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
respectively, where \(\cosh ^{-1}(\sigma )=\log (\sigma +\sqrt{\sigma ^{2}-1})\) is the inverse hyperbolic cosine function.
$$ \begin{aligned} &A(x, y)=\frac{x+y}{2},\qquad Q(x, y)=\sqrt{\frac{x^{2}+y^{2}}{2}}, \qquad C(x, y)= \frac{x^{2}+y^{2}}{x+y}, \\ &\operatorname{SB}(x, y)= \textstyle\begin{cases} \frac{\sqrt{y^{2}-x^{2}}}{\arccos {(x/y)}}, & x< y, \\ x, & x=y, \\ \frac{\sqrt{x^{2}-y^{2}}}{\cosh ^{-1}{(x/y)}}, & x>y, \end{cases}\displaystyle \end{aligned} $$
(1.1)
The Gaussian arithmetic–geometric mean \(\operatorname{AGM}(x, y)\) [5‐7] 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
$$ x_{0}=x,\qquad y_{0}=y, \qquad x_{n+1}=\frac{x_{n}+y_{n}}{2},\qquad y _{n+1}=\sqrt{x_{n}y_{n}}. $$
Anzeige
It is well known that the bivariate means have wide applications in mathematics, physics, engineering, and other natural sciences [8‐55], many special functions can be expressed using bivariate means, for example, the complete elliptic integral
of the first kind [56‐61] 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
of the second kind [64‐70] can be given in terms of the Toader mean [71‐74]
$$ \mathcal{K}(r)= \int _{0}^{\pi /2} \frac{dt}{\sqrt{1-r^{2}\sin ^{2}(t)}} \quad (0< r< 1) $$
$$ \mathcal{E}(r)= \int _{0}^{\pi /2}\sqrt{1-r^{2}\sin ^{2}(t)}\,dt $$
$$ T(a,b)=\frac{2}{\pi } \int _{0}^{\pi /2}\sqrt{a^{2}\cos ^{2}(t)+b^{2} \sin ^{2}(t)}\,dt. $$
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
and provided the formulas
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
hold for \(x, y>0\) with \(x\neq y\).
$$\begin{aligned}& \mathcal{R}_{QA}(x, y)=\frac{1}{2} \biggl[Q(x, y)+ \frac{A^{2}(x, y)}{\operatorname{SB}(Q(x,y), A(x,y))} \biggr], \\& \mathcal{R}_{AQ}(x, y)=\frac{1}{2} \biggl[A(x, y)+ \frac{Q^{2}(x, y)}{\operatorname{SB}(A(x,y), Q(x,y))} \biggr] \end{aligned}$$
$$\begin{aligned}& \mathcal{R}_{QA}(x, y)=\frac{1}{2}A(x, y) \biggl[ \sqrt{1+u^{2}}+\frac{ \sinh ^{-1}(u)}{u} \biggr], \end{aligned}$$
(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)
$$ A(x, y)< \mathcal{R}_{QA}(x, y)< \mathcal{R}_{AQ}(x, y)< Q(x, y) $$
(1.4)
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
hold for \(x, y>0\) with \(x\neq y\).
$$\begin{aligned}& Q\bigl[\alpha _{1}x+(1-\alpha _{1})y, \alpha _{1}y+(1-\alpha _{1})x\bigr] \\& \quad < \mathcal{R}_{QA}(x, y)< Q\bigl[\beta _{1}x+(1-\beta _{1})y, \beta _{1}y+(1-\beta _{1})x\bigr], \end{aligned}$$
(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)
Anzeige
In [77], Yang et al. proved that the double inequalities
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 \(\mu \geq 16/25\).
$$\begin{aligned}& \alpha \biggl[\frac{C(x,y)}{3}+\frac{2A(x,y)}{3} \biggr]+(1-\alpha )C ^{1/3}(x, y)A^{2/3}(x, y) \\& \quad < \mathcal{R}_{AQ}(x, y)< \beta \biggl[\frac{C(x,y)}{3}+\frac{2A(x,y)}{3} \biggr]+(1-\beta )C ^{1/3}(x, y)A^{2/3}(x, y), \\& \lambda \biggl[\frac{C(x,y)}{6}+\frac{5A(x,y)}{6} \biggr]+(1-\lambda )C^{1/6}(x, y)A^{5/6}(x, y) \\& \quad < \mathcal{R}_{QA}(x, y)< \mu \biggl[\frac{C(x,y)}{6}+\frac{5A(x,y)}{6} \biggr]+(1-\mu )C^{1/6}(x, y)A^{5/6}(x, y) \end{aligned}$$
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:
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 \(x\neq y\).
$$ W_{\lambda , \nu }(x, y)=C^{\nu }\bigl[\lambda x+(1-\lambda )y, \lambda y+(1- \lambda )x\bigr]A^{1-\nu }(x, y), $$
(1.7)
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
hold for all \(x, y>0\) with \(x\neq y\) and \(\nu \in [1/2, \infty )\)?
$$\begin{aligned}& W_{\lambda _{1}, \nu }(x,y)< \mathcal{R}_{QA}(x, y)< W_{\mu _{1}, \nu }(x,y), \\& W_{\lambda _{2}, \nu }(x,y)< \mathcal{R}_{AQ}(x, y)< W_{\mu _{2}, \nu }(x,y) \end{aligned}$$
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
are (strictly) increasing (decreasing) on
\((\alpha , \beta )\)
if
\(\varGamma ^{\prime }(\tau )/\varPsi ^{\prime }(\tau )\)
is (strictly) increasing (decreasing) on
\((\alpha , \beta )\).
$$ \frac{\varGamma (\tau )-\varGamma (\alpha )}{\varPsi (\tau )-\varPsi (\alpha )}, \qquad \frac{\varGamma (\tau )-\varGamma (\beta )}{\varPsi (\tau )-\varPsi (\beta )} $$
Lemma 2.2
The function
is strictly increasing from
\((0, 1)\)
onto
\((1, \sqrt{2}\log (1+ \sqrt{2}) )\).
$$ \phi (t)=\frac{\sqrt{1+t^{2}}\sinh ^{-1}(t)}{t} $$
Proof
Differentiating \(\phi (t)\) gives
where
It follows from (2.2) that
for all \(t\in (0, 1)\).
$$ \phi '(t)=\frac{\phi _{1}(t)}{t\sqrt{1+t^{2}}}, $$
(2.1)
$$ \phi _{1}(t)=t\sqrt{1+t^{2}}-\sinh ^{-1}(t). $$
(2.2)
$$\begin{aligned}& \phi _{1}\bigl(0^{+}\bigr)=0, \end{aligned}$$
(2.3)
$$\begin{aligned}& \phi _{1}'(t)=\frac{2t^{2}}{\sqrt{1+t^{2}}}>0 \end{aligned}$$
(2.4)
Note that
$$ \phi \bigl(0^{+}\bigr)=1,\qquad \phi \bigl(1^{-}\bigr)=\sqrt{2} \log (1+\sqrt{2}). $$
(2.5)
Lemma 2.3
The function
is strictly increasing from
\((0, 1)\)
onto
\((3/2, 2/(\pi -2) )\).
$$ \varphi (t)=\frac{t^{3}}{(1+t^{2})\arctan (t)-t} $$
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)
Lemma 2.4
Let
\(\theta \in [0, 1]\), \(\nu \in [1/2, \infty )\), \(t\in (0, 1)\)
and
Then we have the following two conclusions:
$$ f_{\theta , \nu }(t)=\nu \log \bigl(1+\theta t^{2}\bigr)-\log \bigl[t \sqrt{1+t ^{2}}+\sinh ^{-1}(t) \bigr]+\log t+\log 2. $$
(2.9)
(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
where
$$\begin{aligned}& f_{\theta , \nu }\bigl(0^{+}\bigr)=0, \end{aligned}$$
(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)
$$ f_{\nu }(t)=\frac{t\sqrt{1+t^{2}}-\sinh ^{-1}(t)}{(2\nu -1)t^{2}[t\sqrt{1+t ^{2}}-\sinh ^{-1}(t)]+4\nu t^{2}\sinh ^{-1}(t)}. $$
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
where \(\phi (t)\) is defined in Lemma 2.2.
$$\begin{aligned}& \psi _{1}\bigl(0^{+}\bigr)=\psi _{2} \bigl(0^{+}\bigr)=0,\qquad f_{\nu }(t)=\frac{\psi _{1}(t)}{ \psi _{2}(t)}, \end{aligned}$$
(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)
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)
Lemma 2.5
Let
\(\vartheta \in [0, 1]\), \(\nu \in [1/2, \infty )\), \(t\in (0, 1)\)
and
Then the following statements are true:
$$ g_{\vartheta , \nu }(t)=\nu \log \bigl(1+\vartheta t^{2}\bigr)-\log \bigl[t+\bigl(1+t ^{2}\bigr)\arctan (t) \bigr]+\log (t)+\log 2. $$
(2.19)
(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
where
$$\begin{aligned}& g_{\vartheta , \nu }\bigl(0^{+}\bigr)=0, \end{aligned}$$
(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)
$$ g_{\nu }(t)=\frac{t-(1-t^{2})\arctan (t)}{t^{2} [((2\nu -1)t^{2}+2 \nu +1)\arctan (t)+(2\nu -1)t ]}. $$
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
where \(\varphi (t)\) is defined in Lemma 2.3.
$$\begin{aligned}& \omega _{1}\bigl(0^{+}\bigr)=\omega _{2} \bigl(0^{+}\bigr)=0,\qquad g_{\nu }(t)=\frac{\omega _{1}(t)}{\omega _{2}(t)}, \end{aligned}$$
(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)
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)
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
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
\(\mu _{1}\geq 1/2+\sqrt{6\nu }/(12\nu )\).
$$ W_{\lambda _{1}, \nu }(x, y)< \mathcal{R}_{QA}(x, y)< W_{\mu _{1}, \nu }(x, y) $$
(3.1)
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)
Theorem 3.2
Let
\(\lambda _{2}, \mu _{2}\in [1/2, 1]\)
and
\(\nu \in [1/2, \infty )\). Then the double inequality
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
\(\mu _{2}\geq 1/2+\sqrt{3\nu }/(6\nu )\).
$$ W_{\lambda _{2}, \nu }(x, y)< \mathcal{R}_{AQ}(x, y)< W_{\mu _{2}, \nu }(x, y) $$
(3.5)
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)
Remark 3.3
Corollary 3.4
Let
\(\lambda _{1}, \mu _{1}, \lambda _{2}, \mu _{2}\in [1/2, 1]\). Then the double inequalities
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
\(\mu _{2}\geq 1/2+\sqrt{3}/6=0.7886\ldots \) .
$$\begin{aligned}& C\bigl[\lambda _{1}x+(1-\lambda _{1})y, \lambda _{1}y+(1-\lambda _{1})x\bigr]< \mathcal{R}_{QA}(x, y)< C\bigl[\mu _{1}x+(1-\mu _{1})y, \mu _{1}y+(1- \mu _{1})x\bigr], \\& C\bigl[\lambda _{2}x+(1-\lambda _{2})y, \lambda _{2}y+(1-\lambda _{2})x\bigr]< \mathcal{R}_{AQ}(x, y)< C\bigl[\mu _{2}x+(1-\mu _{2})y, \mu _{2}y+(1-\mu _{2})x\bigr] \end{aligned}$$
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
hold for all
\(u\in (0, 1)\)
and
\(\nu \in [1/2, \infty )\).
$$\begin{aligned}& 2 \biggl[\bigl(1-u^{2}\bigr)+ \biggl(\frac{\sqrt{2}+\log (1+\sqrt{2})}{2} \biggr) ^{1/\nu }u^{2} \biggr]^{\nu }-\sqrt{1+u^{2}} \\& \quad < \frac{\sinh ^{-1}(u)}{u}< 2 \biggl(1+\frac{u^{2}}{6\nu } \biggr)^{ \nu }- \sqrt{1+u^{2}}, \\& \frac{2 [ (1-u^{2} )+(\frac{2+\pi }{4})^{1/\nu }u^{2} ] ^{\nu }-1}{1+u^{2}}< \frac{\arctan (u)}{u}< \frac{2 (1+\frac{1}{3 \nu }u^{2} )^{\nu }-1}{1+u^{2}} \end{aligned}$$
4 Results and discussion
In the article, we give the sharp bounds for the Neuman means
and
in terms of the two-parameter contraharmonic and arithmetic mean
and find new bounds for the functions \(\sinh (u)/u\) and \(\arctan (u)/u\) on the interval \((0, 1)\).
$$ \mathcal{R}_{QA}(x, y)=\frac{1}{2} \biggl[Q(x, y)+ \frac{A^{2}(x, y)}{\operatorname{SB}(Q(x,y), A(x,y))} \biggr] $$
$$ \mathcal{R}_{AQ}(x, y)=\frac{1}{2} \biggl[A(x, y)+ \frac{Q^{2}(x, y)}{\operatorname{SB}(A(x,y), Q(x,y))} \biggr] $$
$$ W_{\lambda , \nu }(x, y)=C^{\nu }\bigl[\lambda x+(1-\lambda )y, \lambda y+(1- \lambda )x\bigr]A^{1-\nu }(x, y), $$
5 Conclusion
In the article, we prove that the double inequalities
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.
$$ W_{\lambda _{1}, \nu }(x, y)< \mathcal{R}_{QA}(x, y)< W_{\mu _{1}, \nu }(x, y),\qquad W_{\lambda _{2}, \nu }(x, y)< \mathcal{R}_{AQ}(x, y)< W_{\mu _{2}, \nu }(x, y) $$
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.
Competing interests
The authors declare that they have no competing interests.
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.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.