Sharp bounds for Neuman means in terms of two-parameter contraharmonic and arithmetic mean

Abstract

In the article, we prove that $\lambda _{1}=1/2+\sqrt{ [ (\sqrt{2}+ \log (1+\sqrt{2}) )/2 ]^{1/\nu }-1}/2$, $\mu _{1}=1/2+\sqrt{6 \nu }/(12\nu )$, $\lambda _{2}=1/2+\sqrt{ [(\pi +2)/4 ] ^{1/\nu }-1}/2$ and $\mu _{2}=1/2+\sqrt{3\nu }/(6\nu )$ are the best possible parameters on the interval $[1/2, 1]$ such that the double inequalities

$$\begin{aligned}& C^{\nu }\bigl[\lambda _{1}x+(1-\lambda _{1})y, \lambda _{1}y+(1-\lambda _{1})x\bigr]A ^{1-\nu }(x, y) \\& \quad < \mathcal{R}_{QA}(x, y)< C^{\nu }\bigl[\mu _{1}x+(1-\mu _{1})y, \mu _{1}y+(1-\mu _{1})x\bigr]A^{1-\nu }(x, y), \\& C^{\nu }\bigl[\lambda _{2}x+(1-\lambda _{2})y, \lambda _{2}y+(1-\lambda _{2})x\bigr]A ^{1-\nu }(x, y) \\& \quad < \mathcal{R}_{AQ}(x, y)< C^{\nu }\bigl[\mu _{2}x+(1-\mu _{2})y, \mu _{2}y+(1-\mu _{2})x\bigr]A^{1-\nu }(x, y) \end{aligned}$$

hold for all $x, y>0$ with $x\neq y$ and $\nu \in [1/2, \infty )$, where $A(x, y)$ is the arithmetic mean, $C(x, y)$ is the contraharmonic mean, and $\mathcal{R}_{QA}(x, y)$ and $\mathcal{R}_{AQ}(x, y)$ are two Neuman means.

1 Introduction

Let $x, y>0$. Then the arithmetic mean $A(x, y)$, quadratic mean $Q(x, y)$ [1], contraharmonic mean $C(x, y)$ [2, 3], and Schwab–Borchardt mean $\operatorname{SB}(x, y)$ [4] are given by

$$ \begin{aligned} &A(x, y)=\frac{x+y}{2},\qquad Q(x, y)=\sqrt{\frac{x^{2}+y^{2}}{2}}, \qquad C(x, y)= \frac{x^{2}+y^{2}}{x+y}, \\ &\operatorname{SB}(x, y)= \textstyle\begin{cases} \frac{\sqrt{y^{2}-x^{2}}}{\arccos {(x/y)}}, & x< y, \\ x, & x=y, \\ \frac{\sqrt{x^{2}-y^{2}}}{\cosh ^{-1}{(x/y)}}, & x>y, \end{cases}\displaystyle \end{aligned} $$

(1.1)

respectively, where $\cosh ^{-1}(\sigma )=\log (\sigma +\sqrt{\sigma ^{2}-1})$ is the inverse hyperbolic cosine function.

The Gaussian arithmetic–geometric mean $\operatorname{AGM}(x, y)$ [5,6,7] of two positive real numbers x and y is defined by the common limit of the sequences $\{x_{n}\}_{n=0}^{\infty }$ and $\{y_{n}\}_{n=0}^{\infty }$, which are given by

$$ x_{0}=x,\qquad y_{0}=y, \qquad x_{n+1}=\frac{x_{n}+y_{n}}{2},\qquad y _{n+1}=\sqrt{x_{n}y_{n}}. $$

It is well known that the bivariate means have wide applications in mathematics, physics, engineering, and other natural sciences [8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55], many special functions can be expressed using bivariate means, for example, the complete elliptic integral

$$ \mathcal{K}(r)= \int _{0}^{\pi /2} \frac{dt}{\sqrt{1-r^{2}\sin ^{2}(t)}} \quad (0< r< 1) $$

of the first kind [56,57,58,59,60,61] and the modulus $\mu (r)$ of the plane Grötzsch ring [62, 63] can be expressed by the Gaussian arithmetic–geometric mean $\operatorname{AGM}(x, y)$, the formula of the perimeter of an ellipse and the complete elliptic integral

$$ \mathcal{E}(r)= \int _{0}^{\pi /2}\sqrt{1-r^{2}\sin ^{2}(t)}\,dt $$

of the second kind [64,65,66,67,68,69,70] can be given in terms of the Toader mean [71,72,73,74]

$$ T(a,b)=\frac{2}{\pi } \int _{0}^{\pi /2}\sqrt{a^{2}\cos ^{2}(t)+b^{2} \sin ^{2}(t)}\,dt. $$

Indeed, we have

$$\begin{aligned}& \mathcal{K}(r)=\frac{\pi }{2}\frac{1}{\operatorname{AGM}(1, \sqrt{1-r^{2}})},\qquad \mu (r)=\frac{\pi }{2} \frac{\operatorname{AGM}(1, \sqrt{1-r^{2}})}{\operatorname{AGM}(1, r)}, \\& L(x, y)=2\pi T(x, y),\qquad \mathcal{E}(r)=\frac{\pi }{2}T \bigl(1, \sqrt{1-r ^{2}} \bigr). \end{aligned}$$

Recently, the inequalities for bivariate means have attracted the attention of many mathematicians. Neuman [75] introduced the Neuman means

$$\begin{aligned}& \mathcal{R}_{QA}(x, y)=\frac{1}{2} \biggl[Q(x, y)+ \frac{A^{2}(x, y)}{\operatorname{SB}(Q(x,y), A(x,y))} \biggr], \\& \mathcal{R}_{AQ}(x, y)=\frac{1}{2} \biggl[A(x, y)+ \frac{Q^{2}(x, y)}{\operatorname{SB}(A(x,y), Q(x,y))} \biggr] \end{aligned}$$

and provided the formulas

$$\begin{aligned}& \mathcal{R}_{QA}(x, y)=\frac{1}{2}A(x, y) \biggl[ \sqrt{1+u^{2}}+\frac{ \sinh ^{-1}(u)}{u} \biggr], \end{aligned}$$

(1.2)

$$\begin{aligned}& \mathcal{R}_{AQ}(x, y)=\frac{1}{2}A(x, y) \biggl[1+ \frac{(1+u^{2}) \arctan (u)}{u} \biggr] \end{aligned}$$

(1.3)

if $x>y>0$, where $u=(x-y)/(x+y)$ and $\sinh ^{-1}(\sigma )=\log ( \sigma +\sqrt{\sigma ^{2}+1})$ is the inverse hyperbolic sine function. Neuman [4] proved that the inequalities

$$ A(x, y)< \mathcal{R}_{QA}(x, y)< \mathcal{R}_{AQ}(x, y)< Q(x, y) $$

(1.4)

hold for $x, y>0$ with $x\neq y$.

Zhang et al. [76] proved that $\alpha _{1}=1/2+\sqrt{2\sqrt{2} \log (1+\sqrt{2})+\log ^{2}(1+\sqrt{2})-2}/4=0.7817\ldots$ , $\beta _{1}=1/2+\sqrt{3}/6=0.7886\ldots$ , $\alpha _{2}=1/2+\sqrt{\pi ^{2}+4\pi -12}/8=0.9038\ldots$ and $\beta _{2}=1/2+\sqrt{6}/6=0.9082 \ldots$ are the best possible parameters on the interval $[1/2, 1]$ such that the double inequalities

$$\begin{aligned}& Q\bigl[\alpha _{1}x+(1-\alpha _{1})y, \alpha _{1}y+(1-\alpha _{1})x\bigr] \\& \quad < \mathcal{R}_{QA}(x, y)< Q\bigl[\beta _{1}x+(1-\beta _{1})y, \beta _{1}y+(1-\beta _{1})x\bigr], \end{aligned}$$

(1.5)

$$\begin{aligned}& Q\bigl[\alpha _{2}x+(1-\alpha _{2})y, \alpha _{2}y+(1-\alpha _{2})x\bigr] \\& \quad < \mathcal{R}_{AQ}(x, y)< Q\bigl[\beta _{2}x+(1-\beta _{2})y, \beta _{2}y+(1-\beta _{2})x\bigr] \end{aligned}$$

(1.6)

hold for $x, y>0$ with $x\neq y$.

In [77], Yang et al. proved that the double inequalities

$$\begin{aligned}& \alpha \biggl[\frac{C(x,y)}{3}+\frac{2A(x,y)}{3} \biggr]+(1-\alpha )C ^{1/3}(x, y)A^{2/3}(x, y) \\& \quad < \mathcal{R}_{AQ}(x, y)< \beta \biggl[\frac{C(x,y)}{3}+\frac{2A(x,y)}{3} \biggr]+(1-\beta )C ^{1/3}(x, y)A^{2/3}(x, y), \\& \lambda \biggl[\frac{C(x,y)}{6}+\frac{5A(x,y)}{6} \biggr]+(1-\lambda )C^{1/6}(x, y)A^{5/6}(x, y) \\& \quad < \mathcal{R}_{QA}(x, y)< \mu \biggl[\frac{C(x,y)}{6}+\frac{5A(x,y)}{6} \biggr]+(1-\mu )C^{1/6}(x, y)A^{5/6}(x, y) \end{aligned}$$

hold for for $x, y>0$ with $x\neq y$ if and only if $\alpha \leq (3 \pi +6-12\sqrt[3]{2})/(16-12\sqrt[3]{2})=0.3470\ldots$ , $\beta \geq 2/5$, $\lambda \leq [3\sqrt{2}+3\log (1+\sqrt{2})-6 \sqrt[6]{2}]/(7-6\sqrt[6]{2})=0.5730\ldots$ and $\mu \geq 16/25$.

The main purpose of the article is to generalize inequalities (1.5) and (1.6). To achieve this goal, we define the two-parameter contraharmonic and arithmetic mean $W_{\lambda , \nu }(x, y)$ as follows:

$$ W_{\lambda , \nu }(x, y)=C^{\nu }\bigl[\lambda x+(1-\lambda )y, \lambda y+(1- \lambda )x\bigr]A^{1-\nu }(x, y), $$

(1.7)

where $\lambda \in [1/2, 1]$ and $\nu \in [1/2, \infty )$. We clearly see that the function $\lambda \rightarrow W_{\lambda , \nu }(x, y)$ is strictly increasing on $[1/2, 1]$ for $\nu \in [1/2, \infty )$ and $x, y>0$ with $x\neq y$.

It follows from (1.1), (1.4) and (1.7) that

$$\begin{aligned}& W_{\lambda , 1/2}(x, y)=Q\bigl[\lambda x+(1-\lambda )y, \lambda y+(1- \lambda )x \bigr], \end{aligned}$$

(1.8)

$$\begin{aligned}& W_{\lambda , 1}(x, y)=C\bigl[\lambda x+(1-\lambda )y, \lambda y+(1-\lambda )x \bigr], \end{aligned}$$

(1.9)

$$\begin{aligned}& W_{1/2, \nu }(x, y)=A(x, y), \\& W_{1, \nu }(x, y)=C^{\nu }(x, y)A^{1-\nu }(x, y)=A(x, y) \biggl[\frac{Q(x,y)}{A(x,y)} \biggr] ^{2\nu }\geq Q(x, y), \\& W_{1/2, \nu }(x, y)< \mathcal{R}_{QA}(x,y)< \mathcal{R}_{AQ}(x,y)< W_{1, \nu }(x, y). \end{aligned}$$

(1.10)

Inequalities (1.5), (1.6), and (1.10) give us the motivation to discuss the question: What are the best possible parameters $\lambda _{1}= \lambda _{1}(\nu )$, $\mu _{1}=\mu _{1}(\nu )$, $\lambda _{2}=\lambda _{2}( \nu )$ and $\mu _{2}=\mu _{2}(\nu )$ on the interval $[1/2, 1]$ such that the double inequalities

$$\begin{aligned}& W_{\lambda _{1}, \nu }(x,y)< \mathcal{R}_{QA}(x, y)< W_{\mu _{1}, \nu }(x,y), \\& W_{\lambda _{2}, \nu }(x,y)< \mathcal{R}_{AQ}(x, y)< W_{\mu _{2}, \nu }(x,y) \end{aligned}$$

hold for all $x, y>0$ with $x\neq y$ and $\nu \in [1/2, \infty )$?

2 Lemmas

In order to prove our main results, we need to introduce and establish five lemmas which we present in this section.

Lemma 2.1

([78, Theorem 1.25])

Let $\alpha , \beta \in \mathbb{R}$ with $\alpha <\beta$, $\varGamma , \varPsi : [\alpha , \beta ]\rightarrow \mathbb{R}$ be continuous on $[\alpha , \beta ]$ and differentiable on $(\alpha , \beta )$ with $\varPsi ^{\prime }(\tau )\neq 0$ on $(\alpha , \beta )$. Then the functions

$$ \frac{\varGamma (\tau )-\varGamma (\alpha )}{\varPsi (\tau )-\varPsi (\alpha )}, \qquad \frac{\varGamma (\tau )-\varGamma (\beta )}{\varPsi (\tau )-\varPsi (\beta )} $$

are (strictly) increasing (decreasing) on $(\alpha , \beta )$ if $\varGamma ^{\prime }(\tau )/\varPsi ^{\prime }(\tau )$ is (strictly) increasing (decreasing) on $(\alpha , \beta )$.

Lemma 2.2

The function

$$ \phi (t)=\frac{\sqrt{1+t^{2}}\sinh ^{-1}(t)}{t} $$

is strictly increasing from $(0, 1)$ onto $(1, \sqrt{2}\log (1+ \sqrt{2}) )$.

Proof

Differentiating $\phi (t)$ gives

$$ \phi '(t)=\frac{\phi _{1}(t)}{t\sqrt{1+t^{2}}}, $$

(2.1)

where

$$ \phi _{1}(t)=t\sqrt{1+t^{2}}-\sinh ^{-1}(t). $$

(2.2)

It follows from (2.2) that

$$\begin{aligned}& \phi _{1}\bigl(0^{+}\bigr)=0, \end{aligned}$$

(2.3)

$$\begin{aligned}& \phi _{1}'(t)=\frac{2t^{2}}{\sqrt{1+t^{2}}}>0 \end{aligned}$$

(2.4)

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

Note that

$$ \phi \bigl(0^{+}\bigr)=1,\qquad \phi \bigl(1^{-}\bigr)=\sqrt{2} \log (1+\sqrt{2}). $$

(2.5)

Therefore, Lemma 2.2 follows from (2.1) and (2.3)–(2.5). □

Lemma 2.3

The function

$$ \varphi (t)=\frac{t^{3}}{(1+t^{2})\arctan (t)-t} $$

is strictly increasing from $(0, 1)$ onto $(3/2, 2/(\pi -2) )$.

Proof

Let $\varphi _{1}(t)=t^{3}$ and $\varphi _{2}(t)=(1+t^{2})\arctan (t)-t$. Then we clearly see that

$$\begin{aligned}& \varphi _{1}\bigl(0^{+}\bigr)=\varphi _{2} \bigl(0^{+}\bigr),\qquad \varphi (t)=\frac{\varphi _{1}(t)}{\varphi _{2}(t)}, \end{aligned}$$

(2.6)

$$\begin{aligned}& \frac{\varphi '_{1}(t)}{\varphi '_{2}(t)}=\frac{3t}{2\arctan (t)}. \end{aligned}$$

(2.7)

It is not difficult to verify that the function $t\mapsto t/\arctan (t)$ is strictly increasing from $(0, 1)$ onto $(1, 4/\pi )$. Then equation (2.7) leads to the conclusion that $\varphi '_{1}(t)/\varphi '_{2}(t)$ is strictly increasing on $(0, 1)$.

Note that

$$ \varphi \bigl(0^{+}\bigr)=\frac{3}{2},\qquad \varphi \bigl(1^{-}\bigr)=\frac{2}{\pi -2}. $$

(2.8)

Therefore, Lemma 2.3 follows from Lemma 2.1, (2.6), (2.8), and the monotonicity of $\varphi '_{1}(t)/\varphi '_{2}(t)$. □

Lemma 2.4

Let $\theta \in [0, 1]$, $\nu \in [1/2, \infty )$, $t\in (0, 1)$ and

$$ f_{\theta , \nu }(t)=\nu \log \bigl(1+\theta t^{2}\bigr)-\log \bigl[t \sqrt{1+t ^{2}}+\sinh ^{-1}(t) \bigr]+\log t+\log 2. $$

(2.9)

Then we have the following two conclusions:

• (1) $f_{\theta , \nu }(t)>0$ for all $t\in (0, 1)$ if and only if $\theta \geq 1/(6\nu )$;

• (2) $f_{\theta , \nu }(t)<0$ for all $t\in (0, 1)$ if and only if $\theta \leq [(\sqrt{2}+\log (1+\sqrt{2}))/2 ]^{1/ \nu }-1$.

Proof

It follows from (2.9) that

$$\begin{aligned}& f_{\theta , \nu }\bigl(0^{+}\bigr)=0, \end{aligned}$$

(2.10)

$$\begin{aligned}& f_{\theta , \nu }\bigl(1^{-}\bigr)=\nu \log (1+\theta )-\log \bigl[ \sqrt{2}+ \log (1+\sqrt{2}) \bigr]+\log 2, \end{aligned}$$

(2.11)

$$\begin{aligned}& f^{\prime }_{\theta , \nu }(t)=\frac{t [(2\nu -1)(t\sqrt{1+t ^{2}}-\sinh ^{-1}(t)) +4\nu \sinh ^{-1}(t) ]}{(1+\theta t^{2}) [t\sqrt{1+t^{2}}+\sinh ^{-1}(t) ]} \bigl[\theta -f_{ \nu }(t) \bigr], \end{aligned}$$

(2.12)

where

$$ f_{\nu }(t)=\frac{t\sqrt{1+t^{2}}-\sinh ^{-1}(t)}{(2\nu -1)t^{2}[t\sqrt{1+t ^{2}}-\sinh ^{-1}(t)]+4\nu t^{2}\sinh ^{-1}(t)}. $$

Let $\psi _{1}(t)=t\sqrt{1+t^{2}}-\sinh ^{-1}(t)$ and $\psi _{2}(t)=(2 \nu -1)t^{2}[t\sqrt{1+t^{2}}-\sinh ^{-1}(t)]+4\nu t^{2}\sinh ^{-1}(t)$. Then

$$\begin{aligned}& \psi _{1}\bigl(0^{+}\bigr)=\psi _{2} \bigl(0^{+}\bigr)=0,\qquad f_{\nu }(t)=\frac{\psi _{1}(t)}{ \psi _{2}(t)}, \end{aligned}$$

(2.13)

$$\begin{aligned}& \frac{\psi '_{1}(t)}{\psi '_{2}(t)}=\frac{1}{(2\nu +1)\phi (t)+2(2 \nu -1)t^{2}+4\nu -1}, \end{aligned}$$

(2.14)

where $\phi (t)$ is defined in Lemma 2.2.

Equation (2.14) and Lemma 2.2 imply that $\psi '_{1}(t)/\psi '_{2}(t)$ is strictly decreasing on $(0, 1)$. Therefore, the conclusion that $f_{\nu }(t)$ is strictly decreasing on $(0, 1)$ follows from Lemma 2.1 and (2.13), together with the monotonicity of $\psi '_{1}(t)/\psi '_{2}(t)$ on the interval $(0, 1)$. Moreover, making use of L’Hôpital’s rule, we have that

$$\begin{aligned}& f_{\nu }\bigl(0^{+}\bigr)=\frac{1}{6\nu }, \end{aligned}$$

(2.15)

$$\begin{aligned}& f_{\nu }\bigl(1^{-}\bigr)=\frac{\sqrt{2}-\log (1+\sqrt{2})}{(2\nu -1) \sqrt{2}+(2\nu +1)\log (1+\sqrt{2})}=:\theta _{0}. \end{aligned}$$

(2.16)

We divide the proof into three cases.

Case 1. $\theta \geq 1/(6\nu )$. Then (2.12) and (2.15), together with the monotonicity of $f_{\nu }(t)$ on the interval $(0, 1)$, lead to the conclusion that $f_{\theta , \nu }(t)$ is strictly increasing on $(0, 1)$. Therefore, $f_{\theta , \nu }(t)>0$ for all $t\in (0, 1)$ follows from (2.10) and the monotonicity of $f_{\theta , \nu }(t)$ on the interval $(0, 1)$.

Case 2. $\theta \leq \theta _{0}$. Then from (2.12) and (2.16), together with the monotonicity of $f_{\nu }(t)$ on the interval $(0, 1)$, we clearly see that $f_{\theta , \nu }(t)$ is strictly decreasing on $(0, 1)$. Therefore, $f_{\theta , \nu }(t)<0$ for all $t\in (0, 1)$ follows from (2.10) and the monotonicity of $f_{\theta , \nu }(t)$ on the interval $(0, 1)$.

Case 3. $\theta _{0}<\theta <1/(6\nu )$. Then from (2.12), (2.15), (2.16), and the monotonicity of $f_{\nu }(t)$ on the interval $(0, 1)$, we clearly see that there exists $t_{0}\in (0, 1)$ such that $f_{\theta , \nu }(t)$ is strictly decreasing on $(0, t_{0})$ and strictly increasing on $(t_{0}, 1)$.

We divide the proof into two subcases.

Subcase 3.1. $[(\sqrt{2}+\log (1+\sqrt{2}))/2 ] ^{1/\nu }-1<\theta <1/(6\nu )$. Then (2.11) leads to

$$ f_{\theta , \nu }\bigl(1^{-}\bigr)>0. $$

(2.17)

Therefore, there exists $t^{\ast }\in (t_{0}, 1)$ such that $f_{\theta , \nu }(t)<0$ for $t\in (0, t^{\ast })$ and $f_{\theta , \nu }(t)>0$ for $t\in (t^{\ast }, 1)$ follows from (2.10) and (2.17), together with the piecewise monotonicity of $f_{\theta , \nu }(t)$ on the interval $(0, 1)$.

Subcase 3.2. $\theta _{0}<\theta \leq [(\sqrt{2}+\log (1+ \sqrt{2}))/2 ]^{1/\nu }-1$. Then (2.11) leads to

$$ f_{\theta , \nu }\bigl(1^{-}\bigr)\leq 0. $$

(2.18)

Therefore, $f_{\theta , \nu }(t)<0$ for all $t\in (0, 1)$ follows from (2.10) and (2.18), together with the piecewise monotonicity of $f_{\theta , \nu }(t)$ on the interval $(0, 1)$. □

Lemma 2.5

Let $\vartheta \in [0, 1]$, $\nu \in [1/2, \infty )$, $t\in (0, 1)$ and

$$ g_{\vartheta , \nu }(t)=\nu \log \bigl(1+\vartheta t^{2}\bigr)-\log \bigl[t+\bigl(1+t ^{2}\bigr)\arctan (t) \bigr]+\log (t)+\log 2. $$

(2.19)

Then the following statements are true:

• (1) $g_{\vartheta , \nu }(t)>0$ for all $t\in (0, 1)$ if and only if $\vartheta \geq 1/(3\nu )$;

• (2) $g_{\vartheta , \nu }(t)<0$ for all $t\in (0, 1)$ if and only if $\vartheta \leq [(\pi +2)/4 ]^{1/\nu }-1$.

Proof

It follows from (2.19) that

$$\begin{aligned}& g_{\vartheta , \nu }\bigl(0^{+}\bigr)=0, \end{aligned}$$

(2.20)

$$\begin{aligned}& g_{\vartheta , \nu }\bigl(1^{-}\bigr)=\nu \log (1+\vartheta )-\log \biggl(\frac{ \pi +2}{4} \biggr), \end{aligned}$$

(2.21)

$$\begin{aligned}& g^{\prime }_{\vartheta , \nu }(t)=\frac{t [((2\nu -1)t^{2}+2 \nu +1)\arctan (t)+(2\nu -1)t ]}{(1+\vartheta t^{2}) [t+(1+t ^{2})\arctan (t) ]}\bigl[\vartheta -g_{\nu }(t)\bigr], \end{aligned}$$

(2.22)

where

$$ g_{\nu }(t)=\frac{t-(1-t^{2})\arctan (t)}{t^{2} [((2\nu -1)t^{2}+2 \nu +1)\arctan (t)+(2\nu -1)t ]}. $$

Let $\omega _{1}(t)=[t-(1-t^{2})\arctan (t)]/t^{2}$ and $\omega _{2}(t)=[(2 \nu -1)t^{2}+2\nu +1]\arctan (t)+(2\nu -1)t$. Then elaborate computations lead to

$$\begin{aligned}& \omega _{1}\bigl(0^{+}\bigr)=\omega _{2} \bigl(0^{+}\bigr)=0,\qquad g_{\nu }(t)=\frac{\omega _{1}(t)}{\omega _{2}(t)}, \end{aligned}$$

(2.23)

$$\begin{aligned}& \frac{\omega '_{1}(t)}{\omega '_{2}(t)}=\frac{1}{2[(2\nu -1)t^{2}+ \nu ]\varphi (t)+(2\nu -1)t^{4}} , \end{aligned}$$

(2.24)

where $\varphi (t)$ is defined in Lemma 2.3.

From Lemma 2.3 and (2.24) we know that $\omega '_{1}(t)/\omega '_{2}(t)$ is strictly decreasing on $(0, 1)$. Therefore, the conclusion that $g_{\nu }(t)$ is strictly decreasing on $(0, 1)$ follows from Lemma 2.1 and (2.23), together with the monotonicity of $\omega '_{1}(t)/\omega '_{2}(t)$ on the interval $(0, 1)$. Moreover, making use of L’Hôpital’s rule, we have that

$$\begin{aligned}& g_{\nu }\bigl(0^{+}\bigr)=\frac{1}{3v}, \end{aligned}$$

(2.25)

$$\begin{aligned}& g_{\nu }\bigl(1^{-}\bigr)=\frac{1}{(\pi +2)\nu -1}. \end{aligned}$$

(2.26)

We divide the proof into three cases.

Case 1. $\vartheta \geq 1/(3\nu )$. Then (2.22) and (2.25), together with the monotonicity of $g_{\nu }(t)$ on the interval $(0, 1)$, lead to the conclusion that $g_{\vartheta , \nu }(t)$ is strictly increasing on $(0, 1)$. Therefore, $g_{\vartheta , \nu }(t)>0$ for all $t\in (0, 1)$ follows from (2.20) and the monotonicity of $g_{\vartheta , \nu }(t)$ on the interval $(0, 1)$.

Case 2. $\vartheta \leq 1/[(\pi +2)\nu -1]$. Then from (2.22) and (2.26), together with the monotonicity of $g_{\nu }(t)$ on the interval $(0, 1)$, we clearly see that $g_{\vartheta , \nu }(t)$ is strictly decreasing on $(0, 1)$. Therefore, $g_{\vartheta , \nu }(t)<0$ for all $t\in (0, 1)$ follows from (2.20) and the monotonicity of $g_{\vartheta , \nu }(t)$ on the interval $(0, 1)$.

Case 3. $1/[(\pi +2)\nu -1]<\vartheta <1/(6\nu )$. Then it follows from (2.22), (2.25), (2.26), and the monotonicity of $g_{\nu }(t)$ on the interval $(0, 1)$ that there exists $\rho _{0} \in (0, 1)$ such that $g_{\vartheta , \nu }(t)$ is strictly decreasing on $(0, \rho _{0})$ and strictly increasing on $(\rho _{0}, 1)$.

We divide the proof into two subcases.

Subcase 3.1. $[(\pi +2)/4 ]^{1/\nu }-1<\vartheta <1/(6 \nu )$. Then (2.21) leads to

$$ g_{\vartheta , \nu }\bigl(1^{-}\bigr)>0. $$

(2.27)

Therefore, there exists $\rho ^{\ast }\in (\rho _{0}, 1)$ such that $g_{\vartheta , \nu }(t)<0$ for $t\in (0, \rho ^{\ast })$ and $g_{\vartheta , \nu }(t)>0$ for $t\in (\rho ^{\ast }, 1)$ follows from (2.20) and (2.27), together with the piecewise of $g_{\vartheta , \nu }(t)$ on the interval $(0, 1)$.

Subcase 3.2. $1/[(\pi +2)\nu -1]<\vartheta \leq [(\pi +2)/4 ] ^{1/\nu }-1$. Then (2.21) gives

$$ g_{\vartheta , \nu }\bigl(1^{-}\bigr)\leq 0. $$

(2.28)

Therefore, $g_{\vartheta , \nu }(t)<0$ for all $t\in (0, 1)$ follows from (2.20) and (2.28), together with the piecewise of $g_{\vartheta , \nu }(t)$ on the interval $(0, 1)$. □

3 Main results

Theorem 3.1

Let $\lambda _{1}, \mu _{1}\in [1/2, 1]$ and $\nu \in [1/2, \infty )$. Then the double inequality

$$ W_{\lambda _{1}, \nu }(x, y)< \mathcal{R}_{QA}(x, y)< W_{\mu _{1}, \nu }(x, y) $$

(3.1)

holds for all $x, y>0$ with $x\neq y$ if and only if $\lambda _{1} \leq 1/2+\sqrt{ [ (\sqrt{2}+\log (1+\sqrt{2}) )/2 ] ^{1/\nu }-1}/2$ and $\mu _{1}\geq 1/2+\sqrt{6\nu }/(12\nu )$.

Proof

Since both $W_{\theta , \nu }(x, y)$ and $\mathcal{R}_{QA}(x, y)$ are symmetric and homogenous of degree 1, without loss of generality, we assume that $x>y>0$. Let $t=(x-y)/(x+y)\in (0, 1)$ and $\theta \in [1/2, 1]$. Then from (1.1), (1.2), and (1.7) we get

$$\begin{aligned}& \frac{W_{\theta , \nu }(x, y))}{A(x, y)}= \bigl[1+(2\theta -1)^{2}t ^{2} \bigr]^{\nu }, \end{aligned}$$

(3.2)

$$\begin{aligned}& \frac{\mathcal{R}_{QA}(x, y)}{A(x, y)}=\frac{1}{2} \biggl[\sqrt{1+t ^{2}}+ \frac{\sinh ^{-1}(t)}{t} \biggr]. \end{aligned}$$

(3.3)

It follows from (3.2) and (3.3) that

$$\begin{aligned} \log \biggl[\frac{W_{\theta , \nu }(x, y)}{\mathcal{R}_{QA}(x, y)} \biggr] =& \log \biggl[\frac{W_{\theta , \nu }(x, y)}{A(x, y)} \biggr]- \log \biggl[\frac{\mathcal{R}_{QA}(x, y)}{A(x, y)} \biggr] \\ =&\nu \log \bigl[1+(2\theta -1)^{2}t^{2} \bigr]-\log \bigl[t \sqrt{1+t ^{2}}+\sinh ^{-1}(t) \bigr] \\ &{}+\log (t)+\log 2. \end{aligned}$$

(3.4)

Therefore, Theorem 3.1 follows easily from Lemma 2.4 and (3.4). □

Theorem 3.2

Let $\lambda _{2}, \mu _{2}\in [1/2, 1]$ and $\nu \in [1/2, \infty )$. Then the double inequality

$$ W_{\lambda _{2}, \nu }(x, y)< \mathcal{R}_{AQ}(x, y)< W_{\mu _{2}, \nu }(x, y) $$

(3.5)

holds for all $x, y>0$ with $x\neq y$ if and only if $\lambda _{2} \leq 1/2+\sqrt{ [(\pi +2)/4 ]^{1/\nu }-1}/2$ and $\mu _{2}\geq 1/2+\sqrt{3\nu }/(6\nu )$.

Proof

Since both $W_{\vartheta , \nu }(x, y)$ and $\mathcal{R}_{AQ}(x, y)$ are symmetric and homogenous of degree 1, without loss of generality, we assume that $x>y>0$. Let $t=(x-y)/(x+y)\in (0, 1)$ and $\vartheta \in [1/2, 1]$. Then it follows from (1.1), (1.3), and (1.7) that

$$\begin{aligned}& \frac{W_{\vartheta , \nu }(x, y))}{A(x, y)}= \bigl[1+(2\vartheta -1)^{2}t ^{2} \bigr]^{\nu }, \end{aligned}$$

(3.6)

$$\begin{aligned}& \frac{\mathcal{R}_{AQ}(x,y)}{A(x, y)}=\frac{1}{2} \biggl[1+\frac{(1+t ^{2})\arctan (t)}{t} \biggr]. \end{aligned}$$

(3.7)

From (3.6) and (3.7) we have

$$\begin{aligned} \log \biggl[\frac{W_{\vartheta , \nu }(x, y))}{\mathcal{R}_{AQ}(x,y)} \biggr] =& \log \biggl[\frac{W_{\vartheta , \nu }(x, y)}{A(x, y)} \biggr]- \log \biggl[\frac{\mathcal{R}_{AQ}(x,y)}{A(x,y)} \biggr] \\ =&\nu \log \bigl[1+(2\vartheta -1)^{2} t^{2} \bigr]-\log \bigl[t+\bigl(1+t ^{2}\bigr)\arctan (t) \bigr] \\ &{}+\log (t)+\log 2. \end{aligned}$$

(3.8)

Therefore, Theorem 3.2 follows easily from Lemma 2.5 and (3.8). □

Remark 3.3

Let $\nu =1/2$. Then from (1.8) we clearly see that Theorems 3.1 and 3.2 become (1.5) and (1.6), respectively.

Let $\nu =1$. Then from (1.9) and Theorems 3.1 and 3.2 we get Corollary 3.4 immediately.

Corollary 3.4

Let $\lambda _{1}, \mu _{1}, \lambda _{2}, \mu _{2}\in [1/2, 1]$. Then the double inequalities

$$\begin{aligned}& C\bigl[\lambda _{1}x+(1-\lambda _{1})y, \lambda _{1}y+(1-\lambda _{1})x\bigr]< \mathcal{R}_{QA}(x, y)< C\bigl[\mu _{1}x+(1-\mu _{1})y, \mu _{1}y+(1- \mu _{1})x\bigr], \\& C\bigl[\lambda _{2}x+(1-\lambda _{2})y, \lambda _{2}y+(1-\lambda _{2})x\bigr]< \mathcal{R}_{AQ}(x, y)< C\bigl[\mu _{2}x+(1-\mu _{2})y, \mu _{2}y+(1-\mu _{2})x\bigr] \end{aligned}$$

hold for all $x, y>0$ with $x\neq y$ if and only if $\lambda _{1} \leq 1/2+ \sqrt{ [ (\sqrt{2}+\log (1+\sqrt{2}) )/2 ]-1}/2=0.6922 \ldots$ , $\mu _{1}\geq 1/2+\sqrt{6}/12=0.7041\ldots$ , $\lambda _{2} \leq 1/2+\sqrt{ [(\pi +2)/4 ]-1}/2=0.7671\ldots$ and $\mu _{2}\geq 1/2+\sqrt{3}/6=0.7886\ldots$ .

Let $u\in (0, 1)$, $x=1+u$, $y=1-u$, $\lambda _{1}=1/2+\sqrt{ [ (\sqrt{2}+ \log (1+\sqrt{2}) )/2 ]^{1/\nu }-1}/2$, $\mu _{1}=1/2+\sqrt{6 \nu }/(12\nu )$, $\lambda _{2}=1/2+\sqrt{ [(\pi +2)/4 ] ^{1/\nu }-1}/2$ and $\mu _{2}=1/2+\sqrt{3\nu }/(6\nu )$. Then (1.2), (1.3), and Theorems 3.1 and 3.2 lead to Corollary 3.5.

Corollary 3.5

The double inequalities

$$\begin{aligned}& 2 \biggl[\bigl(1-u^{2}\bigr)+ \biggl(\frac{\sqrt{2}+\log (1+\sqrt{2})}{2} \biggr) ^{1/\nu }u^{2} \biggr]^{\nu }-\sqrt{1+u^{2}} \\& \quad < \frac{\sinh ^{-1}(u)}{u}< 2 \biggl(1+\frac{u^{2}}{6\nu } \biggr)^{ \nu }- \sqrt{1+u^{2}}, \\& \frac{2 [ (1-u^{2} )+(\frac{2+\pi }{4})^{1/\nu }u^{2} ] ^{\nu }-1}{1+u^{2}}< \frac{\arctan (u)}{u}< \frac{2 (1+\frac{1}{3 \nu }u^{2} )^{\nu }-1}{1+u^{2}} \end{aligned}$$

hold for all $u\in (0, 1)$ and $\nu \in [1/2, \infty )$.

4 Results and discussion

In the article, we give the sharp bounds for the Neuman means

$$ \mathcal{R}_{QA}(x, y)=\frac{1}{2} \biggl[Q(x, y)+ \frac{A^{2}(x, y)}{\operatorname{SB}(Q(x,y), A(x,y))} \biggr] $$

and

$$ \mathcal{R}_{AQ}(x, y)=\frac{1}{2} \biggl[A(x, y)+ \frac{Q^{2}(x, y)}{\operatorname{SB}(A(x,y), Q(x,y))} \biggr] $$

in terms of the two-parameter contraharmonic and arithmetic mean

$$ W_{\lambda , \nu }(x, y)=C^{\nu }\bigl[\lambda x+(1-\lambda )y, \lambda y+(1- \lambda )x\bigr]A^{1-\nu }(x, y), $$

and find new bounds for the functions $\sinh (u)/u$ and $\arctan (u)/u$ on the interval $(0, 1)$.

5 Conclusion

In the article, we prove that the double inequalities

$$ W_{\lambda _{1}, \nu }(x, y)< \mathcal{R}_{QA}(x, y)< W_{\mu _{1}, \nu }(x, y),\qquad W_{\lambda _{2}, \nu }(x, y)< \mathcal{R}_{AQ}(x, y)< W_{\mu _{2}, \nu }(x, y) $$

hold for all $x, y>0$ with $x\neq y$ if and only if $\lambda _{1} \leq 1/2+\sqrt{ [ (\sqrt{2}+\log (1+\sqrt{2}) )/2 ] ^{1/\nu }-1}/2$, $\mu _{1}\geq 1/2+\sqrt{6\nu }/(12\nu )$, $\lambda _{2}\leq 1/2+\sqrt{ [(\pi +2)/4 ]^{1/\nu }-1}/2$ and $\mu _{2}\geq 1/2+\sqrt{3\nu }/(6\nu )$ if $\lambda _{1}, \mu _{1}, \lambda _{2}, \mu _{2}\in [1/2, 1]$ and $\nu \in [1/2, \infty )$. Our results are a natural generalization of some previously known results, and our approach may lead to many follow-up studies.