(a,b)>H(a,b)=H_1(a,b)$$

(2.44)

for all $a,b>0$ with $a\ne b$.

Note that

$$\underset{\lambda \to 0^{+}}{lim}\sqrt{\frac{{tanh}^{-1}(\lambda )-\lambda }{{tanh}^{-1}(\lambda )(1-\sqrt{1-{\lambda }^2})}}=\frac{\sqrt{6}}{3},$$

(2.45)

$$\underset{\lambda \to 1}{lim}\sqrt{\frac{{tanh}^{-1}(\lambda )-\lambda }{{tanh}^{-1}(\lambda )(1-\sqrt{1-{\lambda }^2})}}=1.$$

(2.46)

Therefore, Theorem 2.3 follows from (2.43)-(2.46) and the fact that inequality (2.37) is equivalent to

$$q_3<\sqrt{\frac{{tanh}^{-1}(\lambda )-\lambda }{{tanh}^{-1}(\lambda )(1-\sqrt{1-{\lambda }^2})}}<p_3.$$

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Theorem 2.4 Let $p_4,q_4\in [0,1]$. Then the double inequality

$$C_{p_4}(a,b)<S_{AC}(a,b)<C_{q_4}(a,b)$$

(2.47)

holds for all $a,b>0$ with $a\ne b$ if and only if $p_4\le \sqrt{3\sqrt{3}/\pi -1}$ and $q_4\ge \sqrt{6}/3$.

Proof Without loss of generality, we assume that $a>b$. Let $v=(a-b)/(a+b)$, $\mu =v\sqrt{2+v^2}$, $x=\sqrt{1+{\mu }^2}$, and $p\in [0,1]$. Then $v\in (0,1)$, $\mu \in (0,\sqrt{3})$, $x\in (1,2)$, and (1.2) and (1.3) lead to

$$\begin{array}{rl}{S}_{AC}(a,b)-C_p(a,b)& =A(a,b)[\frac{\mu }{arctan(\mu )}-p^2(\sqrt{1+{\mu }^2}-1)-1]\\ =\frac{A(a,b)[1+p^2(\sqrt{1+{\mu }^2}-1)]}{arctan(\mu )}J(x),\end{array}$$

(2.48)

where

$$\begin{array}{r}J(x)=\frac{\sqrt{x^2-1}}{p^{2}x+(1-p^2)}-arctan\left(\sqrt{x^2-1}\right),\\ J(1)=0,J(2)=\frac{\sqrt{3}}{p^2+1}-\frac{\pi }{3},\end{array}$$

(2.49)

$$J^{\prime }(x)=-\frac{x-1}{x\sqrt{x^2-1}{[p^{2}x+(1-p^2)]}^2}g(x),$$

(2.50)

where $g(x)$ is defined by (2.41).

We divide the discussion into two cases.

Case 1 $p=\sqrt{6}/3$. Then (2.41) leads to

$$g(x)=\frac{1}{9}(x-1)>0$$

(2.51)

for $x\in (1,2)$.

Therefore,

$$S_{AC}(a,b)<C_{\sqrt{6}/3}(a,b)$$

(2.52)

for all $a,b>0$ with $a\ne b$ follows easily from (2.48)-(2.51).

Case 2 $p=\sqrt{3\sqrt{3}/\pi -1}$. Then numerical computations lead to

$$p^4+p^2-1=\frac{27-{\pi }^2-3\sqrt{3}\pi }{{\pi }^2}>0,$$

(2.53)

$$g(1)=3p^2-2=\frac{9\sqrt{3}-5\pi }{\pi }<0,$$

(2.54)

$$g(2)=p^4+4p^2-3=\frac{27-6{\pi }^2+6\sqrt{3}\pi }{{\pi }^2}>0.$$

(2.55)

From (2.41) and (2.50) together with (2.53)-(2.55) we clearly see that there exists ${\lambda }_3\in (1,2)$ such that $J(x)$ is strictly increasing on $[1,{\lambda }_3]$ and strictly decreasing on $[{\lambda }_3,2]$.

Note that (2.49) becomes

$$J(1)=J(2)=0.$$

(2.56)

It follows from (2.56) and the piecewise monotonicity of $J(x)$ that

$$J(x)>0$$

(2.57)

for all $x\in (1,2)$.

Therefore,

$$S_{AC}(a,b)>C_{\sqrt{3\sqrt{3}/\pi -1}}(a,b)$$

(2.58)

for all $a,b>0$ with $a\ne b$ follows from (2.48) and (2.58).

Note that

$$\underset{\mu \to 0^{+}}{lim}\sqrt{\frac{\mu -arctan(\mu )}{arctan(\mu )(\sqrt{1+{\mu }^2}-1)}}=\frac{\sqrt{6}}{3},$$

(2.59)

$$\underset{\mu \to 1}{lim}\sqrt{\frac{\mu -arctan(\mu )}{arctan(\mu )(\sqrt{1+{\mu }^2}-1)}}=\sqrt{\frac{3\sqrt{3}}{\pi }-1}.$$

(2.60)

Therefore, Theorem 2.4 follows from (2.52) and (2.58)-(2.60) together with the fact that inequality (2.47) is equivalent to

$$p_4<\sqrt{\frac{\mu -arctan(\mu )}{arctan(\mu )(\sqrt{1+{\mu }^2}-1)}}<q_4.$$

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