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# Optimal bounds for the Neuman-Sándor mean in terms of the first Seiffert and quadratic means

Journal of Inequalities and Applications20132013:552

https://doi.org/10.1186/1029-242X-2013-552

• Accepted: 18 October 2013
• Published:

## Abstract

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

${P}^{\alpha }\left(a,b\right){Q}^{1-\alpha }\left(a,b\right)

holds true for all $a,b>0$ with $a\ne b$, where $P\left(a,b\right)$, $M\left(a,b\right)$ and $Q\left(a,b\right)$ are the first Seiffert, Neuman-Sándor and quadratic means of a and b, respectively.

MSC:26E60.

## Keywords

• Neuman-Sándor mean
• first Seiffert mean

## 1 Introduction

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

For $a,b>0$ with $a\ne b$, the first Seiffert mean $P\left(a,b\right)$ , the Neuman-Sándor mean $M\left(a,b\right)$ , the quadratic mean $Q\left(a,b\right)$ are defined by
$\begin{array}{r}P\left(a,b\right)=\frac{a-b}{4arctan\left(\sqrt{a/b}\right)-\pi },\phantom{\rule{2em}{0ex}}M\left(a,b\right)=\frac{a-b}{2{sinh}^{-1}\left(\frac{a-b}{a+b}\right)},\\ Q\left(a,b\right)=\sqrt{\frac{{a}^{2}+{b}^{2}}{2}},\end{array}$
(1.1)

respectively. In here, ${sinh}^{-1}\left(x\right)=log\left(x+\sqrt{{x}^{2}+1}\right)$ is the inverse hyperbolic sine function.

Recently, the means P, M and Q have been the subject of intensive research. In particular, many remarkable inequalities for these means can be found in the literature . The first Seiffert mean $P\left(a,b\right)$ can be rewritten as (see [, Eq. (2.4)])
$P\left(a,b\right)=\frac{a-b}{2arcsin\left[\left(a-b\right)/\left(a+b\right)\right]}.$
(1.2)
Let $H\left(a,b\right)=2ab/\left(a+b\right)$, $L\left(a,b\right)=\left(b-a\right)/\left(logb-loga\right)$, $A\left(a,b\right)=\left(a+b\right)/2$, $T\left(a,b\right)=\left(a-b\right)/\left[2arctan\left(\left(a-b\right)/\left(a+b\right)\right)\right]$ and $C\left(a,b\right)=\left({a}^{2}+{b}^{2}\right)/\left(a+b\right)$ be the harmonic, logarithmic, arithmetic, second Seiffert and contra-harmonic means of a and b, respectively. Then it is known that the inequalities
$H\left(a,b\right)

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

Neuman and Sándor [2, 15] proved that the inequalities
$\begin{array}{c}\frac{\pi }{4log\left(1+\sqrt{2}\right)}T\left(a,b\right)H\left(M\left(a,b\right),Q\left(a,b\right)\right),\hfill \\ M\left(a,b\right)<\frac{{A}^{2}\left(a,b\right)}{P\left(a,b\right)},\phantom{\rule{2em}{0ex}}{A}^{2/3}\left(a,b\right){Q}^{1/3}\left(a,b\right)

hold for all $a,b>0$ and $x,y\in \left(0,1/2\right]$ with $a\ne b$ and $x\ne y$.

Li et al.  proved that the double inequality ${L}_{{p}_{0}}\left(a,b\right) holds for all $a,b>0$ with $a\ne b$, where ${L}_{p}\left(a,b\right)={\left[\left({b}^{p+1}-{a}^{p+1}\right)/\left(\left(p+1\right)\left(b-a\right)\right)\right]}^{1/p}$ ($p\ne -1,0$), ${L}_{0}\left(a,b\right)=1/e{\left({b}^{b}/{a}^{a}\right)}^{1/\left(b-a\right)}$ and ${L}_{-1}\left(a,b\right)=\left(b-a\right)/\left(logb-loga\right)$ is the p th generalized logarithmic mean of a and b, and ${p}_{0}=1.843\dots$ is the unique solution of the equation ${\left(p+1\right)}^{1/p}=2log\left(1+\sqrt{2}\right)$.

In , Neuman proved that the double inequalities
${Q}^{\alpha }\left(a,b\right){A}^{1-\alpha }\left(a,b\right)
(1.3)
and
${C}^{\lambda }\left(a,b\right){A}^{1-\lambda }\left(a,b\right)
(1.4)

hold for all $a,b>0$ with $a\ne b$ if $\alpha \le 1/3$, $\beta \ge 2\left[log\left(2+\sqrt{2}\right)-log3\right]/log2$, $\lambda \le 1/6$ and $\mu \ge \left[log\left(2+\sqrt{2}\right)-log3\right]/log2$.

Jiang and Qi [17, 18] gave the best possible parameters α, β, ${t}_{1}$ and ${t}_{2}$ in $\left(0,1/2\right)$ such that the inequalities
$\begin{array}{r}Q\left(\alpha a+\left(1-\alpha \right)b,\alpha b+\left(1-\alpha \right)a\right)

hold for all $a,b>0$ with $a\ne b$ and $p\ge 1/2$, where ${Q}_{t,p}\left(a,b\right)={C}^{p}\left(ta+\left(1-t\right)b,tb+\left(1-t\right)a\right){A}^{1-p}\left(a,b\right)$.

Inspired by inequalities (1.3) and (1.4), in this paper, we present the optimal upper and lower bounds for the Neuman-Sándor mean $M\left(a,b\right)$ in terms of the geometric convex combinations of the first Seiffert mean $P\left(a,b\right)$ and the quadratic mean $Q\left(a,b\right)$. All numerical computations are carried out using Mathematica software.

## 2 Lemmas

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

Lemma 2.1 The double inequality
$x+\frac{{x}^{3}}{3}-\frac{2{x}^{5}}{15}<\sqrt{1+{x}^{2}}{sinh}^{-1}\left(x\right)
(2.1)

holds for $x\in \left(0,1\right)$.

Proof To show inequality (2.1), it suffices to prove that
${\omega }_{1}\left(x\right)=\sqrt{1+{x}^{2}}{sinh}^{-1}\left(x\right)-\left(x+\frac{{x}^{3}}{3}-\frac{2{x}^{5}}{15}\right)>0$
(2.2)
and
${\omega }_{2}\left(x\right)=\sqrt{1+{x}^{2}}{sinh}^{-1}\left(x\right)-\left(x+\frac{{x}^{3}}{3}-\frac{2{x}^{5}}{15}+\frac{8{x}^{7}}{105}\right)<0$
(2.3)

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

From the expressions of ${\omega }_{1}\left(x\right)$ and ${\omega }_{2}\left(x\right)$, we get
${\omega }_{1}\left(0\right)={\omega }_{2}\left(0\right)=0,$
(2.4)
${\omega }_{1}^{\prime }\left(x\right)=\frac{x{\omega }_{1}^{\ast }\left(x\right)}{\sqrt{1+{x}^{2}}},\phantom{\rule{2em}{0ex}}{\omega }_{2}^{\prime }\left(x\right)=\frac{x{\omega }_{2}^{\ast }\left(x\right)}{\sqrt{1+{x}^{2}}},$
(2.5)
where
$\begin{array}{c}{\omega }_{1}^{\ast }\left(x\right)={sinh}^{-1}\left(x\right)-\left(x-\frac{2{x}^{3}}{3}\right)\sqrt{1+{x}^{2}},\hfill \\ {\omega }_{2}^{\ast }\left(x\right)={sinh}^{-1}\left(x\right)-\left(x-\frac{2{x}^{3}}{3}+\frac{8{x}^{5}}{15}\right)\sqrt{1+{x}^{2}},\hfill \\ {\omega }_{1}^{\ast }\left(0\right)={\omega }_{2}^{\ast }\left(0\right)=0,\hfill \end{array}$
(2.6)
${{\omega }_{1}^{\ast }}^{\prime }\left(x\right)=\frac{8{x}^{4}}{3\sqrt{1+{x}^{2}}}>0$
(2.7)
and
${{\omega }_{2}^{\ast }}^{\prime }\left(x\right)=-\frac{16{x}^{6}}{5\sqrt{1+{x}^{2}}}<0$
(2.8)

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

Therefore, inequality (2.2) follows from (2.4)-(2.7), and inequality (2.3) follows from (2.4)-(2.6) and (2.8). □

Lemma 2.2 The inequality
$\frac{{x}^{3}}{\sqrt{1+{x}^{2}}}>{\left[{sinh}^{-1}\left(x\right)\right]}^{3}$

holds for $x\in \left(0,1\right)$.

Proof Let $x\in \left(0,1\right)$, then from (1.3) we have
$M\left(1+x,1-x\right)>{A}^{2/3}\left(1+x,1-x\right){Q}^{1/3}\left(1+x,1-x\right).$
(2.9)

Therefore, Lemma 2.2 follows from (2.9). □

Lemma 2.3 The inequality
$\sqrt{1-{x}^{2}}arcsin\left(x\right)>x-\frac{{x}^{3}}{3}-\frac{{x}^{5}}{3}$
(2.10)
holds for $x\in \left(0,0.7\right)$, and the inequality
$\sqrt{1-{x}^{2}}arcsin\left(x\right)
(2.11)

holds for $x\in \left(0,1\right)$, where $arcsin\left(x\right)$ is the inverse sine function.

Proof Let
${\phi }_{1}\left(x\right)=\sqrt{1-{x}^{2}}arcsin\left(x\right)-x+\frac{{x}^{3}}{3}+\frac{{x}^{5}}{3},$
(2.12)
${\phi }_{2}\left(x\right)=\sqrt{1-{x}^{2}}arcsin\left(x\right)-x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}.$
(2.13)
${\phi }_{1}\left(0\right)={\phi }_{2}\left(0\right)=0,$
(2.14)
${\phi }_{1}^{\prime }\left(x\right)=\frac{x{\phi }_{1}^{\ast }\left(x\right)}{\sqrt{1-{x}^{2}}},\phantom{\rule{2em}{0ex}}{\phi }_{2}^{\prime }\left(x\right)=\frac{x{\phi }_{2}^{\ast }\left(x\right)}{\sqrt{1-{x}^{2}}},$
(2.15)
where
$\begin{array}{c}{\phi }_{1}^{\ast }\left(x\right)=\left(x+\frac{5{x}^{3}}{3}\right)\sqrt{1-{x}^{2}}-arcsin\left(x\right),\hfill \\ {\phi }_{2}^{\ast }\left(x\right)=\left(x+\frac{2{x}^{3}}{3}\right)\sqrt{1-{x}^{2}}-arcsin\left(x\right).\hfill \end{array}$
Note that
${\phi }_{1}^{\ast }\left(0\right)={\phi }_{2}^{\ast }\left(0\right)=0,\phantom{\rule{2em}{0ex}}{\phi }_{1}^{\ast }\left(0.7\right)=0.1327\dots ,$
(2.16)
${{\phi }_{1}^{\ast }}^{\prime }\left(x\right)=\frac{{x}^{2}\left(9-20{x}^{2}\right)}{3\sqrt{1-{x}^{2}}},$
(2.17)
${{\phi }_{2}^{\ast }}^{\prime }\left(x\right)=-\frac{8{x}^{4}}{3\sqrt{1-{x}^{2}}}<0$
(2.18)

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

From (2.17) we clearly see that ${\phi }_{1}^{\ast }\left(x\right)$ is strictly increasing on $\left(0,3\sqrt{5}/10\right]$ and strictly decreasing on $\left[3\sqrt{5}/10,0.7\right)$. This in conjunction with (2.16) implies that
${\phi }_{1}^{\ast }\left(x\right)>0$
(2.19)

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

Therefore, inequality (2.10) follows from (2.12), (2.14), (2.15) and (2.19), and inequality (2.11) follows from (2.12) and (2.14)-(2.16) together with (2.18). □

Lemma 2.4 Let
$\mathrm{\Phi }\left(x\right)=\frac{1}{\sqrt{1+{x}^{2}}{sinh}^{-1}\left(x\right)}-\frac{1}{x\left(1+{x}^{2}\right)}.$
Then the inequality
$\mathrm{\Phi }\left(x\right)>\frac{2x}{3}-\frac{34{x}^{3}}{45}+\frac{754{x}^{5}}{945}-{x}^{7}$
(2.20)
holds for $x\in \left(0,0.7\right)$, and
$\mathrm{\Phi }\left(x\right)<\frac{2x}{3}-\frac{34{x}^{3}}{45}+\frac{4{x}^{5}}{5}$
(2.21)

holds for $x\in \left(0,1\right)$.

Proof To show inequalities (2.20) and (2.21), it suffices to prove that
$\begin{array}{rl}{\varphi }_{1}\left(x\right):=& x\left(1+{x}^{2}\right){sinh}^{-1}\left(x\right)\left[\mathrm{\Phi }\left(x\right)-\left(\frac{2x}{3}-\frac{34{x}^{3}}{45}+\frac{754{x}^{5}}{945}-{x}^{7}\right)\right]\\ =& x\sqrt{1+{x}^{2}}-{sinh}^{-1}\left(x\right)\\ -x\left(1+{x}^{2}\right){sinh}^{-1}\left(x\right)\left(\frac{2x}{3}-\frac{34{x}^{3}}{45}+\frac{754{x}^{5}}{945}-{x}^{7}\right)>0\end{array}$
(2.22)
for $x\in \left(0,0.7\right)$, and
$\begin{array}{rl}{\varphi }_{2}\left(x\right):=& x\left(1+{x}^{2}\right){sinh}^{-1}\left(x\right)\left[\mathrm{\Phi }\left(x\right)-\left(\frac{2x}{3}-\frac{34{x}^{3}}{45}+\frac{4{x}^{5}}{5}\right)\right]\\ =& x\sqrt{1+{x}^{2}}-{sinh}^{-1}\left(x\right)\\ -x\left(1+{x}^{2}\right){sinh}^{-1}\left(x\right)\left(\frac{2x}{3}-\frac{34{x}^{3}}{45}+\frac{4{x}^{5}}{5}\right)<0\end{array}$
(2.23)

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

From the expressions of ${\varphi }_{1}\left(x\right)$ and ${\varphi }_{2}\left(x\right)$, one has
${\varphi }_{1}\left(0\right)={\varphi }_{2}\left(0\right)=0,$
(2.24)
${\varphi }_{1}^{\prime }\left(x\right)=\frac{x}{945\sqrt{1+{x}^{2}}}{\varphi }_{1}^{\ast }\left(x\right),\phantom{\rule{2em}{0ex}}{\varphi }_{2}^{\prime }\left(x\right)=-\frac{2x}{45\sqrt{1+{x}^{2}}}{\varphi }_{2}^{\ast }\left(x\right),$
(2.25)
where
$\begin{array}{c}{\varphi }_{1}^{\ast }\left(x\right)=x\left(1\text{,}260+84{x}^{2}-40{x}^{4}+191{x}^{6}+945{x}^{8}\right)\hfill \\ \phantom{{\varphi }_{1}^{\ast }\left(x\right)=}-2\left(630-168{x}^{2}+120{x}^{4}-764{x}^{6}-4\text{,}725{x}^{8}\right)\sqrt{1+{x}^{2}}{sinh}^{-1}\left(x\right),\hfill \end{array}$
(2.26)
$\begin{array}{c}{\varphi }_{2}^{\ast }\left(x\right)=x\left(18{x}^{6}+{x}^{4}-2{x}^{2}-30\right)\hfill \\ \phantom{{\varphi }_{2}^{\ast }\left(x\right)=}+2\left(15-4{x}^{2}+3{x}^{4}+72{x}^{6}\right)\sqrt{1+{x}^{2}}{sinh}^{-1}\left(x\right).\hfill \end{array}$
(2.27)
Note that
$\begin{array}{c}630-168{x}^{2}+120{x}^{4}-764{x}^{6}-4\text{,}725{x}^{8}\hfill \\ \phantom{\rule{1em}{0ex}}>630-168×{\left(0.7\right)}^{2}-764×{\left(0.7\right)}^{6}-4\text{,}725×{\left(0.7\right)}^{8}=185.4\dots >0\hfill \end{array}$
(2.28)

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

Lemma 2.1 and equations (2.26)-(2.28) lead to
$\begin{array}{rl}{\varphi }_{1}^{\ast }\left(x\right)>& x\left(1\text{,}260+84{x}^{2}-40{x}^{4}+191{x}^{6}+945{x}^{8}\right)\\ -2\left(630-168{x}^{2}+120{x}^{4}-764{x}^{6}-4\text{,}725{x}^{8}\right)\left(x+\frac{{x}^{3}}{3}-\frac{2{x}^{5}}{15}+\frac{8{x}^{7}}{105}\right)\\ =& \frac{{x}^{7}}{105}\left(157\text{,}311+1\text{,}151\text{,}003{x}^{2}+307\text{,}438{x}^{4}-120\text{,}076{x}^{6}+75\text{,}600{x}^{8}\right)>0\end{array}$
(2.29)
for $x\in \left(0,0.7\right)$, and
$\begin{array}{rl}{\varphi }_{2}^{\ast }\left(x\right)>& x\left(18{x}^{6}+{x}^{4}-2{x}^{2}-30\right)\\ +2\left(15-4{x}^{2}+3{x}^{4}+72{x}^{6}\right)\left(x+\frac{{x}^{3}}{3}-\frac{2{x}^{5}}{15}\right)\\ =& \frac{{x}^{5}}{15}\left(5+2\text{,}476{x}^{2}+708{x}^{4}-288{x}^{6}\right)>0\end{array}$
(2.30)

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

Therefore, inequality (2.22) follows from (2.24), (2.25) and (2.29), and inequality (2.23) follows from (2.24), (2.25) and (2.30). □

Lemma 2.5 Let
$\mathrm{\Upsilon }\left(x\right)=\frac{1}{x\left(1+{x}^{2}\right)}-\frac{1}{\sqrt{1-{x}^{2}}arcsin\left(x\right)}.$
Then the inequality
$\mathrm{\Upsilon }\left(x\right)>-\frac{4x}{3}+\frac{34{x}^{3}}{45}-\frac{3{x}^{5}}{2}$
(2.31)
holds for $x\in \left(0,0.7\right)$, and
$\mathrm{\Upsilon }\left(x\right)<-\frac{4x}{3}+\frac{34{x}^{3}}{45}-\frac{8{x}^{5}}{9}$
(2.32)

holds for $x\in \left(0,1\right)$.

Proof Let
$\begin{array}{rl}{ϵ}_{1}\left(x\right):=& x\left(1+{x}^{2}\right)\sqrt{1-{x}^{2}}arcsin\left(x\right)\left[\mathrm{\Upsilon }\left(x\right)+\left(\frac{4x}{3}-\frac{34{x}^{3}}{45}+\frac{3{x}^{5}}{2}\right)\right]\\ =& \sqrt{1-{x}^{2}}arcsin\left(x\right)-x\left(1+{x}^{2}\right)\\ +x\left(1+{x}^{2}\right)\sqrt{1-{x}^{2}}arcsin\left(x\right)\left(\frac{4x}{3}-\frac{34{x}^{3}}{45}+\frac{3{x}^{5}}{2}\right)\end{array}$
(2.33)
and
$\begin{array}{rl}{ϵ}_{2}\left(x\right):=& x\left(1+{x}^{2}\right)\sqrt{1-{x}^{2}}arcsin\left(x\right)\left[\mathrm{\Upsilon }\left(x\right)+\left(\frac{4x}{3}-\frac{34{x}^{3}}{45}+\frac{8{x}^{5}}{9}\right)\right]\\ =& \sqrt{1-{x}^{2}}arcsin\left(x\right)-x\left(1+{x}^{2}\right)\\ +x\left(1+{x}^{2}\right)\sqrt{1-{x}^{2}}arcsin\left(x\right)\left(\frac{4x}{3}-\frac{34{x}^{3}}{45}+\frac{8{x}^{5}}{9}\right).\end{array}$
(2.34)
An easy calculation gives rise to
${ϵ}_{1}\left(0\right)={ϵ}_{2}\left(0\right)=0,$
(2.35)
${ϵ}_{1}^{\prime }\left(x\right)=\frac{x}{90\left(1-{x}^{2}\right)}{ϵ}_{1}^{\ast }\left(x\right),\phantom{\rule{2em}{0ex}}{ϵ}_{2}^{\prime }\left(x\right)=\frac{x}{45\left(1-{x}^{2}\right)}{ϵ}_{2}^{\ast }\left(x\right),$
(2.36)
where
$\begin{array}{c}{ϵ}_{1}^{\ast }\left(x\right)=-x\left(150-202{x}^{2}-15{x}^{4}-68{x}^{6}+135{x}^{8}\right)\hfill \\ \phantom{{ϵ}_{1}^{\ast }\left(x\right)=}+\left(150-152{x}^{2}+142{x}^{4}+611{x}^{6}-1\text{,}215{x}^{8}\right)\sqrt{1-{x}^{2}}arcsin\left(x\right),\hfill \end{array}$
(2.37)
$\begin{array}{c}{ϵ}_{2}^{\ast }\left(x\right)=-x\left(1-{x}^{2}\right)\left(75-26{x}^{2}-6{x}^{4}-40{x}^{6}\right)\hfill \\ \phantom{{ϵ}_{2}^{\ast }\left(x\right)=}+\left(75-76{x}^{2}-94{x}^{4}+278{x}^{6}-360{x}^{8}\right)\sqrt{1-{x}^{2}}arcsin\left(x\right).\hfill \end{array}$
(2.38)
Note that
$\begin{array}{c}150-152{x}^{2}+142{x}^{4}+611{x}^{6}-1\text{,}215{x}^{8}\hfill \\ \phantom{\rule{1em}{0ex}}>150-152×{\left(0.7\right)}^{2}-1\text{,}215×{\left(0.7\right)}^{8}=5.477\dots >0\hfill \end{array}$
(2.39)

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

It follows from (2.10), (2.37) and (2.39) that
$\begin{array}{rl}{ϵ}_{1}^{\ast }\left(x\right)>& -x\left(150-202{x}^{2}-15{x}^{4}-68{x}^{6}+135{x}^{8}\right)\\ +\left(150-152{x}^{2}+142{x}^{4}+611{x}^{6}-1\text{,}215{x}^{8}\right)\left(x-\frac{{x}^{3}}{3}-\frac{{x}^{5}}{3}\right)\\ =& \frac{{x}^{5}}{3}\left[\frac{1\text{,}183}{4}+709\left(\frac{1}{4}-{x}^{4}\right)+2\text{,}047{x}^{2}\left(1-2{x}^{2}\right)+604{x}^{6}+1\text{,}215{x}^{8}\right]>0\end{array}$
(2.40)

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

We claim that
${ϵ}_{2}^{\ast }\left(x\right)<0$
(2.41)
for $x\in \left(0,1\right)$. Indeed, let $q\left(x\right)=75-76{x}^{2}-94{x}^{4}+278{x}^{6}-360{x}^{8}$, then $q\left(0.8009\right)=0.000171\dots$ , $q\left(0.80091\right)=-0.00356\dots$ and
${q}^{\prime }\left(x\right)=-4x\left[38+\frac{10\text{,}759{x}^{2}}{320}+720{x}^{2}\left({x}^{2}-\frac{139}{480}\right)\right]<0$
for $x\in \left(0,1\right)$. Therefore, there exists unique ${x}_{0}=0.80090\dots \in \left(0,1\right)$ such that $q\left(x\right)>0$ for $x\in \left(0,{x}_{0}\right)$ and $q\left(x\right)\le 0$ for $\left[{x}_{0},1\right)$. This in conjunction with (2.11) and (2.38) leads to
$\begin{array}{rl}{ϵ}_{2}^{\ast }\left(x\right)<& -x\left(1-{x}^{2}\right)\left(75-26{x}^{2}-6{x}^{4}-40{x}^{6}\right)\\ +\left(75-76{x}^{2}-94{x}^{4}+278{x}^{6}-360{x}^{8}\right)\left(x-\frac{{x}^{3}}{3}-\frac{2{x}^{5}}{15}\right)\\ =& -\frac{2{x}^{5}}{15}\left[\frac{1\text{,}897\text{,}305\text{,}741}{27\text{,}436\text{,}644}+2\text{,}619{\left({x}^{2}-\frac{2\text{,}651}{5\text{,}238}\right)}^{2}+2{x}^{4}\left(1-{x}^{2}\right)\left(491+180{x}^{2}\right)\right]<0\end{array}$

for $x\in \left(0,{x}_{0}\right)$ and ${ϵ}_{2}^{\ast }\left(x\right)\le -x\left(1-{x}^{2}\right)\left(75-26{x}^{2}-6{x}^{4}-40{x}^{6}\right)<0$ for $x\in \left[{x}_{0},1\right)$.

Therefore, inequality (2.31) follows from (2.33), (2.35), (2.36) and (2.40), and inequality (2.32) follows from (2.33)-(2.36) and (2.41). □

Lemma 2.6 Let
$\mu \left(x\right)=\frac{1+3{x}^{2}}{{\left(x+{x}^{3}\right)}^{2}}-\frac{1}{\left(1+{x}^{2}\right){\left[{sinh}^{-1}\left(x\right)\right]}^{2}}-\frac{x}{{\left(1+{x}^{2}\right)}^{3/2}{sinh}^{-1}\left(x\right)}.$

Then $\mu \left(x\right)<0.2$ for $x\in \left[0.7,1\right)$.

Proof Let
${\mu }_{1}\left(x\right)=\frac{1}{{x}^{2}}-\frac{1}{{\left[{sinh}^{-1}\left(x\right)\right]}^{2}},\phantom{\rule{2em}{0ex}}{\mu }_{2}\left(x\right)=\frac{2}{\sqrt{1+{x}^{2}}}-\frac{x}{{sinh}^{-1}\left(x\right)}.$
Then
$\mu \left(x\right)=\frac{{\mu }_{1}\left(x\right)}{1+{x}^{2}}+\frac{{\mu }_{2}\left(x\right)}{{\left(1+{x}^{2}\right)}^{3/2}}.$
(2.42)
Lemma 2.2 together with $x>{sinh}^{-1}\left(x\right)$ gives ${\mu }_{1}\left(x\right)<0$ and
${\mu }_{1}^{\prime }\left(x\right)=\frac{2}{{x}^{3}{\left[{sinh}^{-1}\left(x\right)\right]}^{3}}\left[\frac{{x}^{3}}{\sqrt{1+{x}^{2}}}-{\left({sinh}^{-1}\left(x\right)\right)}^{3}\right]>0$
for $x\in \left(0,1\right)$. This in turn implies that
${\left[\frac{{\mu }_{1}\left(x\right)}{1+{x}^{2}}\right]}^{\prime }=\frac{{\mu }_{1}^{\prime }\left(x\right)\left(1+{x}^{2}\right)-2x{\mu }_{1}\left(x\right)}{{\left(1+{x}^{2}\right)}^{2}}>0$
(2.43)

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

On the other hand, from the expression of ${\mu }_{2}\left(x\right)$, we get
${\mu }_{2}\left(1\right)=0.2796\dots >0,$
(2.44)
${\mu }_{2}^{\prime }\left(x\right)=-\frac{2x}{{\left(1+{x}^{2}\right)}^{3/2}}+\frac{{\mu }_{2}^{\ast }\left(x\right)}{{\left[{sinh}^{-1}\left(x\right)\right]}^{2}},$
(2.45)
where
${\mu }_{2}^{\ast }\left(x\right)=\frac{x}{\sqrt{1+{x}^{2}}}-{sinh}^{-1}\left(x\right),$
(2.46)
${\mu }_{2}^{\ast }\left(0\right)=0,$
(2.47)
${{\mu }_{2}^{\ast }}^{\prime }\left(x\right)=-\frac{{x}^{2}}{{\left(1+{x}^{2}\right)}^{3/2}}<0$
(2.48)

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

From (2.44)-(2.48) we clearly see that ${\mu }_{2}^{\prime }\left(x\right)<0$ and ${\mu }_{2}\left(x\right)>0$ for $x\in \left(0,1\right)$. This in turn implies that
${\left[\frac{{\mu }_{2}\left(x\right)}{{\left(1+{x}^{2}\right)}^{3/2}}\right]}^{\prime }=\frac{{\mu }_{2}^{\prime }\left(x\right){\left(1+{x}^{2}\right)}^{3/2}-3x\sqrt{1+{x}^{2}}{\mu }_{2}\left(x\right)}{{\left(1+{x}^{2}\right)}^{3}}<0$
(2.49)

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

Equation (2.42) and inequalities (2.43) and (2.49) lead to the conclusion that
$\mu \left(x\right)\le \frac{{\mu }_{1}\left(1\right)}{2}+\frac{{\mu }_{2}\left(0.7\right)}{{\left[1+{\left(0.7\right)}^{2}\right]}^{3/2}}=0.167\dots <0.2$

for $x\in \left[0.7,1\right)$. □

Lemma 2.7 Let
$\nu \left(x\right)=-\frac{1+3{x}^{2}}{{\left(x+{x}^{3}\right)}^{2}}+\frac{1}{\left(1-{x}^{2}\right){arcsin}^{2}\left(x\right)}-\frac{x}{{\left(1-{x}^{2}\right)}^{3/2}arcsin\left(x\right)}.$

Then $\nu \left(x\right)<-1.48$ for $x\in \left[0.7,1\right)$.

Proof Differentiating $\nu \left(x\right)$ yields
${\nu }^{\prime }\left(x\right)=\frac{{\left(x+{x}^{3}\right)}^{3}arcsin\left(x\right){\nu }_{1}\left(x\right)+\left(1-{x}^{2}\right){\nu }_{2}\left(x\right)}{{x}^{3}{\left(1-{x}^{2}\right)}^{5/2}{\left(1+{x}^{2}\right)}^{3}{arcsin}^{3}\left(x\right)},$
(2.50)
where
${\nu }_{1}\left(x\right)=3x\sqrt{1-{x}^{2}}-\left(1+2{x}^{2}\right)arcsin\left(x\right),$
(2.51)
${\nu }_{2}\left(x\right)=2\left(1+3{x}^{2}+6{x}^{4}\right){\left[\sqrt{1-{x}^{2}}arcsin\left(x\right)\right]}^{3}-2{\left(x+{x}^{3}\right)}^{3}.$
(2.52)
${\nu }_{1}\left(0.7\right)=-0.03558\dots ,$
(2.53)
${\nu }_{1}^{\prime }\left(x\right)=\frac{2-8{x}^{2}-4x\sqrt{1-{x}^{2}}arcsin\left(x\right)}{\sqrt{1-{x}^{2}}}<0$
(2.54)

for $x\in \left[0.7,1\right)$.

Therefore,
${\nu }_{1}\left(x\right)<0$
(2.55)

for $x\in \left[0.7,1\right)$ follows from (2.53) and (2.54).

It follows from (2.52) and (2.11) that
$\begin{array}{rl}{\nu }_{2}\left(x\right)& <2\left(1+3{x}^{2}+6{x}^{4}\right){\left(x-\frac{{x}^{3}}{3}\right)}^{3}-2{\left(x+{x}^{3}\right)}^{3}\\ =-\frac{2{x}^{5}}{27}\left(27-9{x}^{2}+163{x}^{4}-51{x}^{6}+6{x}^{8}\right)<0\end{array}$
(2.56)

for $x\in \left[0.7,1\right)$.

Equation (2.50) together with inequalities (2.55) and (2.56) leads to the conclusion that $\nu \left(x\right)$ is strictly decreasing on $\left[0.7,1\right)$. This in turn implies that
$\nu \left(x\right)\le \nu \left(0.7\right)=-1.48798\dots <-1.48$

for $x\in \left[0.7,1\right)$. □

Lemma 2.8 Let ${\lambda }_{0}=\left[2log\left(log\left(1+\sqrt{2}\right)\right)+log2\right]/\left[2log\pi -log2\right]=0.2760\dots$ , and $\mathrm{\Theta }\left(x\right)=\mathrm{\Phi }\left(x\right)+{\lambda }_{0}\mathrm{\Upsilon }\left(x\right)$, where $\mathrm{\Phi }\left(x\right)$ and $\mathrm{\Upsilon }\left(x\right)$ are defined as in Lemmas 2.4 and 2.5, respectively. Then the function $\mathrm{\Theta }\left(x\right)$ is strictly decreasing on $\left[0.7,1\right)$.

Proof Let $\mu \left(x\right)$ and $\nu \left(x\right)$ be defined as in Lemmas 2.6 and 2.7, respectively. Then differentiating $\mathrm{\Theta }\left(x\right)$ yields
${\mathrm{\Theta }}^{\prime }\left(x\right)={\mathrm{\Phi }}^{\prime }\left(x\right)+{\lambda }_{0}{\mathrm{\Upsilon }}^{\prime }\left(x\right)=\mu \left(x\right)+{\lambda }_{0}\nu \left(x\right)<0.2-1.48{\lambda }_{0}=-0.208\dots <0$

for $x\in \left[0.7,1\right)$. This in turn implies that $\mathrm{\Theta }\left(x\right)$ is strictly decreasing on $\left[0.7,1\right)$. □

## 3 Main result

Theorem 3.1 The double inequality
${P}^{\alpha }\left(a,b\right){Q}^{1-\alpha }\left(a,b\right)

holds for all $a,b>0$ with $a\ne b$ if and only if $\alpha \ge 1/2$ and $\beta \le \left[2log\left(log\left(1+\sqrt{2}\right)\right)+log2\right]/\left[2log\pi -log2\right]=0.2760\dots$ .

Proof Since $P\left(a,b\right)$, $M\left(a,b\right)$ and $Q\left(a,b\right)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $a>b$. Let $p\in \left(0,1\right)$, ${\lambda }_{0}=\left[2log\left(log\left(1+\sqrt{2}\right)\right)+log2\right]/\left[2log\pi -log2\right]$ and $x=\left(a-b\right)/\left(a+b\right)$. Then $x\in \left(0,1\right)$,
$\begin{array}{c}\frac{P\left(a,b\right)}{A\left(a,b\right)}=\frac{x}{arcsin\left(x\right)},\phantom{\rule{2em}{0ex}}\frac{M\left(a,b\right)}{A\left(a,b\right)}=\frac{x}{{sinh}^{-1}\left(x\right)},\phantom{\rule{2em}{0ex}}\frac{Q\left(a,b\right)}{A\left(a,b\right)}=\sqrt{1+{x}^{2}},\hfill \\ \frac{log\left[Q\left(a,b\right)\right]-log\left[M\left(a,b\right)\right]}{log\left[Q\left(a,b\right)\right]-log\left[P\left(a,b\right)\right]}=\frac{log\left(1+{x}^{2}\right)-2logx+2log\left[{sinh}^{-1}\left(x\right)\right]}{log\left(1+{x}^{2}\right)-2logx+2log\left[arcsin\left(x\right)\right]}\hfill \end{array}$
(3.1)
and
$\underset{x\to {0}^{+}}{lim}\frac{log\left(1+{x}^{2}\right)-2logx+2log\left[{sinh}^{-1}\left(x\right)\right]}{log\left(1+{x}^{2}\right)-2logx+2log\left[arcsin\left(x\right)\right]}=\frac{1}{2},$
(3.2)
$\underset{x\to {1}^{-}}{lim}\frac{log\left(1+{x}^{2}\right)-2logx+2log\left[{sinh}^{-1}\left(x\right)\right]}{log\left(1+{x}^{2}\right)-2logx+2log\left[arcsin\left(x\right)\right]}={\lambda }_{0}.$
(3.3)
The difference between the convex combination of $log\left[P\left(a,b\right)\right]$, $log\left[Q\left(a,b\right)\right]$ and $log\left[M\left(a,b\right)\right]$ is given by
$\begin{array}{c}plog\left[P\left(a,b\right)\right]+\left(1-p\right)log\left[Q\left(a,b\right)\right]-log\left[M\left(a,b\right)\right]\hfill \\ \phantom{\rule{1em}{0ex}}=plog\left[\frac{x}{arcsin\left(x\right)}\right]+\frac{1-p}{2}log\left(1+{x}^{2}\right)-log\left[\frac{x}{{sinh}^{-1}\left(x\right)}\right]:={D}_{p}\left(x\right).\hfill \end{array}$
(3.4)
${D}_{p}\left({0}^{+}\right)=0,\phantom{\rule{2em}{0ex}}{D}_{p}\left({1}^{-}\right)=log\left[\sqrt{2}log\left(1+\sqrt{2}\right)\right]-plog\left(\frac{\pi }{\sqrt{2}}\right),\phantom{\rule{2em}{0ex}}{D}_{{\lambda }_{0}}\left({1}^{-}\right)=0,$
(3.5)
${D}_{p}^{\prime }\left(x\right)=-\frac{p}{\sqrt{1-{x}^{2}}arcsin\left(x\right)}-\frac{\left(1-p\right)}{x\left(1+{x}^{2}\right)}+\frac{1}{\sqrt{1+{x}^{2}}{sinh}^{-1}\left(x\right)}=\mathrm{\Phi }\left(x\right)+p\mathrm{\Upsilon }\left(x\right),$
(3.6)

where $\mathrm{\Phi }\left(x\right)$ and $\mathrm{\Upsilon }\left(x\right)$ are defined as in Lemmas 2.4 and 2.5, respectively.

From Lemmas 2.4 and 2.5, we clearly see that
$\begin{array}{rl}{D}_{1/2}^{\prime }\left(x\right)& =\mathrm{\Phi }\left(x\right)+\frac{1}{2}\mathrm{\Upsilon }\left(x\right)\\ <\frac{2x}{3}-\frac{34{x}^{3}}{45}+\frac{4{x}^{5}}{5}-\frac{1}{2}\left(\frac{4x}{3}-\frac{34{x}^{3}}{45}+\frac{8{x}^{5}}{9}\right)\\ =-\frac{16{x}^{3}}{45}\left(\frac{17}{16}-{x}^{2}\right)<0\end{array}$
(3.7)
for $x\in \left(0,1\right)$, and
$\begin{array}{rl}{D}_{{\lambda }_{0}}^{\prime }\left(x\right)=& \mathrm{\Phi }\left(x\right)+{\lambda }_{0}\mathrm{\Upsilon }\left(x\right)\\ >& \frac{2x}{3}-\frac{34{x}^{3}}{45}+\frac{754{x}^{5}}{945}-{x}^{7}-{\lambda }_{0}\left(\frac{4x}{3}-\frac{34{x}^{3}}{45}+\frac{3{x}^{5}}{2}\right)\\ =& x\left[\frac{2\left(1-2{\lambda }_{0}\right)}{3}-\frac{34\left(1-{\lambda }_{0}\right)}{45}{x}^{2}+\left(\frac{754}{945}-\frac{3{\lambda }_{0}}{2}\right){x}^{4}-{x}^{6}\right]\\ :=& x{F}_{{\lambda }_{0}}\left(x\right)>0\end{array}$
(3.8)

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

Note that
${F}_{{\lambda }_{0}}\left(0\right)=2\left(1-2{\lambda }_{0}\right)/3>0,\phantom{\rule{2em}{0ex}}{F}_{{\lambda }_{0}}\left(0.7\right)=0.00513\dots >0$
(3.9)
and
$\begin{array}{rcl}{F}_{{\lambda }_{0}}^{″}\left(x\right)& =& -30\left[{\left({x}^{2}-\frac{1\text{,}508-2\text{,}835{\lambda }_{0}}{9\text{,}450}\right)}^{2}+\frac{2\text{,}224\text{,}136+4\text{,}052\text{,}160{\lambda }_{0}-8\text{,}037\text{,}225{\lambda }_{0}^{2}}{89\text{,}302\text{,}500}\right]\\ <& 0\end{array}$
(3.10)

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

Inequalities (3.8)-(3.10) lead to the conclusion that
${D}_{{\lambda }_{0}}^{\prime }\left(x\right)>0$
(3.11)

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

It follows from Lemma 2.8 and (3.6) that ${D}_{{\lambda }_{0}}^{\prime }\left(x\right)$ is strictly decreasing in $\left[0.7,1\right)$. Then from (3.11) and ${D}_{{\lambda }_{0}}^{\prime }\left(0.7\right)=0.0626\dots$ together with ${D}_{{\lambda }_{0}}^{\prime }\left({1}^{-}\right)=-\mathrm{\infty }$, we know that there exists ${x}^{\ast }\in \left(0.7,1\right)$ such that ${D}_{{\lambda }_{0}}\left(x\right)$ is strictly increasing on $\left(0,{x}^{\ast }\right]$ and strictly decreasing on $\left[{x}^{\ast },1\right)$. This in conjunction with (3.5) implies that
${D}_{{\lambda }_{0}}\left(x\right)>0$
(3.12)

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

Equations (3.4), (3.5), (3.7) and (3.12) lead to the conclusion that
$M\left(a,b\right)<{P}^{{\lambda }_{0}}\left(a,b\right){Q}^{1-{\lambda }_{0}}\left(a,b\right)$
(3.13)
and
$M\left(a,b\right)>{P}^{1/2}\left(a,b\right){Q}^{1/2}\left(a,b\right).$
(3.14)

Therefore, Theorem 3.1 follows from (3.13) and (3.14) together with the following statements:

• If $\alpha <1/2$, then (3.1) and (3.2) imply that there exists ${\delta }_{1}\in \left(0,1\right)$ such that $M\left(a,b\right)<{P}^{\alpha }\left(a,b\right){Q}^{1-\alpha }\left(a,b\right)$ for all $a,b>0$ with $\left(a-b\right)/\left(a+b\right)\in \left(0,{\delta }_{1}\right)$.

• If $\beta >{\lambda }_{0}$, then (3.1) and (3.3) imply that there exists ${\delta }_{2}\in \left(0,1\right)$ such that $M\left(a,b\right)>{P}^{\beta }\left(a,b\right){Q}^{1-\beta }\left(a,b\right)$ for all $a,b>0$ with $\left(a-b\right)/\left(a+b\right)\in \left(1-{\delta }_{2},1\right)$.

□

## Declarations

### Acknowledgements

This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307, and the Natural Science Foundation of Zhejiang Province under Grants LY13H070004 and LY13A010004.

## Authors’ Affiliations

(1)
School of Mathematics and Computation Science, Hunan City University, Yiyang, 413000, China
(2)
College of Nursing, Huzhou Teachers College, Huzhou, 313000, China

## References 