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

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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.

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 \(\nu \in [1/2, \infty )\)?

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 \((0, 1)\). □

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 \(\nu \in [1/2, \infty )\).

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 \((0, 1)\).

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.

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Acknowledgements

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

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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).

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

Correspondence to Yu-Ming Chu.

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MSC

  • 26E60

Keywords

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