## 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

$$\mathcal{K}(r)= \int_{0}^{\pi/2}\frac{dt}{\sqrt{1-r^{2}\sin^{2}(t)}}, \qquad\mathcal{E}(r)= \int_{0}^{\pi/2}\sqrt{1-r^{2} \sin^{2}(t)}\,dt,$$

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])

$$\begin{gathered} \frac{d{\mathcal{K}(r)}}{dr}=\frac{{\mathcal{E}(r)}-{r'}^{2}{\mathcal {K}}(r)}{r{r'}^{2}},\qquad \frac{d{\mathcal{E}(r)}}{dr}=\frac{{\mathcal{E}(r)}-{\mathcal{K}(r)}}{r}, \\ \mathcal{K} \biggl(\frac{2\sqrt{r}}{1+r} \biggr)=(1+r)\mathcal{K}(r),\qquad \mathcal{E} \biggl(\frac{2\sqrt{r}}{1+r} \biggr)=\frac{2\mathcal {E}(r)-{r'}^{2}\mathcal{K}}{1+r}, \end{gathered}$$

where $$r'=\sqrt{1-r^{2}}$$.

The complete elliptic integrals $$\mathcal{K}(r)$$ and $$\mathcal{E}(r)$$ are the particular cases of the Gaussian hypergeometric function [410]

$$F(a,b;c;x)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}} \frac {x^{n}}{n!}\quad (-1< x< 1),$$

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 [1118]. Indeed,

$$\begin{gathered} \mathcal{K}(r)=\frac{\pi}{2}F \biggl( \frac{1}{2},\frac{1}{2};1;r^{2} \biggr) = \frac{\pi}{2}\sum_{n=0}^{\infty} \frac{ (\frac{1}{2} )_{n}^{2}}{(n!)^{2}}r^{2n}, \\ \mathcal{E}(r)=\frac{\pi}{2}F \biggl(-\frac{1}{2}, \frac{1}{2};1;r^{2} \biggr) =\frac{\pi}{2}\sum _{n=0}^{\infty}\frac{ (-\frac{1}{2} )_{n} (\frac{1}{2} )_{n}}{(n!)^{2}}r^{2n}. \end{gathered}$$

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 [1952].

In 1998, a class of quasi-arithmetic mean was introduced by Toader [53] which is defined by

$$M_{p,n}(a,b)=p^{-1} \biggl(\frac{1}{\pi} \int_{0}^{\pi}p\bigl(r_{n}(\theta )\,d \theta\bigr) \biggr)=p^{-1} \biggl(\frac{2}{\pi} \int_{0}^{\pi/2}p\bigl(r_{n}(\theta )\,d \theta\bigr) \biggr),$$

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,

\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}

is the arithmetic-geometric mean of Gauss [5460],

$$M_{x,2}(a,b))=\frac{2}{\pi} \int_{0}^{\pi/2}\sqrt{a^{2}\cos^{2} \theta +b^{2}\sin^{2}\theta}\,d\theta = \textstyle\begin{cases}2a{\mathcal{E}} (\sqrt{1-(b/a)^{2}} )/\pi,&a\geq b,\\ 2b{\mathcal{E}} (\sqrt{1-(a/b)^{2}} )/\pi,&a< b, \end{cases}$$

is the Toader mean [6170], and

$$M_{x,0}(a,b))=\frac{2}{\pi} \int_{0}^{\pi/2}a^{\cos^{2}\theta}b^{\sin ^{2}\theta}\,d \theta$$

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

$$\begin{gathered} A(a,b)=\frac{a+b}{2}, \qquad G(a,b)=\sqrt{ab}, \\ M_{p}(a,b)= \biggl(\frac{a^{p}+b^{p}}{2} \biggr)^{1/p} (p\neq0), \qquad M_{0}(a,b)=\sqrt{ab}, \end{gathered}$$

be the arithmetic, geometric and pth power means of a and b, respectively. Then it is well known that the inequality

$$G(a,b)=M_{0}(a,b)< A(a,b)=M_{1}(a,b)$$
(1.2)

holds for all $$a, b>0$$ with $$a\neq b$$, and the double inequality

$$\frac{\pi}{2}M_{3/2}\bigl(1, r^{\prime}\bigr)< \mathcal{E}(r)< \frac{\pi }{2}M_{2}\bigl(1, r^{\prime}\bigr)$$
(1.3)

holds for all $$r\in(0, 1)$$ (see [75, 19.9.4]).

From (1.1)-(1.3) we clearly see that

$$G(a,b)< E(a,b)< A(a,b)$$

for all $$a, b>0$$ with $$a\neq b$$.

Let $$p\in[1, \infty)$$ and

$$f(x; p; a, b)=G^{p}\bigl[xa+(1-x)b, xb+(1-x)a\bigr]A^{1-p}(a,b).$$

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

\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)

for all $$p\in[1, \infty)$$ and $$a, b>0$$ with $$a\neq b$$.

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

\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}

holds for any $$p\in[1, \infty)$$ and all $$a, b>0$$ with $$a\neq b$$.

## 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

$$\frac{f(x)-f(a)}{g(x)-g(a)}, \qquad\frac{f(x)-f(b)}{g(x)-g(b)}.$$

If $$f^{\prime}(x)/g^{\prime}(x)$$ is strictly monotone, then the monotonicity in the conclusion is also strict.

### Lemma 2.2

The inequality

$$\frac{1}{4p}+ \biggl(\frac{2\sqrt{2}}{\pi} \biggr)^{4/p}< 1$$

holds for all $$p\in[1, \infty)$$.

### Proof

Let

$$f(p)=\frac{1}{4p}+ \biggl(\frac{2\sqrt{2}}{\pi} \biggr)^{4/p}.$$
(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)

for $$p\in[1, \infty)$$.

Therefore, Lemma 2.2 follows easily from (2.1)-(2.3). □

### Lemma 2.3

The following statements are true:

1. (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. (2)

The function $$r\mapsto[\mathcal{K}(r)-\mathcal {E}(r)]/r^{2}$$ is strictly increasing from $$(0, 1)$$ onto $$(\pi/4, \infty)$$.

3. (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. (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. (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

\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)

It follows from part (1) and (2.5) that

$$f^{\prime}_{1}(r)>0$$
(2.6)

for all $$r\in(0, 1)$$. Therefore, part (3) follows from (2.4) and (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

\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)

From part (1) and (2.8) we clearly see that

$$f^{\prime}_{2}(r)< -\frac{r}{(1+r^{2})}< 0$$
(2.9)

for all $$r\in(0, 1)$$. Therefore, part (4) follows from (2.7) and (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

$$f_{3}(r)=\frac{2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r)}{1+r^{2}} \times\frac{1}{\frac{\mathcal{K}(r)-\mathcal{E}(r)}{r^{2}}}\times \frac {1}{1+r^{2}}.$$
(2.10)

Therefore, part (5) follows easily from parts (2) and (4) together with (2.10). □

### Lemma 2.4

The function

$$g(r)=\frac{r^{2}\mathcal{K}(r)}{(1+r^{2})[\mathcal{K}(r)-\mathcal{E}(r)]}$$

is strictly decreasing from $$(0, 1)$$ onto $$(1/2, 2)$$.

### 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)

From Lemma 2.3(3), (2.11) and (2.13) we know that

$$g\bigl(0^{+}\bigr)=\lim_{r\rightarrow0^{+}}\frac{g^{\prime}_{1}(r)}{g^{\prime}_{2}(r)}=2$$
(2.14)

and the function $$g^{\prime}_{1}(r)/g^{\prime}_{2}(r)$$ is strictly decreasing on $$(0, 1)$$.

Therefore, Lemma 2.4 follows easily from Lemma 2.1, (2.11), (2.12) and (2.14) together with the monotonicity of the function $$g^{\prime}_{1}(r)/g^{\prime}_{2}(r)$$. □

### Lemma 2.5

Let $$u\in[0, 1]$$, $$r\in(0, 1)$$, $$p\in[1, \infty)$$ and

$$h(u, p; r)=\frac{1}{2}p\log \biggl[1-\frac{4ur^{2}}{(1+r^{2})^{2}} \biggr] -\log \biggl[\frac{4 (2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r) )^{2}}{\pi^{2}(1+r^{2})} \biggr].$$
(2.15)

Then one has

1. (1)

$$h(u, p; r)>0$$ for all $$r\in(0, 1)$$ if and only if $$u\leq1/4p$$;

2. (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

\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)

where

\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)

where $$f_{3}(r)$$ and $$g(r)$$ are defined by (2.10) and Lemma 2.4, respectively.

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

\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)

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.

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

$$h\bigl(u, p; 1^{-}\bigr)\leq0.$$
(2.22)

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 $$(0, 1)$$.

Case 3 $$1/(4p)< u<1-(2\sqrt{2}/\pi)^{4/p}$$. Then (2.17) leads to

$$h\bigl(u, p; 1^{-}\bigr)>0.$$
(2.23)

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 $$r\in(\lambda, 1)$$. □

## 3 Main result

### Theorem 3.1

Let $$\lambda, \mu\in[0, 1/2]$$. Then the double inequality

$$\begin{gathered} G^{p}\bigl[\lambda a+(1-\lambda)b, \lambda b+(1-\lambda)a\bigr]A^{1-p}(a,b) \\ \quad< E(a,b)< G^{p}\bigl[\mu a+(1-\mu)b, \mu b+(1-\mu)a \bigr]A^{1-p}(a,b) \end{gathered}$$

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 $$\mu\geq1/2-\sqrt{p}/(4p)$$.

### 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

$$\begin{gathered} E(a,b)=\frac{4(1+r)^{2}}{\pi^{2}(1+r^{2})}A(a,b){\mathcal{E}}^{2} \biggl( \frac{2\sqrt{r}}{1+r} \biggr) =\frac{4}{\pi^{2}}A(a,b)\frac{ [2\mathcal{E}(r)-(1-r^{2})\mathcal {K}(r) ]^{2}}{1+r^{2}}, \\ \log \bigl[G^{p}\bigl(ta+(1-t)b, tb+(1-t)a \bigr)A^{1-p}(a,b) \bigr]-\log E(a,b) \\ \quad=\log \biggl[\frac{G^{p}(ta+(1-t)b, tb+(1-t)a)A^{1-p}(a,b)}{A(a,b)} \biggr]-\log \biggl[\frac {E(a,b)}{A(a,b)} \biggr] \\ \quad=\frac{1}{2}p\log \biggl[1-\frac {4(1-2t)^{2}r^{2}}{(1+r^{2})^{2}} \biggr] -\log \biggl[ \frac{4 (2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r) )^{2}}{\pi^{2}(1+r^{2})} \biggr]. \end{gathered}$$
(3.1)

Therefore, Theorem 3.1 follows easily from Lemma 2.5 and (3.1). □

Let $$p=1, 2$$, then Theorem 3.1 leads to Corollary 3.2 immediately.

### Corollary 3.2

Let $$\lambda_{1}, \mu_{1}, \lambda_{2}, \mu_{2}\in[0, 1/2]$$. Then the double inequalities

$$\begin{gathered} H\bigl[\lambda_{1}a+(1- \lambda_{1})b, \lambda_{1}b+(1-\lambda _{1})a \bigr]< E(a,b)< H\bigl[\mu_{1}a+(1-\mu_{1})b, \mu_{1}b+(1-\mu_{1})a\bigr], \\ G\bigl[\lambda_{2}a+(1-\lambda_{2})b, \lambda_{2}b+(1- \lambda _{2})a\bigr]< E(a,b)< G\bigl[\mu_{2}a+(1- \mu_{2})b, \mu_{2}b+(1-\mu_{2})a\bigr] \end{gathered}$$

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 $$\mu_{2}\geq1/4$$.

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

$$\begin{gathered} \frac{\sqrt{2}\pi}{4} \bigl(1+{r^{\prime}}^{2} \bigr)^{(1-p)/2} \biggl[4{r^{\prime}}^{2} + \biggl( \frac{8}{\pi^{2}} \biggr)^{2/p}r^{4} \biggr]^{p/4} \\ \quad< \mathcal{E}(r) < \frac{\sqrt{2}\pi}{4} \bigl(1+{r^{\prime }}^{2} \bigr)^{(1-p)/2} \biggl[\bigl(1+{r^{\prime}}^{2} \bigr)^{2}-\frac {r^{4}}{4p} \biggr]^{p/4} \end{gathered}$$

holds for all $$r\in(0, 1)$$ and $$p\in[1, \infty)$$.

## 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.