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

Abstract

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

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

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

MSC:26E60.

1 Introduction

Let u, v and w be the bivariate means such that $u(a,b)<w(a,b)<v(a,b)$ for all $a,b>0$ with $a\ne b$. The problems of finding the best possible parameters α and β such that the inequalities $\alpha u(a,b)+(1-\alpha )v(a,b)<w(a,b)<\beta u(a,b)+(1-\beta )v(a,b)$ and $u^{\alpha }(a,b)v^{1-\alpha }(a,b)<w(a,b)<u^{\beta }(a,b)v^{1-\beta }(a,b)$ 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 $P(a,b)$ [1], the Neuman-Sándor mean $M(a,b)$ [2], the quadratic mean $Q(a,b)$ are defined by

$$\begin{array}{r}P(a,b)=\frac{a-b}{4arctan(\sqrt{a/b})-\pi },M(a,b)=\frac{a-b}{2{sinh}^{-1}(\frac{a-b}{a+b})},\\ Q(a,b)=\sqrt{\frac{a^2+b^2}{2}},\end{array}$$

(1.1)

respectively. In here, ${sinh}^{-1}(x)=log(x+\sqrt{x^2+1})$ 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 [3–14]. The first Seiffert mean $P(a,b)$ can be rewritten as (see [[2], Eq. (2.4)])

$$P(a,b)=\frac{a-b}{2arcsin[(a-b)/(a+b)]}.$$

(1.2)

Let $H(a,b)=2ab/(a+b)$, $L(a,b)=(b-a)/(logb-loga)$, $A(a,b)=(a+b)/2$, $T(a,b)=(a-b)/[2arctan((a-b)/(a+b))]$ and $C(a,b)=(a^2+b^2)/(a+b)$ be the harmonic, logarithmic, arithmetic, second Seiffert and contra-harmonic means of a and b, respectively. Then it is known that the inequalities

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

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

Neuman and Sándor [2, 15] proved that the inequalities

$$\begin{array}{c}\frac{\pi }{4log(1+\sqrt{2})}T(a,b)<M(a,b)<\frac{A(a,b)}{log(1+\sqrt{2})},\hfill \\ \sqrt{2T^2(a,b)-Q^2(a,b)}<M(a,b)<\frac{T^2(a,b)}{Q(a,b)},\hfill \\ H(T(a,b),A(a,b))<M(a,b)<L(A(a,b),Q(a,b)),T(a,b)>H(M(a,b),Q(a,b)),\hfill \\ M(a,b)<\frac{A^2(a,b)}{P(a,b)},A^{2/3}(a,b)Q^{1/3}(a,b)<M(a,b)<\frac{2A(a,b)+Q(a,b)}{3},\hfill \\ \sqrt{A(a,b)T(a,b)}<M(a,b)<\sqrt{A^2(a,b)+T^2(a,b)},\hfill \\ \frac{A(x,y)}{A(1-x,1-y)}<\frac{M(x,y)}{M(1-x,1-y)}<\frac{T(x,y)}{T(1-x,1-y)},\hfill \\ \frac{1}{A(1-x,1-y)}-\frac{1}{A(x,y)}<\frac{1}{M(1-x,1-y)}-\frac{1}{M(x,y)}<\frac{1}{T(1-x,1-y)}-\frac{1}{T(x,y)},\hfill \\ A(x,y)A(1-x,1-y)<M(x,y)M(1-x,1-y)<T(x,y)T(1-x,1-y)\hfill \end{array}$$

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

Li et al. [16] proved that the double inequality $L_{p_0}(a,b)<M(a,b)<L_2(a,b)$ holds for all $a,b>0$ with $a\ne b$, where $L_p(a,b)={[(b^{p+1}-a^{p+1})/((p+1)(b-a))]}^{1/p}$ ($p\ne -1,0$), $L_0(a,b)=1/e{(b^b/a^a)}^{1/(b-a)}$ and $L_{-1}(a,b)=(b-a)/(logb-loga)$ is the p th generalized logarithmic mean of a and b, and $p_0=1.843\dots$ is the unique solution of the equation ${(p+1)}^{1/p}=2log(1+\sqrt{2})$.

In [13], Neuman proved that the double inequalities

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

(1.3)

and

$$C^{\lambda }(a,b)A^{1-\lambda }(a,b)<M(a,b)<C^{\mu }(a,b)A^{1-\mu }(a,b)$$

(1.4)

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

Jiang and Qi [17, 18] gave the best possible parameters α, β, $t_1$ and $t_2$ in $(0,1/2)$ such that the inequalities

$$\begin{array}{r}Q(\alpha a+(1-\alpha )b,\alpha b+(1-\alpha )a)<M(a,b)<Q(\beta a+(1-\beta )b,\beta b+(1-\beta )a),\\ Q_{t_1,p}(a,b)<M(a,b)<Q_{t_2,p}(a,b)\end{array}$$

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

Inspired by inequalities (1.3) and (1.4), in this paper, we present the optimal upper and lower bounds for the Neuman-Sándor mean $M(a,b)$ in terms of the geometric convex combinations of the first Seiffert mean $P(a,b)$ and the quadratic mean $Q(a,b)$. 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

$$x+\frac{x^3}{3}-\frac{2x^5}{15}<\sqrt{1+x^2}{sinh}^{-1}(x)<x+\frac{x^3}{3}-\frac{2x^5}{15}+\frac{8x^7}{105}$$

(2.1)

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

Proof To show inequality (2.1), it suffices to prove that

$${\omega }_1(x)=\sqrt{1+x^2}{sinh}^{-1}(x)-(x+\frac{x^3}{3}-\frac{2x^5}{15})>0$$

(2.2)

and

$${\omega }_2(x)=\sqrt{1+x^2}{sinh}^{-1}(x)-(x+\frac{x^3}{3}-\frac{2x^5}{15}+\frac{8x^7}{105})<0$$

(2.3)

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

From the expressions of ${\omega }_1(x)$ and ${\omega }_2(x)$, we get

$${\omega }_1(0)={\omega }_2(0)=0,$$

(2.4)

$${\omega }_1^{\prime }(x)=\frac{x{\omega }_1^{\ast }(x)}{\sqrt{1+x^2}},{\omega }_2^{\prime }(x)=\frac{x{\omega }_2^{\ast }(x)}{\sqrt{1+x^2}},$$

(2.5)

where

$$\begin{array}{c}{\omega }_1^{\ast }(x)={sinh}^{-1}(x)-(x-\frac{2x^3}{3})\sqrt{1+x^2},\hfill \\ {\omega }_2^{\ast }(x)={sinh}^{-1}(x)-(x-\frac{2x^3}{3}+\frac{8x^5}{15})\sqrt{1+x^2},\hfill \\ {\omega }_1^{\ast }(0)={\omega }_2^{\ast }(0)=0,\hfill \end{array}$$

(2.6)

$${{\omega }_1^{\ast }}^{\prime }(x)=\frac{8x^4}{3\sqrt{1+x^2}}>0$$

(2.7)

and

$${{\omega }_2^{\ast }}^{\prime }(x)=-\frac{16x^6}{5\sqrt{1+x^2}}<0$$

(2.8)

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

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

$$\frac{x^3}{\sqrt{1+x^2}}>{[{sinh}^{-1}(x)]}^3$$

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

Proof Let $x\in (0,1)$, then from (1.3) we have

$$M(1+x,1-x)>A^{2/3}(1+x,1-x)Q^{1/3}(1+x,1-x).$$

(2.9)

Therefore, Lemma 2.2 follows from (2.9). □

Lemma 2.3 The inequality

$$\sqrt{1-x^2}arcsin(x)>x-\frac{x^3}{3}-\frac{x^5}{3}$$

(2.10)

holds for $x\in (0,0.7)$, and the inequality

$$\sqrt{1-x^2}arcsin(x)<x-\frac{x^3}{3}-\frac{2x^5}{15}$$

(2.11)

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

Proof Let

$${\phi }_1(x)=\sqrt{1-x^2}arcsin(x)-x+\frac{x^3}{3}+\frac{x^5}{3},$$

(2.12)

$${\phi }_2(x)=\sqrt{1-x^2}arcsin(x)-x+\frac{x^3}{3}+\frac{2x^5}{15}.$$

(2.13)

Then simple computations lead to

$${\phi }_1(0)={\phi }_2(0)=0,$$

(2.14)

$${\phi }_1^{\prime }(x)=\frac{x{\phi }_1^{\ast }(x)}{\sqrt{1-x^2}},{\phi }_2^{\prime }(x)=\frac{x{\phi }_2^{\ast }(x)}{\sqrt{1-x^2}},$$

(2.15)

where

$$\begin{array}{c}{\phi }_1^{\ast }(x)=(x+\frac{5x^3}{3})\sqrt{1-x^2}-arcsin(x),\hfill \\ {\phi }_2^{\ast }(x)=(x+\frac{2x^3}{3})\sqrt{1-x^2}-arcsin(x).\hfill \end{array}$$

Note that

$${\phi }_1^{\ast }(0)={\phi }_2^{\ast }(0)=0,{\phi }_1^{\ast }(0.7)=0.1327\dots ,$$

(2.16)

$${{\phi }_1^{\ast }}^{\prime }(x)=\frac{x^2(9-20x^2)}{3\sqrt{1-x^2}},$$

(2.17)

$${{\phi }_2^{\ast }}^{\prime }(x)=-\frac{8x^4}{3\sqrt{1-x^2}}<0$$

(2.18)

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

From (2.17) we clearly see that ${\phi }_1^{\ast }(x)$ is strictly increasing on $(0,3\sqrt{5}/10]$ and strictly decreasing on $[3\sqrt{5}/10,0.7)$. This in conjunction with (2.16) implies that

$${\phi }_1^{\ast }(x)>0$$

(2.19)

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

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

$$\mathrm{\Phi }(x)=\frac{1}{\sqrt{1+x^2}{sinh}^{-1}(x)}-\frac{1}{x(1+x^2)}.$$

Then the inequality

$$\mathrm{\Phi }(x)>\frac{2x}{3}-\frac{34x^3}{45}+\frac{754x^5}{945}-x^7$$

(2.20)

holds for $x\in (0,0.7)$, and

$$\mathrm{\Phi }(x)<\frac{2x}{3}-\frac{34x^3}{45}+\frac{4x^5}{5}$$

(2.21)

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

Proof To show inequalities (2.20) and (2.21), it suffices to prove that

$$\begin{array}{rl}{\varphi }_1(x):=& x(1+x^2){sinh}^{-1}(x)[\mathrm{\Phi }(x)-(\frac{2x}{3}-\frac{34x^3}{45}+\frac{754x^5}{945}-x^7)]\\ =& x\sqrt{1+x^2}-{sinh}^{-1}(x)\\ -x(1+x^2){sinh}^{-1}(x)(\frac{2x}{3}-\frac{34x^3}{45}+\frac{754x^5}{945}-x^7)>0\end{array}$$

(2.22)

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

$$\begin{array}{rl}{\varphi }_2(x):=& x(1+x^2){sinh}^{-1}(x)[\mathrm{\Phi }(x)-(\frac{2x}{3}-\frac{34x^3}{45}+\frac{4x^5}{5})]\\ =& x\sqrt{1+x^2}-{sinh}^{-1}(x)\\ -x(1+x^2){sinh}^{-1}(x)(\frac{2x}{3}-\frac{34x^3}{45}+\frac{4x^5}{5})<0\end{array}$$

(2.23)

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

From the expressions of ${\varphi }_1(x)$ and ${\varphi }_2(x)$, one has

$${\varphi }_1(0)={\varphi }_2(0)=0,$$

(2.24)

$${\varphi }_1^{\prime }(x)=\frac{x}{945\sqrt{1+x^2}}{\varphi }_1^{\ast }(x),{\varphi }_2^{\prime }(x)=-\frac{2x}{45\sqrt{1+x^2}}{\varphi }_2^{\ast }(x),$$

(2.25)

where

$$\begin{array}{c}{\varphi }_1^{\ast }(x)=x(1\text{,}260+84x^2-40x^4+191x^6+945x^8)\hfill \\ \phantom{{\varphi }_1^{\ast }(x)=}-2(630-168x^2+120x^4-764x^6-4\text{,}725x^8)\sqrt{1+x^2}{sinh}^{-1}(x),\hfill \end{array}$$

(2.26)

$$\begin{array}{c}{\varphi }_2^{\ast }(x)=x(18x^6+x^4-2x^2-30)\hfill \\ \phantom{{\varphi }_2^{\ast }(x)=}+2(15-4x^2+3x^4+72x^6)\sqrt{1+x^2}{sinh}^{-1}(x).\hfill \end{array}$$

(2.27)

Note that

$$\begin{array}{c}630-168x^2+120x^4-764x^6-4\text{,}725x^8\hfill \\ >630-168\times {(0.7)}^2-764\times {(0.7)}^6-4\text{,}725\times {(0.7)}^8=185.4\dots >0\hfill \end{array}$$

(2.28)

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

Lemma 2.1 and equations (2.26)-(2.28) lead to

$$\begin{array}{rl}{\varphi }_1^{\ast }(x)>& x(1\text{,}260+84x^2-40x^4+191x^6+945x^8)\\ -2(630-168x^2+120x^4-764x^6-4\text{,}725x^8)(x+\frac{x^3}{3}-\frac{2x^5}{15}+\frac{8x^7}{105})\\ =& \frac{x^7}{105}(157\text{,}311+1\text{,}151\text{,}003x^2+307\text{,}438x^4-120\text{,}076x^6+75\text{,}600x^8)>0\end{array}$$

(2.29)

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

$$\begin{array}{rl}{\varphi }_2^{\ast }(x)>& x(18x^6+x^4-2x^2-30)\\ +2(15-4x^2+3x^4+72x^6)(x+\frac{x^3}{3}-\frac{2x^5}{15})\\ =& \frac{x^5}{15}(5+2\text{,}476x^2+708x^4-288x^6)>0\end{array}$$

(2.30)

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

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

$$\mathrm{\Upsilon }(x)=\frac{1}{x(1+x^2)}-\frac{1}{\sqrt{1-x^2}arcsin(x)}.$$

Then the inequality

$$\mathrm{\Upsilon }(x)>-\frac{4x}{3}+\frac{34x^3}{45}-\frac{3x^5}{2}$$

(2.31)

holds for $x\in (0,0.7)$, and

$$\mathrm{\Upsilon }(x)<-\frac{4x}{3}+\frac{34x^3}{45}-\frac{8x^5}{9}$$

(2.32)

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

Proof Let

$$\begin{array}{rl}{ϵ}_1(x):=& x(1+x^2)\sqrt{1-x^2}arcsin(x)[\mathrm{\Upsilon }(x)+(\frac{4x}{3}-\frac{34x^3}{45}+\frac{3x^5}{2})]\\ =& \sqrt{1-x^2}arcsin(x)-x(1+x^2)\\ +x(1+x^2)\sqrt{1-x^2}arcsin(x)(\frac{4x}{3}-\frac{34x^3}{45}+\frac{3x^5}{2})\end{array}$$

(2.33)

and

$$\begin{array}{rl}{ϵ}_2(x):=& x(1+x^2)\sqrt{1-x^2}arcsin(x)[\mathrm{\Upsilon }(x)+(\frac{4x}{3}-\frac{34x^3}{45}+\frac{8x^5}{9})]\\ =& \sqrt{1-x^2}arcsin(x)-x(1+x^2)\\ +x(1+x^2)\sqrt{1-x^2}arcsin(x)(\frac{4x}{3}-\frac{34x^3}{45}+\frac{8x^5}{9}).\end{array}$$

(2.34)

An easy calculation gives rise to

$${ϵ}_1(0)={ϵ}_2(0)=0,$$

(2.35)

$${ϵ}_1^{\prime }(x)=\frac{x}{90(1-x^2)}{ϵ}_1^{\ast }(x),{ϵ}_2^{\prime }(x)=\frac{x}{45(1-x^2)}{ϵ}_2^{\ast }(x),$$

(2.36)

where

$$\begin{array}{c}{ϵ}_1^{\ast }(x)=-x(150-202x^2-15x^4-68x^6+135x^8)\hfill \\ \phantom{{ϵ}_1^{\ast }(x)=}+(150-152x^2+142x^4+611x^6-1\text{,}215x^8)\sqrt{1-x^2}arcsin(x),\hfill \end{array}$$

(2.37)

$$\begin{array}{c}{ϵ}_2^{\ast }(x)=-x(1-x^2)(75-26x^2-6x^4-40x^6)\hfill \\ \phantom{{ϵ}_2^{\ast }(x)=}+(75-76x^2-94x^4+278x^6-360x^8)\sqrt{1-x^2}arcsin(x).\hfill \end{array}$$

(2.38)

Note that

$$\begin{array}{c}150-152x^2+142x^4+611x^6-1\text{,}215x^8\hfill \\ >150-152\times {(0.7)}^2-1\text{,}215\times {(0.7)}^8=5.477\dots >0\hfill \end{array}$$

(2.39)

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

It follows from (2.10), (2.37) and (2.39) that

$$\begin{array}{rl}{ϵ}_1^{\ast }(x)>& -x(150-202x^2-15x^4-68x^6+135x^8)\\ +(150-152x^2+142x^4+611x^6-1\text{,}215x^8)(x-\frac{x^3}{3}-\frac{x^5}{3})\\ =& \frac{x^5}{3}[\frac{1\text{,}183}{4}+709(\frac{1}{4}-x^4)+2\text{,}047x^2(1-2x^2)+604x^6+1\text{,}215x^8]>0\end{array}$$

(2.40)

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

We claim that

$${ϵ}_2^{\ast }(x)<0$$

(2.41)

for $x\in (0,1)$. Indeed, let $q(x)=75-76x^2-94x^4+278x^6-360x^8$, then $q(0.8009)=0.000171\dots$ , $q(0.80091)=-0.00356\dots$ and

$$q^{\prime }(x)=-4x[38+\frac{10\text{,}759x^2}{320}+720x^2(x^2-\frac{139}{480})]<0$$

for $x\in (0,1)$. Therefore, there exists unique $x_0=0.80090\dots \in (0,1)$ such that $q(x)>0$ for $x\in (0,x_0)$ and $q(x)\le 0$ for $[x_0,1)$. This in conjunction with (2.11) and (2.38) leads to

$$\begin{array}{rl}{ϵ}_2^{\ast }(x)<& -x(1-x^2)(75-26x^2-6x^4-40x^6)\\ +(75-76x^2-94x^4+278x^6-360x^8)(x-\frac{x^3}{3}-\frac{2x^5}{15})\\ =& -\frac{2x^5}{15}[\frac{1\text{,}897\text{,}305\text{,}741}{27\text{,}436\text{,}644}+2\text{,}619{(x^2-\frac{2\text{,}651}{5\text{,}238})}^2+2x^4(1-x^2)(491+180x^2)]<0\end{array}$$

for $x\in (0,x_0)$ and ${ϵ}_2^{\ast }(x)\le -x(1-x^2)(75-26x^2-6x^4-40x^6)<0$ for $x\in [x_0,1)$.

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

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

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

Proof Let

$${\mu }_1(x)=\frac{1}{x^2}-\frac{1}{{[{sinh}^{-1}(x)]}^2},{\mu }_2(x)=\frac{2}{\sqrt{1+x^2}}-\frac{x}{{sinh}^{-1}(x)}.$$

Then

$$\mu (x)=\frac{{\mu }_1(x)}{1+x^2}+\frac{{\mu }_2(x)}{{(1+x^2)}^{3/2}}.$$

(2.42)

Lemma 2.2 together with $x>{sinh}^{-1}(x)$ gives ${\mu }_1(x)<0$ and

$${\mu }_1^{\prime }(x)=\frac{2}{x^3{[{sinh}^{-1}(x)]}^3}[\frac{x^3}{\sqrt{1+x^2}}-{({sinh}^{-1}(x))}^3]>0$$

for $x\in (0,1)$. This in turn implies that

$${\left[\frac{{\mu }_1(x)}{1+x^2}\right]}^{\prime }=\frac{{\mu }_1^{\prime }(x)(1+x^2)-2x{\mu }_1(x)}{{(1+x^2)}^2}>0$$

(2.43)

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

On the other hand, from the expression of ${\mu }_2(x)$, we get

$${\mu }_2(1)=0.2796\dots >0,$$

(2.44)

$${\mu }_2^{\prime }(x)=-\frac{2x}{{(1+x^2)}^{3/2}}+\frac{{\mu }_2^{\ast }(x)}{{[{sinh}^{-1}(x)]}^2},$$

(2.45)

where

$${\mu }_2^{\ast }(x)=\frac{x}{\sqrt{1+x^2}}-{sinh}^{-1}(x),$$

(2.46)

$${\mu }_2^{\ast }(0)=0,$$

(2.47)

$${{\mu }_2^{\ast }}^{\prime }(x)=-\frac{x^2}{{(1+x^2)}^{3/2}}<0$$

(2.48)

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

From (2.44)-(2.48) we clearly see that ${\mu }_2^{\prime }(x)<0$ and ${\mu }_2(x)>0$ for $x\in (0,1)$. This in turn implies that

$${\left[\frac{{\mu }_2(x)}{{(1+x^2)}^{3/2}}\right]}^{\prime }=\frac{{\mu }_2^{\prime }(x){(1+x^2)}^{3/2}-3x\sqrt{1+x^2}{\mu }_2(x)}{{(1+x^2)}^3}<0$$

(2.49)

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

Equation (2.42) and inequalities (2.43) and (2.49) lead to the conclusion that

$$\mu (x)\le \frac{{\mu }_1(1)}{2}+\frac{{\mu }_2(0.7)}{{[1+{(0.7)}^2]}^{3/2}}=0.167\dots <0.2$$

for $x\in [0.7,1)$. □

Lemma 2.7 Let

$$\nu (x)=-\frac{1+3x^2}{{(x+x^3)}^2}+\frac{1}{(1-x^2){arcsin}^2(x)}-\frac{x}{{(1-x^2)}^{3/2}arcsin(x)}.$$

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

Proof Differentiating $\nu (x)$ yields

$${\nu }^{\prime }(x)=\frac{{(x+x^3)}^{3}arcsin(x){\nu }_1(x)+(1-x^2){\nu }_2(x)}{x^3{(1-x^2)}^{5/2}{(1+x^2)}^3{arcsin}^3(x)},$$

(2.50)

where

$${\nu }_1(x)=3x\sqrt{1-x^2}-(1+2x^2)arcsin(x),$$

(2.51)

$${\nu }_2(x)=2(1+3x^2+6x^4){[\sqrt{1-x^2}arcsin(x)]}^3-2{(x+x^3)}^3.$$

(2.52)

Equation (2.51) leads to

$${\nu }_1(0.7)=-0.03558\dots ,$$

(2.53)

$${\nu }_1^{\prime }(x)=\frac{2-8x^2-4x\sqrt{1-x^2}arcsin(x)}{\sqrt{1-x^2}}<0$$

(2.54)

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

Therefore,

$${\nu }_1(x)<0$$

(2.55)

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

It follows from (2.52) and (2.11) that

$$\begin{array}{rl}{\nu }_2(x)& <2(1+3x^2+6x^4){(x-\frac{x^3}{3})}^3-2{(x+x^3)}^3\\ =-\frac{2x^5}{27}(27-9x^2+163x^4-51x^6+6x^8)<0\end{array}$$

(2.56)

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

Equation (2.50) together with inequalities (2.55) and (2.56) leads to the conclusion that $\nu (x)$ is strictly decreasing on $[0.7,1)$. This in turn implies that

$$\nu (x)\le \nu (0.7)=-1.48798\dots <-1.48$$

for $x\in [0.7,1)$. □

Lemma 2.8 Let ${\lambda }_0=[2log(log(1+\sqrt{2}))+log2]/[2log\pi -log2]=0.2760\dots$ , and $\mathrm{\Theta }(x)=\mathrm{\Phi }(x)+{\lambda }_0\mathrm{\Upsilon }(x)$, where $\mathrm{\Phi }(x)$ and $\mathrm{\Upsilon }(x)$ are defined as in Lemmas 2.4 and 2.5, respectively. Then the function $\mathrm{\Theta }(x)$ is strictly decreasing on $[0.7,1)$.

Proof Let $\mu (x)$ and $\nu (x)$ be defined as in Lemmas 2.6 and 2.7, respectively. Then differentiating $\mathrm{\Theta }(x)$ yields

$${\mathrm{\Theta }}^{\prime }(x)={\mathrm{\Phi }}^{\prime }(x)+{\lambda }_0{\mathrm{\Upsilon }}^{\prime }(x)=\mu (x)+{\lambda }_0\nu (x)<0.2-1.48{\lambda }_0=-0.208\dots <0$$

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

3 Main result

Theorem 3.1 The double inequality

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

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

Proof Since $P(a,b)$, $M(a,b)$ and $Q(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $a>b$. Let $p\in (0,1)$, ${\lambda }_0=[2log(log(1+\sqrt{2}))+log2]/[2log\pi -log2]$ and $x=(a-b)/(a+b)$. Then $x\in (0,1)$,

$$\begin{array}{c}\frac{P(a,b)}{A(a,b)}=\frac{x}{arcsin(x)},\frac{M(a,b)}{A(a,b)}=\frac{x}{{sinh}^{-1}(x)},\frac{Q(a,b)}{A(a,b)}=\sqrt{1+x^2},\hfill \\ \frac{log[Q(a,b)]-log[M(a,b)]}{log[Q(a,b)]-log[P(a,b)]}=\frac{log(1+x^2)-2logx+2log[{sinh}^{-1}(x)]}{log(1+x^2)-2logx+2log[arcsin(x)]}\hfill \end{array}$$

(3.1)

and

$$\underset{x\to 0^{+}}{lim}\frac{log(1+x^2)-2logx+2log[{sinh}^{-1}(x)]}{log(1+x^2)-2logx+2log[arcsin(x)]}=\frac{1}{2},$$

(3.2)

$$\underset{x\to 1^{-}}{lim}\frac{log(1+x^2)-2logx+2log[{sinh}^{-1}(x)]}{log(1+x^2)-2logx+2log[arcsin(x)]}={\lambda }_0.$$

(3.3)

The difference between the convex combination of $log[P(a,b)]$, $log[Q(a,b)]$ and $log[M(a,b)]$ is given by

$$\begin{array}{c}plog[P(a,b)]+(1-p)log[Q(a,b)]-log[M(a,b)]\hfill \\ =plog\left[\frac{x}{arcsin(x)}\right]+\frac{1-p}{2}log(1+x^2)-log\left[\frac{x}{{sinh}^{-1}(x)}\right]:=D_p(x).\hfill \end{array}$$

(3.4)

Equation (3.4) leads to

$$D_p\left(0^{+}\right)=0,D_p\left(1^{-}\right)=log[\sqrt{2}log(1+\sqrt{2})]-plog\left(\frac{\pi }{\sqrt{2}}\right),D_{{\lambda }_0}\left(1^{-}\right)=0,$$

(3.5)

$$D_p^{\prime }(x)=-\frac{p}{\sqrt{1-x^2}arcsin(x)}-\frac{(1-p)}{x(1+x^2)}+\frac{1}{\sqrt{1+x^2}{sinh}^{-1}(x)}=\mathrm{\Phi }(x)+p\mathrm{\Upsilon }(x),$$

(3.6)

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

From Lemmas 2.4 and 2.5, we clearly see that

$$\begin{array}{rl}{D}_{1/2}^{\prime }(x)& =\mathrm{\Phi }(x)+\frac{1}{2}\mathrm{\Upsilon }(x)\\ <\frac{2x}{3}-\frac{34x^3}{45}+\frac{4x^5}{5}-\frac{1}{2}(\frac{4x}{3}-\frac{34x^3}{45}+\frac{8x^5}{9})\\ =-\frac{16x^3}{45}(\frac{17}{16}-x^2)<0\end{array}$$

(3.7)

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

$$\begin{array}{rl}{D}_{{\lambda }_0}^{\prime }(x)=& \mathrm{\Phi }(x)+{\lambda }_0\mathrm{\Upsilon }(x)\\ >& \frac{2x}{3}-\frac{34x^3}{45}+\frac{754x^5}{945}-x^7-{\lambda }_0(\frac{4x}{3}-\frac{34x^3}{45}+\frac{3x^5}{2})\\ =& x[\frac{2(1-2{\lambda }_0)}{3}-\frac{34(1-{\lambda }_0)}{45}{x}^2+(\frac{754}{945}-\frac{3{\lambda }_0}{2})x^4-x^6]\\ :=& x{F}_{{\lambda }_0}(x)>0\end{array}$$

(3.8)

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

Note that

$$F_{{\lambda }_0}(0)=2(1-2{\lambda }_0)/3>0,F_{{\lambda }_0}(0.7)=0.00513\dots >0$$

(3.9)

and

$$\begin{array}{rcl}{F}_{{\lambda }_0}^{″}(x)& =& -30[{(x^2-\frac{1\text{,}508-2\text{,}835{\lambda }_0}{9\text{,}450})}^2+\frac{2\text{,}224\text{,}136+4\text{,}052\text{,}160{\lambda }_0-8\text{,}037\text{,}225{\lambda }_0^2}{89\text{,}302\text{,}500}]\\ <& 0\end{array}$$

(3.10)

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

Inequalities (3.8)-(3.10) lead to the conclusion that

$$D_{{\lambda }_0}^{\prime }(x)>0$$

(3.11)

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

It follows from Lemma 2.8 and (3.6) that $D_{{\lambda }_0}^{\prime }(x)$ is strictly decreasing in $[0.7,1)$. Then from (3.11) and $D_{{\lambda }_0}^{\prime }(0.7)=0.0626\dots$ together with $D_{{\lambda }_0}^{\prime }(1^{-})=-\mathrm{\infty }$, we know that there exists $x^{\ast }\in (0.7,1)$ such that $D_{{\lambda }_0}(x)$ is strictly increasing on $(0,x^{\ast }]$ and strictly decreasing on $[x^{\ast },1)$. This in conjunction with (3.5) implies that

$$D_{{\lambda }_0}(x)>0$$

(3.12)

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

Equations (3.4), (3.5), (3.7) and (3.12) lead to the conclusion that

$$M(a,b)<P^{{\lambda }_0}(a,b)Q^{1-{\lambda }_0}(a,b)$$

(3.13)

and

$$M(a,b)>P^{1/2}(a,b)Q^{1/2}(a,b).$$

(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 ${\delta }_1\in (0,1)$ such that $M(a,b)<P^{\alpha }(a,b)Q^{1-\alpha }(a,b)$ for all $a,b>0$ with $(a-b)/(a+b)\in (0,{\delta }_1)$.

• If $\beta >{\lambda }_0$, then (3.1) and (3.3) imply that there exists ${\delta }_2\in (0,1)$ such that $M(a,b)>P^{\beta }(a,b)Q^{1-\beta }(a,b)$ for all $a,b>0$ with $(a-b)/(a+b)\in (1-{\delta }_2,1)$.

 □