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

• Wei-Mao Qian
• Zai-Yin He
• Hong-Wei Zhang
• Yu-Ming Chu
Open Access
Research
Part of the following topical collections:
1. Recent Advances in General Inequalities on Pure & Applied Mathematics and Related Areas

## 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.

## Keywords

Arithmetic mean Quadratic mean Contraharmonic mean Schwab–Borchardt mean Neuman mean Two-parameter contraharmonic and arithmetic mean

26E60

## 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. (1)

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

2. (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. (1)

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

2. (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.

## Notes

### Acknowledgements

The authors would like to express their sincere thanks to the editor and the anonymous reviewers for their helpful comments and suggestions.

Not applicable.

### Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

### Funding

The work was supported by the Natural Science Foundation of China (Grant Nos. 61673169, 11301127, 11701176, 11626101, 11601485) and the Natural Science Foundation of Huzhou City (Grant No. 2018YZ07).

### Competing interests

The authors declare that they have no competing interests.

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## Authors and Affiliations

• Wei-Mao Qian
• 1
• Zai-Yin He
• 2
• Hong-Wei Zhang
• 3
• Yu-Ming Chu
• 4
1. 1.School of Continuing EducationHuzhou Vocational & Technical CollegeHuzhouChina
2. 2.College of Mathematics and EconometricsHunan UniversityChangshaChina
3. 3.School of Mathematics and StatisticsChangsha University of Science & TechnologyChangshaChina
4. 4.Department of MathematicsHuzhou UniversityHuzhouChina