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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 ChuEmail author
Open Access
Research
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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 

MSC

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.

Availability of data and materials

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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© The Author(s) 2019

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  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

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