1 Introduction
Let \(r\in(0,1)\). Then the Legendre complete elliptic integrals \(\mathcal {K}(r)\) and \(\mathcal{E}(r)\) [1, 2] of the first and second kinds are defined as
respectively. It is well known that the function \(r\rightarrow\mathcal {K}(r)\) is strictly increasing from \((0, 1)\) onto \((\pi/2, \infty)\) and the function \(r\rightarrow\mathcal{E}(r)\) is strictly decreasing from \((0, 1)\) onto \((1, \pi/2)\), and they satisfy the formulas (see [3, Appendix E, pp. 474,475])
where \(r'=\sqrt{1-r^{2}}\).
$$ \mathcal{K}(r)= \int_{0}^{\pi/2}\frac{dt}{\sqrt{1-r^{2}\sin^{2}(t)}}, \qquad\mathcal{E}(r)= \int_{0}^{\pi/2}\sqrt{1-r^{2} \sin^{2}(t)}\,dt, $$
$$ \begin{gathered} \frac{d{\mathcal{K}(r)}}{dr}=\frac{{\mathcal{E}(r)}-{r'}^{2}{\mathcal {K}}(r)}{r{r'}^{2}},\qquad \frac{d{\mathcal{E}(r)}}{dr}=\frac{{\mathcal{E}(r)}-{\mathcal{K}(r)}}{r}, \\ \mathcal{K} \biggl(\frac{2\sqrt{r}}{1+r} \biggr)=(1+r)\mathcal{K}(r),\qquad \mathcal{E} \biggl(\frac{2\sqrt{r}}{1+r} \biggr)=\frac{2\mathcal {E}(r)-{r'}^{2}\mathcal{K}}{1+r}, \end{gathered} $$
The complete elliptic integrals \(\mathcal{K}(r)\) and \(\mathcal{E}(r)\) are the particular cases of the Gaussian hypergeometric function [4‐10]
where \((a)_{0}=1\) for \(a\neq0\), \((a)_{n}=a(a+1)(a+2)\cdots (a+n-1)=\Gamma(a+n)/\Gamma(a)\) is the shifted factorial function and \(\Gamma(x)=\int_{0}^{\infty }t^{x-1}e^{-t}\,dt\) (\(x>0\)) is the gamma function [11‐18]. Indeed,
$$ F(a,b;c;x)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}} \frac {x^{n}}{n!}\quad (-1< x< 1), $$
$$ \begin{gathered} \mathcal{K}(r)=\frac{\pi}{2}F \biggl( \frac{1}{2},\frac{1}{2};1;r^{2} \biggr) = \frac{\pi}{2}\sum_{n=0}^{\infty} \frac{ (\frac{1}{2} )_{n}^{2}}{(n!)^{2}}r^{2n}, \\ \mathcal{E}(r)=\frac{\pi}{2}F \biggl(-\frac{1}{2}, \frac{1}{2};1;r^{2} \biggr) =\frac{\pi}{2}\sum _{n=0}^{\infty}\frac{ (-\frac{1}{2} )_{n} (\frac{1}{2} )_{n}}{(n!)^{2}}r^{2n}. \end{gathered} $$
Anzeige
Recently, the bounds for the complete elliptic integrals have attracted the attention of many researchers. In particular, many remarkable inequalities and properties for \(\mathcal{K}(r)\), \(\mathcal{E}(r)\) and \(F(a,b;c;x)\) can be found in the literature [19‐52].
In 1998, a class of quasi-arithmetic mean was introduced by Toader [53] which is defined by
where \(r_{n}(\theta)=(a^{n}\cos^{2}\theta+b^{n}\sin^{2}\theta)^{1/n}\) for \(n\neq0\), \(r_{0}(\theta)=a^{\cos^{2}\theta}b^{\sin^{2}\theta}\), and p is a strictly monotonic function. It is well known that many important means are the special cases of the quasi-arithmetic mean. For example,
is the arithmetic-geometric mean of Gauss [54‐60],
is the Toader mean [61‐70], and
is the Toader-Qi mean [71‐74].
$$ M_{p,n}(a,b)=p^{-1} \biggl(\frac{1}{\pi} \int_{0}^{\pi}p\bigl(r_{n}(\theta )\,d \theta\bigr) \biggr)=p^{-1} \biggl(\frac{2}{\pi} \int_{0}^{\pi/2}p\bigl(r_{n}(\theta )\,d \theta\bigr) \biggr), $$
$$\begin{aligned} M_{1/x,2}(a,b)= \frac{\pi}{2 \int_{0}^{\pi /2}{\frac{d\theta}{\sqrt{a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}}}} = \textstyle\begin{cases} {\pi a} / [2{\mathcal{K}} (\sqrt{1-(b/a)^{2}} ) ],&a\geq b,\\ {\pi b} / [2{\mathcal{K}} (\sqrt{1-(a/b)^{2}} ) ],&a< b, \end{cases}\displaystyle \end{aligned}$$
$$ M_{x,2}(a,b))=\frac{2}{\pi} \int_{0}^{\pi/2}\sqrt{a^{2}\cos^{2} \theta +b^{2}\sin^{2}\theta}\,d\theta = \textstyle\begin{cases}2a{\mathcal{E}} (\sqrt{1-(b/a)^{2}} )/\pi,&a\geq b,\\ 2b{\mathcal{E}} (\sqrt{1-(a/b)^{2}} )/\pi,&a< b, \end{cases} $$
$$ M_{x,0}(a,b))=\frac{2}{\pi} \int_{0}^{\pi/2}a^{\cos^{2}\theta}b^{\sin ^{2}\theta}\,d \theta $$
Let \(p=\sqrt{x}\) and \(n=1\). Then \(M_{p,n}(a,b)\) reduces to a special quasi-arithmetic mean
$$ E(a,b)=M_{\sqrt{x},1}(a,b))= \textstyle\begin{cases}4a [{\mathcal{E}} (\sqrt{1-b/a} ) ]^{2}/\pi ^{2},&a\geq b,\\ 4b [{\mathcal{E}} (\sqrt{1-a/b} ) ]^{2}/\pi^{2},&a< b. \end{cases} $$
(1.1)
Let
be the arithmetic, geometric and pth power means of a and b, respectively. Then it is well known that the inequality
holds for all \(a, b>0\) with \(a\neq b\), and the double inequality
holds for all \(r\in(0, 1)\) (see [75, 19.9.4]).
$$ \begin{gathered} A(a,b)=\frac{a+b}{2}, \qquad G(a,b)=\sqrt{ab}, \\ M_{p}(a,b)= \biggl(\frac{a^{p}+b^{p}}{2} \biggr)^{1/p} (p\neq0), \qquad M_{0}(a,b)=\sqrt{ab}, \end{gathered} $$
$$ G(a,b)=M_{0}(a,b)< A(a,b)=M_{1}(a,b) $$
(1.2)
$$ \frac{\pi}{2}M_{3/2}\bigl(1, r^{\prime}\bigr)< \mathcal{E}(r)< \frac{\pi }{2}M_{2}\bigl(1, r^{\prime}\bigr) $$
(1.3)
Anzeige
From (1.1)-(1.3) we clearly see that
for all \(a, b>0\) with \(a\neq b\).
$$ G(a,b)< E(a,b)< A(a,b) $$
Let \(p\in[1, \infty)\) and
Then it is not difficult to verify that the function \(x\rightarrow f(x; p; a, b)\) is strictly increasing on \([0, 1/2]\) for fixed \(p\in[1, \infty)\) and \(a, b>0\) with \(a\neq b\). Note that
for all \(p\in[1, \infty)\) and \(a, b>0\) with \(a\neq b\).
$$ f(x; p; a, b)=G^{p}\bigl[xa+(1-x)b, xb+(1-x)a\bigr]A^{1-p}(a,b). $$
$$ \begin{aligned}[b] f(0; p; a, b)&=G^{p}(a,b)A^{1-p}(a,b) \leq G(a,b) \\ &< E(a,b)< A(a,b)=f(1/2; p; a, b) \end{aligned} $$
(1.4)
Motivated by inequalities (1.4) and the monotonicity of the function \(x\rightarrow f(x; p; a, b)\) on the interval \([0, 1/2]\), in the article, we shall find the best possible parameters \(\lambda=\lambda(p), \mu=\mu(p)\) on the interval \([0, 1/2]\) such that the double inequality
holds for any \(p\in[1, \infty)\) and all \(a, b>0\) with \(a\neq b\).
$$ \begin{aligned} &G^{p}\bigl[\lambda a+(1-\lambda)b, \lambda b+(1-\lambda)a\bigr]A^{1-p}(a,b) \\ &\quad< E(a,b)< G^{p}\bigl[\mu a+(1-\mu)b, \mu b+(1-\mu)a \bigr]A^{1-p}(a,b) \end{aligned} $$
2 Lemmas
Lemma 2.1
(see [3, Theorem 1.25])
Let
\(-\infty< a< b<+\infty\), \(f, g: [a, b]\rightarrow\mathbb{R}\)
be continuous on
\([a, b]\)
and differentiable on
\((a,b)\), and
\(g^{\prime}(x)\neq0\)
on
\((a, b)\). If
\(f^{\prime}(x)/g^{\prime}(x)\)
is increasing (decreasing) on
\((a,b)\), then so are the functions
If
\(f^{\prime}(x)/g^{\prime}(x)\)
is strictly monotone, then the monotonicity in the conclusion is also strict.
$$ \frac{f(x)-f(a)}{g(x)-g(a)}, \qquad\frac{f(x)-f(b)}{g(x)-g(b)}. $$
Lemma 2.2
The inequality
holds for all
\(p\in[1, \infty)\).
$$ \frac{1}{4p}+ \biggl(\frac{2\sqrt{2}}{\pi} \biggr)^{4/p}< 1 $$
Proof
Let
Then simple computations lead to
for \(p\in[1, \infty)\).
$$ f(p)=\frac{1}{4p}+ \biggl(\frac{2\sqrt{2}}{\pi} \biggr)^{4/p}. $$
(2.1)
$$\begin{aligned}& \lim_{p\rightarrow\infty}f(p)=1, \end{aligned}$$
(2.2)
$$\begin{aligned}& \begin{aligned}[b] f^{\prime}(p)&=\frac{4}{p^{2}}\log \biggl(\frac{\sqrt{2}\pi}{4} \biggr) \biggl[ \biggl(\frac{2\sqrt{2}}{\pi} \biggr)^{4/p}-\frac{1}{16\log (\frac{\sqrt{2}\pi}{4} )} \biggr] \\ &\geq\frac{4}{p^{2}}\log \biggl(\frac{\sqrt{2}\pi}{4} \biggr) \biggl[ \biggl( \frac{2\sqrt{2}}{\pi} \biggr)^{4}-\frac{1}{16\log (\frac {\sqrt{2}\pi}{4} )} \biggr] \\ &=\frac{1024\log (\frac{\sqrt{2}\pi}{4} )-\pi^{4}}{4\pi^{4}p^{2}}>0 \end{aligned} \end{aligned}$$
(2.3)
Lemma 2.3
The following statements are true:
(1)
The function
\(r\mapsto[\mathcal{E}(r)-(1-r^{2})\mathcal {K}(r)]/r^{2}\)
is strictly increasing from
\((0, 1)\)
onto
\((\pi/4, 1)\).
(2)
The function
\(r\mapsto[\mathcal{K}(r)-\mathcal {E}(r)]/r^{2}\)
is strictly increasing from
\((0, 1)\)
onto
\((\pi/4, \infty)\).
(3)
The function
\(r\mapsto[\mathcal{E}(r)+(1-r^{2})\mathcal {K}(r)]/(1-r^{2})\)
is strictly increasing from
\((0, 1)\)
onto
\((\pi, \infty)\).
(4)
The function
\(r\mapsto[2\mathcal{E}(r)-(1-r^{2})\mathcal {K}(r)]/(1+r^{2})\)
is strictly decreasing from
\((0, 1)\)
onto
\((1, \pi/2)\).
(5)
The function
\(r\mapsto r^{2}[2\mathcal {E}(r)-(1-r^{2})\mathcal{K}(r)]/ [(1+r^{2})^{2}(\mathcal {K}(r)-\mathcal{E}(r)) ]\)
is strictly decreasing from
\((0, 1)\)
onto
\((0, 2)\).
Proof
Parts (1) and (2) can be found in the literature [3, Theorem 3.21(1) and Exercise 3.43(11)].
For part (3), let \(f_{1}(r)=[\mathcal{E}(r)+(1-r^{2})\mathcal {K}(r)]/(1-r^{2})\). Then simple computations lead to
It follows from part (1) and (2.5) that
for all \(r\in(0, 1)\). Therefore, part (3) follows from (2.4) and (2.6).
$$\begin{aligned}& f_{1}\bigl(0^{+}\bigr)=\pi, \qquad f_{1} \bigl(1^{-}\bigr)=\infty, \end{aligned}$$
(2.4)
$$\begin{aligned}& f^{\prime}_{1}(r)=\frac{r}{(1-r^{2})^{2}} \biggl[\frac{2}{r^{2}} \bigl(\mathcal {E}(r)-\bigl(1-r^{2}\bigr)\mathcal{K}(r)\bigr)+ \bigl(1-r^{2}\bigr)\mathcal{K}(r) \biggr]. \end{aligned}$$
(2.5)
$$ f^{\prime}_{1}(r)>0 $$
(2.6)
For part (4), let \(f_{2}(r)=[2\mathcal{E}(r)-(1-r^{2})\mathcal {K}(r)]/(1+r^{2})\), then one has
From part (1) and (2.8) we clearly see that
for all \(r\in(0, 1)\). Therefore, part (4) follows from (2.7) and (2.9).
$$\begin{aligned}& f_{2}\bigl(0^{+}\bigr)=\frac{\pi}{2}, \qquad f_{1}\bigl(1^{-}\bigr)=1, \end{aligned}$$
(2.7)
$$\begin{aligned}& f^{\prime}_{2}(r)=\frac{r}{(1+r^{2})^{2}} \biggl[ \bigl(1-r^{2}\bigr)\frac{\mathcal {E}(r)-(1-r^{2})\mathcal{K}(r)}{r^{2}}-2\mathcal{E}(r) \biggr]. \end{aligned}$$
(2.8)
$$ f^{\prime}_{2}(r)< -\frac{r}{(1+r^{2})}< 0 $$
(2.9)
For part (5), let \(f_{3}(r)=r^{2}[2\mathcal{E}(r)-(1-r^{2})\mathcal {K}(r)]/ [(1+r^{2})^{2}(\mathcal{K}(r)-\mathcal{E}(r)) ]\), then \(f_{3}(r)\) can be rewritten as
Therefore, part (5) follows easily from parts (2) and (4) together with (2.10). □
$$ f_{3}(r)=\frac{2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r)}{1+r^{2}} \times\frac{1}{\frac{\mathcal{K}(r)-\mathcal{E}(r)}{r^{2}}}\times \frac {1}{1+r^{2}}. $$
(2.10)
Lemma 2.4
The function
is strictly decreasing from
\((0, 1)\)
onto
\((1/2, 2)\).
$$ g(r)=\frac{r^{2}\mathcal{K}(r)}{(1+r^{2})[\mathcal{K}(r)-\mathcal{E}(r)]} $$
Proof
Let \(g_{1}(r)=r^{2}\mathcal{K}(r)\) and \(g_{2}(r)=(1+r^{2})[\mathcal {K}(r)-\mathcal{E}(r)]\). Then we clearly see that
$$\begin{aligned}& g_{1}\bigl(0^{+}\bigr)=g_{2} \bigl(0^{+}\bigr)=0, \qquad g(r)=\frac{g_{1}(r)}{g_{2}(r)}, \end{aligned}$$
(2.11)
$$\begin{aligned}& g\bigl(1^{-}\bigr)=\frac{1}{2}, \end{aligned}$$
(2.12)
$$\begin{aligned}& \frac{g^{\prime}_{1}(r)}{g^{\prime}_{2}(r)}=\frac{1}{2-\frac{3\mathcal {E}(r)}{\frac{\mathcal{E}(r)+(1-r^{2})\mathcal{K}(r)}{1-r^{2}}}}. \end{aligned}$$
(2.13)
Lemma 2.5
Let
\(u\in[0, 1]\), \(r\in(0, 1)\), \(p\in[1, \infty)\)
and
Then one has
$$ h(u, p; r)=\frac{1}{2}p\log \biggl[1-\frac{4ur^{2}}{(1+r^{2})^{2}} \biggr] -\log \biggl[\frac{4 (2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r) )^{2}}{\pi^{2}(1+r^{2})} \biggr]. $$
(2.15)
(1)
\(h(u, p; r)>0\)
for all
\(r\in(0, 1)\)
if and only if
\(u\leq1/4p\);
(2)
\(h(u, p; r)<0\)
for all
\(r\in(0, 1)\)
if and only if
\(u\geq 1-(2\sqrt{2}/\pi)^{4/p}\).
Proof
It follows from (2.15) that
where
where \(f_{3}(r)\) and \(g(r)\) are defined by (2.10) and Lemma 2.4, respectively.
$$\begin{aligned}& h\bigl(u, p; 0^{+}\bigr)=0, \end{aligned}$$
(2.16)
$$\begin{aligned}& h\bigl(u, p; 1^{-}\bigr)=\frac{p}{2}\log(1-u)+\log \biggl( \frac{\pi^{2}}{8} \biggr), \end{aligned}$$
(2.17)
$$\begin{aligned}& \begin{aligned}[b] \frac{\partial h(u, p; r)}{\partial r}&=\frac{2(1-r^{2})[\mathcal {K}(r)-\mathcal{E}(r)]}{ r(1+r^{2})[2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r)]} - \frac{4pur(1-r^{2})}{(1+r^{2}) [(1+r^{2})^{2}-4ur^{2} ]} \\ &=\frac{2(1-r^{2}) [2(\mathcal{K}(r)-\mathcal {E}(r))+p(2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r)) ]}{(1+r^{2}) [(1+r^{2})^{2}-4ur^{2} ][2\mathcal {E}(r)-(1-r^{2})\mathcal{K}(r)]}\bigl[h_{1}(p; r)-2u\bigr], \end{aligned} \end{aligned}$$
(2.18)
$$ \begin{aligned}[b] h_{1}(p; r)&=\frac{(1+r^{2})^{2}[\mathcal{K}(r)-\mathcal{E}(r)]}{r^{2} [2(\mathcal{K}(r)-\mathcal{E}(r))+p(2\mathcal {E}(r)-(1-r^{2})\mathcal{K}(r)) ]} \\ &=\frac{1}{g(r)+(p-1)f_{3}(r)}, \end{aligned} $$
(2.19)
From Lemma 2.3(5) and Lemma 2.4 together with (2.19) we clearly see that the function \(r\rightarrow h_{1}(p; r)\) is strictly increasing on \((0, 1)\) and
From Lemma 2.2 we know that \(1-(2\sqrt{2}/\pi)^{4/p}>1/(4p)\). Therefore, we only need to divide the proof into three cases as follows.
$$\begin{aligned}& h_{1}\bigl(p; 0^{+}\bigr)=\frac{1}{2p}, \end{aligned}$$
(2.20)
$$\begin{aligned}& h_{1}\bigl(p; 1^{-}\bigr)=2. \end{aligned}$$
(2.21)
Case 1
\(u\leq1/(4p)\). Then Lemma 2.3(4), (2.18), (2.20) and the monotonicity of the function \(r\rightarrow h_{1}(p; r)\) on the interval \((0, 1)\) lead to the conclusion that the function \(r\rightarrow h(u, p; r)\) is strictly increasing on \((0, 1)\). Therefore, \(h(u, p; r)>0\) for all \(r\in(0, 1)\) follows from (2.16) and the monotonicity of the function \(r\rightarrow h(u, p; r)\).
Case 2
\(u\geq1-(2\sqrt{2}/\pi)^{4/p}\). Then from Lemma 2.2, Lemma 2.3(5), (2.17), (2.18), (2.20), (2.21) and the monotonicity of the function \(r\rightarrow h_{1}(p; r)\) on the interval \((0, 1)\) we clearly see that there exists \(r_{0}\in(0, 1)\) such that the function \(r\rightarrow h(u, p; r)\) is strictly decreasing on \((0, r_{0})\) and strictly increasing on \((r_{0}, 1)\), and
Therefore, \(h(u, p; r)<0\) for all \(r\in(0, 1)\) follows from (2.16) and (2.22) together with the piecewise monotonicity of the function \(r\rightarrow h(u, p; r)\) on the interval \((0, 1)\).
$$ h\bigl(u, p; 1^{-}\bigr)\leq0. $$
(2.22)
Case 3
\(1/(4p)< u<1-(2\sqrt{2}/\pi)^{4/p}\). Then (2.17) leads to
It follows from Lemma 2.3(5), (2.18), (2.20), (2.21) and the monotonicity of the function \(r\rightarrow h_{1}(p; r)\) on the interval \((0, 1)\) that there exists \(r^{\ast}\in(0, 1)\) such that the function \(r\rightarrow h(u, p; r)\) is strictly decreasing on \((0, r^{\ast})\) and strictly increasing on \((r^{\ast}, 1)\). Therefore, there exists \(\lambda \in(0, 1)\) such that \(h(u, p; r)<0\) for \(r\in(0, \lambda)\) and \(h(u, p; r)>0\) for \(r\in(\lambda, 1)\). □
$$ h\bigl(u, p; 1^{-}\bigr)>0. $$
(2.23)
3 Main result
Theorem 3.1
Let
\(\lambda, \mu\in[0, 1/2]\). Then the double inequality
holds for any
\(p\in[1, \infty)\)
and all
\(a, b>0\)
with
\(a\neq b\)
if and only if
\(\lambda\leq1/2-\sqrt{1-(2\sqrt{2}/\pi)^{4/p}}/2\)
and
\(\mu\geq1/2-\sqrt{p}/(4p)\).
$$ \begin{gathered} G^{p}\bigl[\lambda a+(1-\lambda)b, \lambda b+(1-\lambda)a\bigr]A^{1-p}(a,b) \\ \quad< E(a,b)< G^{p}\bigl[\mu a+(1-\mu)b, \mu b+(1-\mu)a \bigr]A^{1-p}(a,b) \end{gathered} $$
Proof
Let \(t\in[0, 1/2]\), since \(G^{p}[ta+(1-t)b, tb+(1-t)a]A^{1-p}(a,b)\) and \(E(a,b)\) are symmetric and homogeneous of degree one, without loss of generality, we assume that \(a>b>0\). Let \(r\in(0, 1)\) and \(b/a=(1-r)^{2}/(1+r)^{2}\). Then (1.1) leads to
Therefore, Theorem 3.1 follows easily from Lemma 2.5 and (3.1). □
$$ \begin{gathered} E(a,b)=\frac{4(1+r)^{2}}{\pi^{2}(1+r^{2})}A(a,b){\mathcal{E}}^{2} \biggl( \frac{2\sqrt{r}}{1+r} \biggr) =\frac{4}{\pi^{2}}A(a,b)\frac{ [2\mathcal{E}(r)-(1-r^{2})\mathcal {K}(r) ]^{2}}{1+r^{2}}, \\ \log \bigl[G^{p}\bigl(ta+(1-t)b, tb+(1-t)a \bigr)A^{1-p}(a,b) \bigr]-\log E(a,b) \\ \quad=\log \biggl[\frac{G^{p}(ta+(1-t)b, tb+(1-t)a)A^{1-p}(a,b)}{A(a,b)} \biggr]-\log \biggl[\frac {E(a,b)}{A(a,b)} \biggr] \\ \quad=\frac{1}{2}p\log \biggl[1-\frac {4(1-2t)^{2}r^{2}}{(1+r^{2})^{2}} \biggr] -\log \biggl[ \frac{4 (2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r) )^{2}}{\pi^{2}(1+r^{2})} \biggr]. \end{gathered} $$
(3.1)
Corollary 3.2
Let
\(\lambda_{1}, \mu_{1}, \lambda_{2}, \mu_{2}\in[0, 1/2]\). Then the double inequalities
hold for all
\(a, b>0\)
with
\(a\neq b\)
if and only if
\(\lambda_{1}\leq 1/2-\sqrt{1-8/\pi^{2}}/2=0.2823\ldots\) , \(\mu_{1}\geq1/2-\sqrt{2}/8=0.3232\ldots\) , \(\lambda_{2}\leq1/2-\sqrt {1-64/\pi^{4}}/2=0.2071\ldots\)
and
\(\mu_{2}\geq1/4\).
$$ \begin{gathered} H\bigl[\lambda_{1}a+(1- \lambda_{1})b, \lambda_{1}b+(1-\lambda _{1})a \bigr]< E(a,b)< H\bigl[\mu_{1}a+(1-\mu_{1})b, \mu_{1}b+(1-\mu_{1})a\bigr], \\ G\bigl[\lambda_{2}a+(1-\lambda_{2})b, \lambda_{2}b+(1- \lambda _{2})a\bigr]< E(a,b)< G\bigl[\mu_{2}a+(1- \mu_{2})b, \mu_{2}b+(1-\mu_{2})a\bigr] \end{gathered} $$
Let \(p\in[1, \infty)\), \(r\in(0, 1)\), \(a=r\), \(b=1-r^{2}={r^{\prime }}^{2}\), \(\lambda=1/2-\sqrt{1-(2\sqrt{2}/\pi)^{4/p}}/2\) and \(\mu=1/2-\sqrt{p}/(4p)\). Then (1.1) and Theorem 3.1 lead to Corollary 3.3 immediately.
Corollary 3.3
The double inequality
holds for all
\(r\in(0, 1)\)
and
\(p\in[1, \infty)\).
$$ \begin{gathered} \frac{\sqrt{2}\pi}{4} \bigl(1+{r^{\prime}}^{2} \bigr)^{(1-p)/2} \biggl[4{r^{\prime}}^{2} + \biggl( \frac{8}{\pi^{2}} \biggr)^{2/p}r^{4} \biggr]^{p/4} \\ \quad< \mathcal{E}(r) < \frac{\sqrt{2}\pi}{4} \bigl(1+{r^{\prime }}^{2} \bigr)^{(1-p)/2} \biggl[\bigl(1+{r^{\prime}}^{2} \bigr)^{2}-\frac {r^{4}}{4p} \biggr]^{p/4} \end{gathered} $$
4 Results and discussion
In this paper, we provide the sharp bounds for the special quasi-arithmetic mean \(E(a,b)\) in terms of the arithmetic mean \(A(a,b)\) and geometric mean \(G(a,b)\) with two parameters. As consequences, we present the best possible one-parameter harmonic and geometric means bounds for \(E(a,b)\) and find new bounds for the complete elliptic integral of the second kind.
5 Conclusion
In the article, we derive a new bivariate mean \(E(a,b)\) from the quasi-arithmetic mean and provide its sharp upper and lower bounds in terms of the concave combination of arithmetic and geometric means.
Acknowledgements
The research was supported by the Natural Science Foundation of China (Grants Nos. 61673169, 61374086, 11371125, 11401191) and the Tianyuan Special Funds of the National Natural Science Foundation of China (Grant No. 11626101).
Competing interests
The authors declare that they have no competing interests.
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