Optimal bounds for Neuman-Sándor mean in terms of the geometric convex combination of two Seiffert means

Abstract

In this paper, we find the least value α and the greatest value β such that the double inequality

$$P^{\alpha}(a,b)T^{1-\alpha}(a,b)< M(a,b)< P^{\beta}(a,b)T^{1-\beta}(a,b) $$

holds for all $a,b>0$ with $a\neq b$, where $M(a,b)$, $P(a,b)$, and $T(a,b)$ are the Neuman-Sándor, the first and second Seiffert means of two positive numbers a and b, respectively.

1 Introduction

For $a,b>0$ with $a\neq b$, the Neuman-Sándor mean $M(a,b)$ [1], the first Seiffert mean $P(a,b)$ [2], and the second Seiffert mean $T(a,b)$ [3] are defined by

$$\begin{aligned}& M(a,b)=\frac{a-b}{2\sinh^{-1}(\frac{a-b}{a+b})}, \end{aligned}$$

(1.1)

$$\begin{aligned}& P(a,b)=\frac{a-b}{4\tan^{-1}(\sqrt{a/b})-\pi}, \\& T(a,b)=\frac{a-b}{2\tan^{-1}(\frac{a-b}{a+b})}, \end{aligned}$$

(1.2)

respectively. It can be observed that the first Seiffert mean $P(a,b)$ can be rewritten as (see [1])

$$ P(a,b)= \frac{a-b}{2\sin^{-1}(\frac{a-b}{a+b})}, $$

(1.3)

where $\sinh^{-1}(x)=\log(x+\sqrt{x^{2}+1})$, $\tan^{-1}(x)=\arctan (x)$, and $\sin^{-1}(x)=\arcsin(x)$ are the inverse hyperbolic sine, inverse tangent, inverse sine functions, respectively.

Recently, the means M, P, and T and other means have been the subject of intensive research. Many remarkable inequalities for means can be found in the literature [4–10].

Let $H(a,b)=2ab/(a+b)$, $G(a,b)=\sqrt{ab}$, $L(a, b)=(b-a)/(\log b-\log a)$, $I(a,b)=1/e(b^{b}/ a^{a})^{1/(b-a)}$, $A(a,b)=(a+b)/2$, $Q(a,b)=\sqrt{(a^{2}+b^{2})/2}$ and

$$ M_{p}(a,b)=\left \{ \textstyle\begin{array}{l@{\quad}l} (\frac{a^{p}+b^{p}}{2})^{1/p}, & p\neq0, \\ \sqrt{ab}, & p=0, \end{array}\displaystyle \right . $$

denote the harmonic, geometric, logarithmic, identric, arithmetic, root-square, and the pth power means of two positive numbers a and b with $a\neq b$, respectively. Then it is well known that the inequalities

$$H(a,b)< G(a,b)< L(a,b)< P(a,b)< I(a,b)< A(a,b)< M(a,b)< T(a,b)< Q(a,b) $$

hold for $a,b>0$ with $a\neq b$.

Neuman and Sándor [1] established

$$\frac{\pi}{2}P(a,b)>\sinh^{-1}(1)M(a,b)>\frac{\pi}{4}T(a,b) $$

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

Gao [11] proved that the optimal double inequalities

$$\frac{e}{\pi}I(a,b)< P(a,b)< I(a,b), \qquad I(a,b)< T(a,b)< \frac{2e}{\pi}I(a,b) $$

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

The following bounds for the Seiffert means $P(a,b)$ and $T(a,b)$ in terms of the power mean were presented by Jagers in [12]:

$$M_{\frac{1}{2}}< P(a,b)< M_{\frac{2}{3}}(a,b) $$

for all $a,b>0$ with $a\neq b$. Hästö [13] improved the results of [12] and found the sharp lower power mean bound for the Seiffert mean $P(a,b)$ as follows:

$$P(a,b)>M_{\frac{\log2}{\log\pi}}(a,b) $$

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

In [14], the authors proved that the sharp double inequality

$$M_{\frac{\log2}{\log\pi-\log2}}< T(a,b)< M_{\frac{5}{3}}(a,b) $$

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

Let $\overline{L}_{p}(a,b)=(a^{p+1}+b^{p+1})/(a^{p}+b^{p})$ be the Lehmer mean of two positive numbers a and b with $a\neq b$. In [15], the authors presented the following best possible Lehmer mean bounds for the Seiffert means $P(a,b)$ and $T(a,b)$:

$$ \overline{L}_{-1/6}(a,b)< P(a,b)< \overline {L}_{0}(a,b) \quad \mbox{and}\quad \overline{L}_{0}(a,b)< T(a,b)< \overline{L}_{1/3}(a,b) $$

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

Let u, v, and w be bivariate means such that $u(a,b)< v(a,b)< w(a,b)$ for all $a,b>0$ with $a\neq b$. The problems of finding the best possible parameters α and β such that the inequalities $\alpha u(a,b)+(1-\alpha)v(a,b)< w(a,b)<\beta u(a,b)+(1-\beta)v(a,b)$ and $u(a,b)^{\alpha}v^{1-\alpha}(a,b)< w(a,b)< u(a,b)^{\beta}v^{1-\beta }(a,b)$ hold for all $a,b>0$ with $a\neq b$ have attracted the interest of many mathematicians.

In [16] and [17], the authors proved that the double inequalities

$$\begin{aligned}& \alpha_{1}Q(a,b)+(1-\alpha_{1})A(a,b)< T(a,b)< \beta _{1}Q(a,b)+(1-\beta_{1})A(a,b), \\& Q^{\alpha_{2}}(a,b)A^{1-\alpha_{2}}(a,b)< T(a,b)< Q^{\beta _{2}}(a,b)A^{1-\beta_{2}}(a,b) \end{aligned}$$

hold for all $a,b>0$ with $a\neq b$ if and only if $\alpha_{1}\leq (4-\pi)/[(\sqrt{2}-1)\pi]$, $\beta_{1}\geq2/3$, $\alpha_{2}\leq2/3$, $\beta_{2}\geq4-2\log\pi/\log2$.

In [1], Neuman and Sándor gave the inequality

$$Q(a,b)^{\frac{1}{3}}A(a,b)^{\frac{2}{3}}< M(a,b)< {\frac {1}{3}}Q(a,b)+{ \frac{2}{3}}A(a,b). $$

In [8], Sándor proved the inequality

$$G(a,b)^{\frac{1}{3}}A(a,b)^{\frac{2}{3}}< P(a,b)< {\frac {1}{3}}G(a,b)+{ \frac{2}{3}}A(a,b). $$

In [18] and [19], the authors proved that the double inequalities

$$\begin{aligned}& Q(a,b)^{\alpha_{3}}A^{1-\alpha_{3}}(a,b)< M(a,b)< Q(a,b)^{\beta _{3}}A^{1-\beta_{3}}(a,b), \\& \alpha_{4}Q(a,b)+(1-\alpha_{4})G(a,b)< M(a,b)< \beta _{4}Q(a,b)+(1-\beta_{4})G(a,b) \end{aligned}$$

hold for all $a,b>0$ with $a\neq b$ if and only if $\alpha_{3}\leq 1/3$, $\beta_{3}\geq2(\log(2+\sqrt{2})-\log3)/\log2$, $\alpha _{4}\leq2/3$, $\beta_{4}\geq1/[\sqrt{2}\log(1+\sqrt{2})]$.

In [20], the authors proved that the double inequality

$$\alpha_{5}A(a,b)+(1-\alpha_{5})G(a,b)< P(a,b)< \beta _{5}A(a,b)+(1-\beta_{5})G(a,b) $$

holds for all $a,b>0$ with $a\neq b$ if and only if $\alpha_{5}\leq \pi/2$, $\beta_{5}\geq2/3$.

The main purpose of this paper is to find the least value α and the greatest value β such that the double inequality

$$P^{\alpha}(a,b)T^{1-\alpha}(a,b)< M(a,b)< P^{\beta}(a,b)T^{1-\beta}(a,b) $$

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

Theorem 3.1 and Theorem 3.3 in [21] provide the inequality

$$A(a,b)T(a,b)\leq M^{2}(a,b), P(a,b)M(a,b)\leq A^{2}(a,b), $$

following which one can get $P^{\frac{1}{3}}(a,b)T^{\frac{2}{3}}\leq M(a,b)$. Then the lower bound of α in Theorem 3.1 of Section 3 is achieved.

2 Lemmas

To establish our main result, we need several lemmas, which we present in this section.

For $x\in(0,1)$, the power series expansions of the functions $\tan ^{-1}(x)$ and $\sinh^{-1} (x)$ are presented as follows:

$$\begin{aligned}& \sinh^{-1} (x) =x-\frac{x^{3}}{6}+\frac{3x^{5}}{40}- \frac {15x^{7}}{336}+\cdots=\sum^{\infty}_{n=0} \frac {(-1)^{n}(2n)!}{(2n+1)2^{2n}(n!)^{2}}x^{2n+1}, \end{aligned}$$

(2.1)

$$\begin{aligned}& \tan^{-1}(x)=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}- \frac{x^{7}}{7}+\cdots =\sum^{\infty}_{n=0} \frac{(-1)^{n}}{2n+1}x^{2n+1}. \end{aligned}$$

(2.2)

Lemma 2.1

If $x\in(0,1)$, then one has

$$\begin{aligned}& 1+\frac{6x^{2}}{15}< \sqrt{1+x^{2}}< 1+\frac{x^{2}}{2}, \end{aligned}$$

(2.3)

$$\begin{aligned}& \sin^{-1}(x)\sqrt{1-x^{2}}< x-\frac{x^{3}}{3}- \frac{2x^{5}}{15}, \end{aligned}$$

(2.4)

$$\begin{aligned}& \sinh^{-1}(x)>x-\frac{x^{3}}{6}, \end{aligned}$$

(2.5)

$$\begin{aligned}& x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}< \tan^{-1}(x)< x- \frac {x^{3}}{3}+\frac{x^{5}}{5} , \end{aligned}$$

(2.6)

and

$$ \tan^{-1}(x)>x-\frac{x^{3}}{3}+\frac{x^{5}}{9}. $$

(2.7)

Proof

Square every terms of inequality (2.3) at the same time, then it is easy to prove it. Inequality (2.4) was proved in Lemma 2.3 of [22]. Inequalities (2.5) and (2.6) follow immediately from equations (2.1) and (2.2), respectively.

Let $\Phi(x)=\tan^{-1}(x)-(x-\frac{x^{3}}{3}+\frac{x^{5}}{9})$. Then $\Phi'(x)=\frac{x^{4}(4-5x^{2})}{9(1+x^{2})}$. Thus, $\Phi(x)$ is strictly increasing on $(0,\frac{2\sqrt{5}}{5}]$ and strictly decreasing on $[\frac{2\sqrt{5}}{5},1)$. Considering $\Phi(0)=0$ and $\Phi(1)=0.0076\ldots>0$, we can get $\Phi(x)>0$ for $x \in(0,1)$. Therefore, inequality (2.7) holds. □

Lemma 2.2

If $x\in(0,0.7)$, then one has

$$\begin{aligned}& \sin^{-1}(x)\sqrt{1-x^{2}}>x-\frac{x^{3}}{3}- \frac{x^{5}}{5}, \end{aligned}$$

(2.8)

$$\begin{aligned}& \tan^{-1}(x)>x-\frac{x^{3}}{3}+\frac{x^{5}}{7}, \end{aligned}$$

(2.9)

and

$$ \sinh^{-1}(x)< x-\frac{2x^{3}}{15}. $$

(2.10)

Proof

Let

$$\begin{aligned}& \gamma_{1}(x)=\sin^{-1}(x)\sqrt{1-x^{2}}- \biggl(x-\frac{x^{3}}{3}-\frac{x^{5}}{5}\biggr), \end{aligned}$$

(2.11)

$$\begin{aligned}& \gamma_{2}(x)=\tan^{-1}(x)-\biggl(x-\frac{x^{3}}{3}+ \frac{x^{5}}{7}\biggr), \end{aligned}$$

(2.12)

$$\begin{aligned}& \gamma_{3}(x)=\biggl(x-\frac{2x^{3}}{15}\biggr)- \sinh^{-1}(x). \end{aligned}$$

(2.13)

Then

$$\begin{aligned}& \gamma_{1}'(x)=\frac{x\gamma_{1}^{*}(x)}{\sqrt{1-x^{2}}}, \end{aligned}$$

(2.14)

$$\begin{aligned}& \gamma_{2}'(x)=\frac{x^{4}(2-5x^{2})}{7+7x^{2}}, \end{aligned}$$

(2.15)

$$\begin{aligned}& \gamma_{3}'(x)=\frac{\gamma_{3}^{*}(x)}{5\sqrt{1+x^{2}}}, \end{aligned}$$

(2.16)

where

$$\begin{aligned}& \gamma_{1}^{*}(x)=\bigl(x+x^{3}\bigr) \sqrt{1-x^{2}}-\sin^{-1}(x), \end{aligned}$$

(2.17)

$$\begin{aligned}& \gamma_{3}^{*}(x)=\sqrt{1+x^{2}} \bigl(5-2x^{2}\bigr)-5. \end{aligned}$$

(2.18)

Differentiating $\gamma_{1}^{*}(x)$ and $\gamma_{3}^{*}(x)$, we have

$$\begin{aligned}& \gamma_{1}^{*\prime}(x)=\frac{x^{2}(1-4x^{2})}{\sqrt{1-x^{2}}}, \end{aligned}$$

(2.19)

$$\begin{aligned}& \gamma_{3}^{*\prime}(x)=\frac{x}{\sqrt{1+x^{2}}}\bigl(1-6x^{2} \bigr). \end{aligned}$$

(2.20)

Furthermore, direct or numerical computations lead to

$$\begin{aligned}& \gamma_{1}(0)=0, \qquad \gamma_{1}(0.7)=0.0017 \ldots>0, \end{aligned}$$

(2.21)

$$\begin{aligned}& \gamma_{1}^{*}(0)=0,\qquad \gamma_{1}^{*}(1)=-1.5708 \ldots< 0, \end{aligned}$$

(2.22)

$$\begin{aligned}& \gamma_{2}(0)=0, \qquad \gamma_{2}(0.7)=0.0010 \ldots>0, \end{aligned}$$

(2.23)

$$\begin{aligned}& \gamma_{3}(0)=0,\qquad \gamma_{3}(0.7)=0.0016 \ldots>0, \end{aligned}$$

(2.24)

and

$$ \gamma_{3}^{*}(0)=0, \qquad \gamma_{3}^{*}(1)=-0.7574\ldots< 0. $$

(2.25)

From (2.19), we can easy to see that $\gamma_{1}^{*}(x)$ is strictly increasing on $(0,\frac{1}{2}]$ and strictly decreasing on $[\frac {1}{2},1)$. This fact and (2.22) together with (2.14) imply that there exists $x_{0} \in(\frac{1}{2},1)$, such that $\gamma _{1}'(x)>0$ on $(0,x_{0})$ and $\gamma_{1}'(x)<0$ on $(x_{0},1)$. The monotonicity of $\gamma_{1}(x)$ and (2.21) lead to

$$ \gamma_{1}(x)>0 $$

for x $\in(0,0.7)$. Therefore, inequality (2.8) holds.

Equation (2.15) shows that $\gamma_{2}(x)>0$ on $(0,\frac{\sqrt {10}}{5})$ and $\gamma_{2}(x)<0$ on $(\frac{\sqrt{10}}{5},1)$. This fact and (2.23) lead to

$$ \gamma_{2}(x)>0 $$

for x $\in(0,0.7)$. That is to say inequality (2.9) holds.

By (2.20), we know that $\gamma_{3}^{*}(x)$ is strictly increasing on $(0,\frac{\sqrt{6}}{6}]$ and strictly decreasing on $[\frac{\sqrt {6}}{6},1)$. This fact and (2.25) together with (2.16) imply that there must exist $x_{1} \in(\frac{\sqrt{6}}{6},1)$, such that $\gamma_{3}'(x)>0$ on $(0,x_{1})$ and $\gamma_{3}'(x)<0$ on $(x_{1},1)$. It follows from the monotonicity of $\gamma_{3}(x)$ and (2.24) that

$$ \gamma_{3}(x)>0 $$

for x $\in(0,0.7)$. This means the inequality (2.10) holds. □

Lemma 2.3

If x $\in(0.7,1)$, the double inequality

$$ x-\frac{x^{3}}{3}+\frac{2x^{5}}{17}< \tan^{-1}(x)< x- \frac{x^{3}}{3}+\frac {2x^{5}}{13} $$

(2.26)

holds.

Proof

Let

$$\begin{aligned}& \xi_{1}(x)=\biggl(x-\frac{x^{3}}{3}+\frac{2x^{5}}{13}\biggr)- \tan^{-1}(x), \\& \xi_{2}(x)=\biggl(x-\frac{x^{3}}{3}+\frac{2x^{5}}{17}\biggr)- \tan^{-1}(x). \end{aligned}$$

Then

$$\begin{aligned}& \xi_{1}'(x)=\frac{x^{4}(10x^{2}-3)}{13(1+x^{2})}, \end{aligned}$$

(2.27)

$$\begin{aligned}& \xi_{2}'(x)=\frac{x^{4}(10x^{2}-7)}{17(1+x^{2})}. \end{aligned}$$

(2.28)

Equality (2.27) implies that $\xi_{1}(x)$ is strictly increasing on $[\frac{\sqrt{30}}{10},1)$. Additional numerical computations lead to $\frac{\sqrt{30}}{10}<0.7$ and $\xi_{1}(0.7)=0.0007976\ldots>0$. Therefore, we can get $\xi_{1}(x)>0$ for x $\in(0.7,1)$. This implies the right hand side of the double inequality (2.26) holds.

Equality (2.28) implies $\xi_{2}(x)$ is strictly decreasing on $(0,\frac{\sqrt{70}}{10}]$ and strictly increasing on $[\frac{\sqrt {70}}{10},1)$. Because of $\xi_{2}(0)=0$ and $\xi _{2}(1)=-0.0011\ldots<0$, it leads to $\xi_{2}(x)<0$ for $x \in(0,1)$. Specially, for $x\in(0.7,1)$. This means the left hand side of the double inequality (2.26) holds. □

Lemma 2.4

Let

$$\begin{aligned}& \mu_{1}(x)=\frac{1+3x^{2}}{(x+x^{3})^{2}}-\frac {1}{(1+x^{2})[\sinh^{-1}(x)]^{2}}- \frac{x}{(1+x^{2})^{\frac{3}{2}}\sinh^{-1}(x)}, \end{aligned}$$

(2.29)

$$\begin{aligned}& \mu_{2}(x)=-\frac{1+3x^{2}}{(x+x^{3})^{2}}+\frac {1}{(1-x^{2})[\sin^{-1}(x)]^{2}}- \frac{x}{(1-x^{2})^{\frac{3}{2}}\sin^{-1}(x)}, \end{aligned}$$

(2.30)

and

$$ \mu_{3}(x)=-\frac{2x}{(1+x^{2})^{2}\tan^{-1}(x)}-\frac{1}{[\tan ^{-1}(x)]^{2}(1+x^{2})^{2}}+ \frac{1+3x^{2}}{(x+x^{3})^{2}}. $$

(2.31)

Then, for any $x\in(0.7,1)$, we have

$$ \mu_{1}(x)< 0.17, \qquad \mu_{2}(x)< -1.48, $$

(2.32)

and

$$ \mu_{3}(x)>-0.05. $$

(2.33)

Proof

From Lemmas 2.6 and 2.7 of [22], for any $x\in [0.7,1)$, we can get $\mu_{1}'(x)\leq0.167\ldots<0.17$ and $\mu _{2}'(x)\leq-1.48798\ldots<-1.48$, respectively.

Differentiating $\mu_{3}(x)$, we have

$$ \mu_{3}'(x)=\frac{2\eta_{1}(x)+6x^{2}\tan^{-1}(x)\eta _{2}(x)}{[\tan^{-1}(x)(x+x^{3})]^{3}}, $$

(2.34)

where

$$ \eta_{1}(x)=x^{3}-x^{3}\bigl[ \tan^{-1}(x)\bigr]^{2}-\bigl[\tan^{-1}(x) \bigr]^{3} $$

and

$$ \eta_{2}(x)=x^{2}+x^{3}\tan^{-1}(x)- \bigl(2x^{2}+1\bigr)\bigl[\tan^{-1}(x)\bigr]^{2}. $$

For any $x \in[0.7,1)$,

$$\begin{aligned} \eta_{1}(x) < &x^{3}-x^{3}\biggl(x- \frac{x^{3}}{3}+\frac{x^{5}}{9}\biggr)^{2}-\biggl(x- \frac{x^{3}}{3}+\frac {x^{5}}{9}\biggr)^{3} \\ =&-x^{9}\bigl(54+x^{6}\bigr)< 0 \end{aligned}$$

(2.35)

and

$$\begin{aligned} \eta_{2}(x) < &x^{2}+x^{3}\biggl(x- \frac{x^{3}}{3}+\frac{2x^{5}}{13}\biggr)-\bigl(2x^{2}+1\bigr) \biggl(x-\frac {x^{3}}{3}+\frac{2x^{5}}{17}\biggr)^{2} \\ =&\frac{x^{4}}{1\text{,}989}\bigl(-663+1\text{,}300x^{2}-637x^{4} \bigr)+\frac {x^{8}}{33\text{,}813}\bigl(-4\text{,}743+4\text{,}836x^{2}-936x^{4} \bigr) \end{aligned}$$

follow from inequalities (2.7) and (2.26), respectively. Because $-663+1\text{,}300x^{2}-637x^{4}<0$ and $-4\text{,}743+4\text{,}836x^{2}-936x^{4}<0$ for x $\in (0.7,1)$, we have

$$ \eta_{2}(x)< 0 $$

(2.36)

for $x\in(0.7,1)$. Thus $\mu_{3}'(x)<0$ for $x\in(0.7,1)$ follows from (2.34), (2.35), and (2.36). Therefore, we obtain $\mu_{3}(x)>\mu_{3}(1)=-0.0419\ldots>-0.05$ for $x\in(0.7,1)$. □

Lemma 2.5

Let $f(x)=\frac{1}{\sqrt{1+x^{2}}\sinh^{-1}(x)}-(1-\lambda_{0})\frac {1}{(1+x^{2})\tan^{-1}(x)}-\lambda_{0}\frac{1}{\sqrt {1-x^{2}}\sin^{-1}(x)}$, where $\lambda_{0}=\frac{\log(\frac{4\log (1+\sqrt{2})}{\pi})}{\log2}=0.1663\ldots$ . Then the function $f(x)$ is strictly decreasing on $(0.7,1)$.

Proof

It is obvious that

$$\begin{aligned} f(x) =&\biggl[\frac{1}{\sqrt{1+x^{2}}\sinh^{-1}(x)}-\frac{1}{x(1+x^{2})}\biggr] + \lambda_{0}\biggl[\frac{1}{x(1+x^{2})}-\frac{1}{\sqrt{1-x^{2}}\sin^{-1}(x)}\biggr] \\ &{}+(\lambda_{0}-1)\biggl[\frac{1}{(1+x^{2})\tan^{-1}(x)}-\frac{1}{x(1+x^{2})} \biggr] \\ :=&U_{1}(x)+\lambda_{0}U_{2}(x)+( \lambda_{0}-1)U_{3}(x). \end{aligned}$$

Differentiating $f(x)$, we have

$$\begin{aligned} f'(x) =&U'_{1}(x)+\lambda_{0}U'_{2}(x)+( \lambda_{0}-1)U'_{3}(x) \\ =&\mu_{1}(x)+\lambda_{0}\mu_{2}(x)+( \lambda_{0}-1)\mu_{3}(x), \end{aligned}$$

(2.37)

where $\mu_{1}(x)$, $\mu_{2}(x)$, and $\mu_{3}(x)$ are defined as in Lemma 2.4. Therefore, Lemma 2.4 and equation (2.37) yield

$$\begin{aligned} f'(x) < &0.17+\lambda_{0}(-1.48)+(\lambda_{0}-1) (-0.05) \\ =&-0.0345\ldots< 0 \end{aligned}$$

for $x\in(0.7,1)$. The proof is completed. □

3 Main results

Theorem 3.1

The double inequality

$$ P^{\alpha}(a,b)T^{1-\alpha}(a,b)< M(a,b)< P^{\beta}(a,b)T^{1-\beta}(a,b) $$

holds for all $a,b > 0$ with $a\neq b$ if and only if $a\geq1/3$ and $\beta\leq\frac{\log(\frac{4\log(1+\sqrt{2})}{\pi})}{\log 2}=0.1663\ldots$ .

Proof

Because $P(a,b)$, $M(a,b)$, and $T(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $a>b$. Let $p\in(0,1)$, $\lambda_{0}=\frac{\log(\frac{4\log (1+\sqrt{2})}{\pi})}{\log2}$, and $x=(a-b)/(a+b)$. Then $x\in(0,1)$ and

$$\begin{aligned}& p\log\bigl[P(a,b)\bigr]+(1-p)\log\bigl[T(a,b)\bigr]-\log \bigl[M(a,b)\bigr] \\& \quad =p\log\biggl[\frac{x}{\sin^{-1}(x)}\biggr]+(1-p)\log\biggl[\frac {x}{\tan^{-1}(x)} \biggr]-\log\biggl[\frac{x}{\sinh^{-1}(x)}\biggr] \\& \quad =\log\bigl[\sinh^{-1}(x)\bigr]-(1-p)\log\bigl[ \tan^{-1}(x)\bigr]-p\log\bigl[\sin^{-1}(x) \bigr]:=D_{p}(x). \end{aligned}$$

(3.1)

It follows that

$$ D_{p}\bigl(0^{+}\bigr)=0 \quad \mbox{and} \quad D_{\lambda_{0}}\bigl(1^{-}\bigr)=0. $$

(3.2)

Differentiating $D_{p}(x)$, we have

$$\begin{aligned} D_{p}'(x) =&\frac{1}{\sinh^{-1}(x)\sqrt{1+x^{2}}}-(1-p) \frac {1}{\tan^{-1}(x)(1+x^{2})}-p\frac{1}{\sin^{-1}(x)\sqrt{1-x^{2}}} \\ =&\frac{g_{p}(x)}{\sinh^{-1}(x)(1+x^{2})\tan^{-1}(x)\sin^{-1}(x)\sqrt {1-x^{2}}}, \end{aligned}$$

(3.3)

where

$$\begin{aligned} g_{p}(x) =&\bigl[\sqrt{1+x^{2}} \tan^{-1}(x)-(1-p)\sinh^{-1}(x)\bigr]\sin ^{-1}(x) \sqrt{1-x^{2}} \\ &{}-p\sinh^{-1}(x) \bigl(1+x^{2}\bigr)\tan^{-1}(x). \end{aligned}$$

(3.4)

On one hand, when $p=\frac{1}{3}$, Lemma 2.1 and equation (3.4) lead to

$$\begin{aligned} g_{\frac{1}{3}}(x) =&\biggl[\sqrt{1+x^{2}} \tan^{-1}(x)-\frac{2}{3}\sinh ^{-1}(x)\biggr] \sin^{-1}(x)\sqrt{1-x^{2}} \\ &{}-\frac{1}{3}\sinh^{-1}(x) \bigl(1+x^{2}\bigr) \tan^{-1}(x) \\ < &\biggl[\biggl(1+\frac{x^{2}}{2}\biggr) \biggl(x-\frac{x^{3}}{3}+ \frac{x^{5}}{5}\biggr)-\frac {2}{3}\biggl(x-\frac{x^{3}}{6}\biggr) \biggr] \biggl(x-\frac{x^{3}}{3}-\frac{2x^{5}}{15}\biggr) \\ &{}-\frac{1}{3}\biggl(x-\frac{x^{3}}{6}\biggr) \bigl(1+x^{2} \bigr) \biggl(x-\frac{x^{3}}{3}+\frac {x^{5}}{5}-\frac{x^{7}}{7}\biggr) \\ =&\frac{x^{6}}{3\text{,}150}\bigl(-70+80x^{2}+41x^{4}-67x^{6} \bigr)< 0 \end{aligned}$$

(3.5)

for $x\in(0,1)$. According to (3.3) and (3.5), we can see that

$$ D_{\frac{1}{3}}'(x)< 0 $$

(3.6)

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

On the other hand, when $p=\lambda_{0}$, the inequalities (2.3) and (2.6) and Lemma 2.2 together with equation (3.4) lead to

$$\begin{aligned} g_{\lambda_{0}}(x) =&\bigl[\sqrt{1+x^{2}} \tan^{-1}(x)-(1-\lambda_{0})\sinh ^{-1}(x)\bigr] \sin^{-1}(x)\sqrt{1-x^{2}} \\ &{}-\lambda_{0}\sinh^{-1}(x) \bigl(1+x^{2}\bigr) \tan^{-1}(x) \\ >&\biggl[\biggl(1+\frac{6x^{2}}{15}\biggr) \biggl(x-\frac{x^{3}}{3}+ \frac{x^{5}}{7}\biggr)-(1-\lambda _{0}) \biggl(x-\frac{2x^{3}}{15} \biggr)\biggr]\biggl(x-\frac{x^{3}}{3}-\frac{x^{5}}{5}\biggr) \\ &{}-\lambda_{0}\biggl(x-\frac{2x^{3}}{15}\biggr) \bigl(1+x^{2}\bigr) \biggl(x-\frac{x^{3}}{3}+\frac {x^{5}}{5} \biggr) \\ =&\frac{x^{4}}{1\text{,}575}F_{\lambda_{0}}(x) \end{aligned}$$

(3.7)

for $x\in(0,0.7)$, where

$$F_{\lambda_{0}}(x)=(315-1\text{,}575\lambda_{0})+(105\lambda _{0}-90)x^{2}+(22-301\lambda_{0})x^{4}+(42 \lambda_{0}-33)x^{6}-18x^{8}. $$

Because of $42\lambda_{0}-33=-26.0142\ldots<0$ and $105\lambda _{0}-90=-72.5354\ldots<0$, it follows that

$$\begin{aligned} F_{\lambda_{0}}(x) >&(315-1\text{,}575\lambda_{0})+(105 \lambda _{0}-90)x^{2}+(22-301\lambda_{0})x^{4}+(42 \lambda _{0}-33)x^{4}-18x^{4} \\ =&(315-1\text{,}575\lambda_{0})+(105\lambda_{0}-90)x^{2}+(-29-259 \lambda _{0})x^{4} \\ :=&F^{*}(x) \end{aligned}$$

(3.8)

and

$$F^{*\prime}(x)=2(105\lambda_{0}-90)x+4(-29-259 \lambda_{0})x^{3}< 0 $$

for $x\in(0,0.7)$. Thus, we can get

$$ F_{\lambda_{0}}(x)>F^{*}(x)>F^{*}(0.7)=0.1825 \ldots>0 $$

(3.9)

for $x\in(0,0.7)$. Therefore, equation (3.3) and inequalities (3.7)-(3.9) imply

$$ D_{\lambda_{0}}'(x)>0 $$

(3.10)

for $x\in(0,0.7)$.

It follows from equation (3.3) and Lemma 2.5 that $D_{\lambda _{0}}'(x)$ is strictly decreasing on $(0.7,1)$. Then from equation (3.10) and $D_{\lambda_{0}}'(1^{-})=-\infty$, we know that there exists $x_{*} \in(0.7,1)$ such that $D_{\lambda_{0}}(x)$ is strictly increasing on $(0,x_{*}]$ and strictly decreasing on $[x_{*},1)$. This in conjunction with (3.2) means that

$$ D_{\lambda_{0}}(x)>0 $$

(3.11)

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

Therefore, for all $a,b>0$ with $a\neq b$,

$$ M(a,b)>P^{\frac{1}{3}}(a,b)T^{\frac{2}{3}}(a,b), $$

(3.12)

follows from equations (3.1), (3.2), and (3.6) as well as

$$ M(a,b)< P^{\lambda_{0}}(a,b)T^{1-\lambda_{0}}(a,b) $$

(3.13)

follows from equations (3.1), (3.2), and (3.11).

Finally, by easy computations, equations (1.1), (1.2), and (1.3) lead to

$$\begin{aligned}& \frac{\log[T(a,b)]-\log[M(a,b)]}{\log[T(a,b)]-\log[P(a,b)]}=\frac {\log[\sinh^{-1}(x)]-\log[\tan^{-1}(x)]}{\log[\sin^{-1}(x)]-\log [\tan^{-1}(x)]}, \end{aligned}$$

(3.14)

$$\begin{aligned}& \lim_{x\rightarrow0^{+}}\frac{\log[\sinh^{-1}(x)]-\log [\tan^{-1}(x)]}{\log[\sin^{-1}(x)]-\log[\tan^{-1}(x)]}=\frac{1}{3} \end{aligned}$$

(3.15)

and

$$ \lim_{x\rightarrow1^{-}}\frac{\log[\sinh^{-1}(x)]-\log [\tan^{-1}(x)]}{\log[\sin^{-1}(x)]-\log[\tan^{-1}(x)]}= \lambda_{0}. $$

(3.16)

Thus, we have the following two claims.

Claim 1

If $\alpha<\frac{1}{3}$, then from (3.14) and (3.15), there must exist $\delta_{1}\in(0,1)$ such that $M(a,b)< P^{\alpha}(a,b)T^{1-\alpha}(a,b)$ for all $a,b >0$ with $(a-b)/(a+b)\in(0,\delta_{1})$.

Claim 2

If $\beta>\lambda_{0}$, then from (3.14) and (3.16), there must exist $\delta_{2} \in(0,1)$ such that $M(a,b)>P^{\beta}(a,b)T^{1-\beta}(a,b)$ for all $a,b >0$ with $(a-b)/(a+b)\in(1-\delta_{2},1)$.

Inequalities (3.12) and (3.13) in conjunction with the above two claims mean the proof is completed.  □