On approximating the quasi-arithmetic mean

Abstract

In this article, we prove that the double inequalities

$$\begin{aligned} &\alpha_{1} \biggl[\frac{7C(a,b)}{16}+\frac{9H(a,b)}{16} \biggr]+(1- \alpha_{1}) \biggl[\frac{3A(a,b)}{4}+\frac{G(a, b)}{4} \biggr]\\ &\quad< E(a,b) \\ &\quad< \beta_{1} \biggl[\frac{7C(a,b)}{16}+\frac{9H(a,b)}{16} \biggr]+(1- \beta_{1}) \biggl[\frac{3A(a,b)}{4}+\frac{G(a, b)}{4} \biggr], \\ &\biggl[\frac{7C(a,b)}{16}+\frac{9H(a,b)}{16} \biggr]^{\alpha _{2}} \biggl[ \frac{3A(a,b)}{4}+\frac{G(a, b)}{4} \biggr]^{1-\alpha_{2}}\\ &\quad< E(a,b) \\ &\quad< \biggl[\frac{7C(a,b)}{16}+\frac{9H(a,b)}{16} \biggr]^{\beta _{2}} \biggl[ \frac{3A(a,b)}{4}+\frac{G(a, b)}{4} \biggr]^{1-\beta_{2}} \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$ if and only if $\alpha_{1}\leq 3/16=0.1875$, $\beta_{1}\geq64/\pi^{2}-6= 0.484555\dots$, $\alpha_{2}\leq3/16=0.1875$ and $\beta_{2}\geq(5\log2-\log3-2\log \pi)/(\log7-\log6)= 0.503817\dots$, where $E(a,b)= (\frac{2}{\pi}\int^{\pi/2}_{0}\sqrt{a\cos^{2}\theta +b\sin^{2}\theta}\,d\theta )^{2}$, $H(a,b)=2ab/(a+b)$, $G(a,b)=\sqrt{ab}$, $A(a,b)=(a+b)/2$ and $C(a,b)=(a^{2}+b^{2})/(a+b)$ are the quasi-arithmetic, harmonic, geometric, arithmetic and contra-harmonic means of a and b, respectively.

1 Introduction

Let $a, b>0$, $p:(0,\infty)\mapsto(0, \infty)$ be a strictly monotone real-valued function, $\theta\in(0,2\pi)$ and

$$ r_{n}(\theta)= \textstyle\begin{cases} (a^{n}\cos^{2}\theta+b^{n}\sin^{2}\theta)^{1/n},&n\neq0,\\ a^{\cos^{2}\theta}b^{\sin^{2}\theta},&n=0. \end{cases} $$

(1.1)

Then the class of quasi-arithmetic mean [1] is defined by

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

(1.2)

where $p^{-1}$ is the inverse function of p.

Many important means are the special cases of the quasi-arithmetic mean $M_{p,n}(a,b)$. For example, from (1.1) and (1.2) we clearly see that

$$ M_{1/x,2}(a,b)=\frac{\pi}{2\int^{\pi/2}_{0}(a^{2}\cos^{2}\theta+b^{2}\sin ^{2}\theta)^{-1/2}\,d\theta}=\mathit{AGM}(a,b) $$

is the Gaussian arithmetic–geometric mean [2,3,4,5,6,7,8,9], which is related to the complete elliptic integral of the first kind $\mathcal{K}=\mathcal {K}(r)=\int^{\pi/2}_{0}(1-r^{2}\sin^{2}\theta)^{-1/2}\,d\theta$ ($0< r<1$),

$$ T(a,b)=M_{x,2}(a,b)=\frac{2}{\pi} \int^{\pi/2}_{0}\sqrt{a^{2}\cos ^{2}\theta+b^{2}\sin^{2}\theta}\,d\theta $$

is the Toader mean [10,11,12], which can be expressed in terms of the complete elliptic integral of the second kind $\mathcal{E}=\mathcal {E}(r)=\int^{\pi/2}_{0}\sqrt{1-r^{2}\sin^{2}\theta}\,d\theta$ ($0< r<1$), and

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

is the Toader–Qi mean [13,14,15], which is related to the modified Bessel function of the first kind $I_{0}(x)=\sum_{n=0}^{\infty}(x/2)^{2n}/(n!)^{2}$ ($x>0$).

It is well-known that $\mathcal{K}(r)$ is strictly increasing from $(0,1)$ onto $(\pi/2, \infty)$ and $\mathcal{E}(r)$ is strictly decreasing from $(0,1)$ onto $(1,\pi/2)$. Moreover, $\mathcal{K}(r)$ and $\mathcal{E}(r)$ satisfy the following Landen identities and derivative formulas (see [16, Appendix E, pp. 474–475])

$$\begin{aligned} &\mathcal{K} \biggl(\frac{2\sqrt{r}}{1+r} \biggr)=(1+r)\mathcal{K}, \qquad\mathcal{E} \biggl(\frac{2\sqrt{r}}{1+r} \biggr)=\frac{2\mathcal {E}-r^{\prime 2}\mathcal{K}}{1+r}, \\ &\frac{d\mathcal{K}}{dr}=\frac{\mathcal{E}-r^{\prime 2}\mathcal{K}}{rr^{\prime 2}}, \qquad \frac{d\mathcal{E}}{dr}=\frac{\mathcal{E}-\mathcal{K}}{r}, \\ &\frac{d(\mathcal{E}-r^{\prime 2}\mathcal{K})}{dr}=r\mathcal{K}, \qquad\frac {d(\mathcal{K}-\mathcal{E})}{dr}=\frac{r\mathcal{E}}{r^{\prime 2}}. \end{aligned}$$

In particular, $\mathcal{K}(r)$ and $\mathcal{E}(r)$ are the special cases of the Gaussian hypergeometric function [17,18,19,20,21,22,23,24,25,26] as follows:

$$ \mathcal{K}(r)=\frac{\pi}{2}F \biggl(\frac{1}{2}, \frac {1}{2};1;r^{2} \biggr), \qquad\mathcal{E}(r)=\frac{\pi}{2}F \biggl(-\frac {1}{2},\frac{1}{2};1;r^{2} \biggr), $$

(1.3)

and the Gaussian hypergeometric function $F(a,b;c;x)$ with real parameters $a,b$, and c $(c\neq0,-1,-2,\dots)$ is defined by

$$ F(a,b;c;x)={}_{2}F_{1}(a,b;c;x)=\sum _{n=0}^{\infty}\frac {(a,n)(b,n)}{(c,n)}\frac{x^{n}}{n!} $$

(1.4)

for $x\in(-1,1)$, where $(a)_{0}=1$ for $a\neq0$, $(a)_{n}=a(a+1)(a+2)\cdots(a+n-1)=\varGamma(a+n)/\varGamma(a)$ is the shifted factorial function and $\varGamma(x)=\int^{\infty}_{0}t^{x-1}e^{-t}\,dt\ (x>0)$ is the classical gamma function [27,28,29,30,31,32,33,34,35].

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 [36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66].

In this article, we focus on the special quasi-arithmetic mean $E(a,b)$ obtained by substituting $p=\sqrt{x}$ and $n=1$ into (1.2), more explicitly,

$$ E(a,b)=M_{\sqrt{x},1}(a,b)= \biggl(\frac{2}{\pi} \int^{\pi/2}_{0}\sqrt {a\cos^{2}\theta+b \sin^{2}\theta}\,d\theta \biggr)^{2}, $$

(1.5)

which can be rewritten in terms of complete elliptic integral of the second kind as

$$ E(a,b)= \textstyle\begin{cases} \frac{4a\mathcal{E}(\sqrt{1-b/a})^{2}}{\pi^{2}},&a\geq b,\\ \frac{4b\mathcal{E}(\sqrt{1-a/b})^{2}}{\pi^{2}},&a< b. \end{cases} $$

(1.6)

Very recently, Meng [67], and Yuan, Yu and Wang [68] proved that the double inequalities

$$\begin{aligned} & \lambda_{1}A(a,b)+(1-\lambda_{1})G(a,b)< E(a,b)< \mu_{1}A(a,b)+(1-\mu _{1})G(a,b), \end{aligned}$$

(1.7)

$$\begin{aligned} & \lambda_{2}C(a,b)+(1-\lambda_{2})H(a,b)< E(a,b)< \mu_{2}C(a,b)+(1-\mu_{2})H(a,b) \end{aligned}$$

(1.8)

hold for $a,b>0$ with $a\neq b$ if and only if $\lambda_{1}\leq3/4$, $\mu_{1}\geq8/\pi^{2}$, $\lambda_{2}\leq4/\pi^{2}$ and $\mu_{2}\geq7/16$, where $A(a,b)=(a+b)/2$, $G(a,b)=\sqrt{ab}$, $H(a,b)=2ab/(a+b)$ and $C(a,b)=(a^{2}+b^{2})/(a+b)$ are the arithmetic, geometric, harmonic and contra-harmonic means of a and b, respectively.

Qian and Chu [69] showed that the double inequality

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

holds for any $p\in[1,\infty)$ and all $a,b>0$ with $a\neq b$ if and only if $\lambda\leq1/2-\sqrt{1-(2\sqrt{2}/\pi)^{4/p}}/2$ and $\mu \geq1/2-\sqrt{p}/(4p)$.

From (1.7) and (1.8) we clearly see that

$$ \frac{3A(a,b)}{4}+\frac{G(a,b)}{4}< E(a,b)< \frac{7C(a,b)}{16}+ \frac {9H(a,b)}{16} $$

(1.9)

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

We define

$$ M_{1}(a,b)=\frac{3A(a,b)+G(a,b)}{4},\qquad M_{2}(a,b)= \frac {7C(a,b)+9H(a,b)}{16} . $$

(1.10)

Motivated by inequality (1.9), it is natural to ask what are the best possible parameters $\alpha_{i},\beta_{i}\in(0,1)$ ($i=1,2$) such that the double inequalities

$$\begin{aligned} &\alpha_{1}M_{2}(a,b)+(1-\alpha_{1})M_{1}(a,b)< E(a,b)< \beta _{1}M_{2}(a,b)+(1-\beta_{1})M_{1}(a,b), \\ &M_{2}(a,b)^{\alpha_{2}}M_{1}(a,b)^{1-\alpha_{2}}< E(a,b)< M_{2}(a,b)^{\beta _{2}}M_{1}(a,b)^{1-\beta_{2}} \end{aligned}$$

hold for all $a,b>0$ with $a\neq b$? The main purpose of this article is to answer this question.

2 Lemmas

In order to prove our main results we need several lemmas, which we present in this section.

Lemma 2.1

(See [16, 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 $g'(x)\neq0$ on $(a, b)$. If $f'(x)/g'(x)$ is increasing (decreasing) on $(a, b)$, then so are the functions

$$ \frac{f(x)-f(a)}{g(x)-g(a)} \quad\textit{and}\quad \frac{f(x)-f(b)}{g(x)-g(b)}. $$

If $f '(x)/g'(x)$ is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2.2

(See [70])

Suppose that the power series $f(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$ and $g(x)=\sum_{n=0}^{\infty}b_{n}x^{n}$ have the radius of convergence $r>0$ with $b_{n}>0$ for all $n\in\{0,1,2,\dots\}$. If the non-constant sequence $\{a_{n}/b_{n}\}_{n=0}^{\infty}$ is increasing (decreasing) for all $n>0$, then $f(x)/g(x)$ is strictly increasing (decreasing) on $(0,r)$.

Lemma 2.3

The following assertions hold true:

• (1) The function $r\rightarrow(\mathcal{E}-r^{\prime 2}\mathcal{K})/r^{2}$ is strictly increasing from $(0,1)$ onto $(\pi/4,1)$;

• (2) The function $r\rightarrow2\mathcal{E}-r^{\prime 2}\mathcal{K}$ is strictly increasing from $(0,1)$ onto $(\pi/2,2)$;

• (3) The function $r\rightarrow[\mathcal{K}-\mathcal{E}-(\mathcal {E}-r^{\prime 2}\mathcal{K})]/r^{4}$ is strictly increasing from $(0,1)$ onto $(\pi/16,+\infty)$.

Proof

Parts (1) and (2) can be found in the literature [16, Theorem 3.21(1) and Exercise 3.43(13)].

For part (3), we clearly see that

$$ \frac{\mathcal{K}-\mathcal{E}-(\mathcal{E}-r^{\prime 2}\mathcal {K})}{r^{4}}=\frac{\mathcal{K}-\mathcal{E}- (\mathcal{E}-r^{\prime 2}\mathcal{K})}{(\mathcal{E}-r^{\prime 2}\mathcal {K})^{2}}\cdot \biggl(\frac{\mathcal{E}-r^{\prime 2}\mathcal{K}}{r^{2}} \biggr)^{2}. $$

Therefore, part (3) follows easily from part (1) and [16, Exercise 3.43(25)]. □

Lemma 2.4

The function

$$ f(r)=\frac{8/\pi^{2}(1+r^{2})(2\mathcal{E}-r^{\prime 2}\mathcal {K})^{2}-(r^{2}+1)(r^{2}+2)}{r^{4}} $$

is strictly increasing from $(0,1)$ onto $(3/16,64/\pi^{2}-6)$.

Proof

Let $f_{1}(r)=8/\pi^{2}(1+r^{2})(2\mathcal{E}-r^{\prime 2}\mathcal {K})^{2}-(r^{2}+1)(r^{2}+2)$ and $f_{2}(r)=r^{4}$, then $f_{1}(0^{+})=f_{2}(0^{+})=0$ and $f(r)=f_{1}(r)/f_{2}(r)$.

A simple calculation yields

$$ \frac{f'_{1}(r)}{f'_{2}(r)}=\frac{f_{11}(r)}{f_{22}(r)}, $$

(2.1)

where

$$\begin{aligned} &f_{11}(r)=16\bigl(2\mathcal{E}-r^{\prime 2} \mathcal{K}\bigr)^{2}+16\bigl(1+r^{2}\bigr) \bigl(2\mathcal {E}-r^{\prime 2}\mathcal{K}\bigr) \bigl(\mathcal{E}-r^{\prime 2} \mathcal{K}\bigr)/r^{2}-\bigl(4r^{2}+6\bigr), \\ &f_{22}(r)=4r^{2}. \end{aligned}$$

Moreover,

$$\begin{aligned} &f_{11}\bigl(0^{+}\bigr)=f_{22}\bigl(0^{+} \bigr)=0, \end{aligned}$$

(2.2)

$$\begin{aligned} &\frac{f'_{11}(r)}{f'_{22}(r)}=8\bigl(2\mathcal{E}-r^{\prime 2} \mathcal{K}\bigr)\frac {\mathcal{E}-r^{\prime 2}\mathcal{K}}{r^{2}}+2\bigl(1+r^{2}\bigr) \biggl( \frac{\mathcal {E}-r^{\prime 2}\mathcal{K}}{r^{2}} \biggr)^{2} \\ &\phantom{\frac{f'_{11}(r)}{f'_{22}(r)}=}{}+2\bigl(1+r^{2}\bigr) \bigl(2\mathcal{E}-r^{\prime 2} \mathcal{K}\bigr)\frac{\mathcal {K}-\mathcal{E}-(\mathcal{E}-r^{\prime 2}\mathcal{K})}{r^{4}}-1. \end{aligned}$$

(2.3)

From Lemma 2.3 and (2.3), we clearly see that $f'_{11}(r)/f'_{22}(r)$ is strictly increasing on $(0,1)$. Equations (2.1)–(2.2) and Lemma 2.1 lead to the conclusion that $f(r)$ is strictly increasing on $(0,1)$.

Therefore, Lemma 2.4 follows from the monotonicity of $f(r)$, together with the facts that $f(0^{+})=3/16$ and $f(1^{-})=64/\pi^{2}-6$. □

Lemma 2.5

The function

$$ g(r)=\frac{(2r^{6}+5r^{4}+5r^{2}+2)[2(\mathcal{E}-r^{\prime 2}\mathcal {K})-r^{2}\mathcal{E}]}{r^{4}(3r^{2}+4)(2\mathcal{E}-r^{\prime 2}\mathcal{K})} $$

is strictly increasing from $(0,1)$ onto $(3/16,1)$.

Proof

Let $g_{1}(r)=(2r^{6}+5r^{4}+5r^{2}+2)[2(\mathcal{E}-r^{\prime 2}\mathcal {K})-r^{2}\mathcal{E}]$ and $g_{2}(r)=r^{4}(3r^{2}+4)(2\mathcal {E}-r^{\prime 2}\mathcal{K})$, then $g(r)=g_{1}(r)/g_{2}(r)$.

Making use of (1.3) and (1.4), we get

$$\begin{aligned} &\frac{2}{\pi}\bigl[2\bigl(\mathcal{E}-r^{\prime 2} \mathcal{K}\bigr)-r^{2}\mathcal{E}\bigr]=\sum _{n=0}^{\infty}\frac{3 (\frac{1}{2},n ) (\frac {1}{2},n+1 )}{2n!(n+2)!}r^{2n+4}, \end{aligned}$$

(2.4)

$$\begin{aligned} & \frac{2}{\pi}\bigl(2\mathcal{E}-r^{\prime 2} \mathcal{K}\bigr)=1+\sum_{n=0}^{\infty } \frac{ (\frac{1}{2},n )^{2}}{4[(n+1)!]^{2}}r^{2n+2}. \end{aligned}$$

(2.5)

It follows from (2.4) and (2.5) that

$$\begin{aligned} \frac{2}{\pi}g_{1}(r)={}&\bigl(2r^{6}+5r^{4}+5r^{2}+2 \bigr)\sum_{n=0}^{\infty}\frac {3 (\frac{1}{2},n ) (\frac{1}{2},n+1 )}{2n!(n+2)!}r^{2n+4} \\ ={}&\sum_{n=0}^{\infty} \frac{3 (\frac{1}{2},n ) (\frac{1}{2},n+1 )}{n!(n+2)!}r^{2n+4}+\sum_{n=0}^{\infty} \frac {15 (\frac{1}{2},n ) (\frac{1}{2},n+1 )}{2n!(n+2)!}r^{2n+6} \\ & {}+\sum_{n=0}^{\infty} \frac{15 (\frac{1}{2},n ) (\frac{1}{2},n+1 )}{2n!(n+2)!}r^{2n+8}+\sum_{n=0}^{\infty } \frac{3 (\frac{1}{2},n ) (\frac{1}{2},n+1 )}{n!(n+2)!}r^{2n+10} \\ ={}&r^{4} \Biggl(\frac{3}{4}+\frac{33}{16}r^{2}+ \frac{1245}{512}r^{4}+\sum_{n=0}^{\infty} \widetilde{A}_{n}r^{2n+6} \Biggr) \\ ={}&r^{4}\sum_{n=0}^{\infty}A_{n}r^{2n} \end{aligned}$$

(2.6)

and

$$\begin{aligned} \frac{2}{\pi}g_{2}(r)&=r^{4} \bigl(3r^{2}+4\bigr) \Biggl(1+\sum_{n=0}^{\infty} \frac { (\frac{1}{2},n )^{2}}{4[(n+1)!]^{2}}r^{2n+2} \Biggr) \\ &=r^{4} \Biggl(4+3r^{2}+\sum _{n=0}^{\infty}\frac{ (\frac {1}{2},n )^{2}}{[(n+1)!]^{2}}r^{2n+2}+\sum _{n=0}^{\infty}\frac {3 (\frac{1}{2},n )^{2}}{4[(n+1)!]^{2}}r^{2n+4} \Biggr) \\ &=r^{4} \Biggl(4+4r^{2}+\frac{13}{16}r^{4}+ \sum_{n=0}^{\infty}\widetilde {B}_{n}r^{2n+6} \Biggr) \\ &=r^{4}\sum_{n=0}^{\infty}B_{n}r^{2n}, \end{aligned}$$

(2.7)

where

$$\begin{aligned} &A_{0}=\frac{3}{4},\qquad A_{1}=\frac{33}{16},\qquad A_{2}=\frac{1245}{512},\qquad A_{n}=\widetilde{A}_{n-3}\quad (n\geq3), \\ &B_{0}=4,\qquad B_{1}=4,\qquad B_{2}=\frac{13}{16},\qquad B_{n}=\widetilde{B}_{n-3}\quad (n\geq3), \\ &\widetilde{A}_{n}=\frac{3(\frac{1}{2},n)(\frac {1}{2},n+1)}{64(n+3)!(n+5)!}\bigl(45{,}765+152{,}928n+192{,}838n^{2} \\ &\phantom{\widetilde{A}_{n}=}{}+120{,}672n^{3}+40{,}024n^{4}+6720n^{5}+448n^{6} \bigr), \\ &\widetilde{B}_{n}=\frac{(\frac{1}{2},n+1)^{2}(7n^{2}+30n+36)}{[4(n+3)!]^{2}} \end{aligned}$$

for $n\geq0$.

It follows from (2.6) and (2.7) that

$$ g(r)=\frac{\sum_{n=0}^{\infty}A_{n}r^{2n}}{\sum_{n=0}^{\infty}B_{n}r^{2n}} $$

(2.8)

for $r\in(0,1)$.

In order to prove the monotonicity of $g(r)$, Lemma 2.2 and (2.8) enable us to conclude that it suffices to show the monotonicity of $\{A_{n}/B_{n}\}_{n=0}^{\infty}$.

A simple calculation leads to

$$ \frac{A_{0}}{B_{0}}=\frac{3}{16},\qquad \frac{A_{1}}{B_{1}}= \frac{33}{64},\qquad \frac {A_{2}}{B_{2}}=\frac{1245}{416} ,\qquad \frac{A_{3}}{B_{3}}= \frac{3051}{128} $$

(2.9)

and

$$\begin{aligned} &\frac{A_{n+3}}{B_{n+3}}=\frac{\widetilde{A}_{n}}{\widetilde {B}_{n}}=\frac{3}{8(n+4)(n+5)(2n+1)(36+30n+7 n^{2})} \bigl(45{,}765+152{,}928n \\ &\phantom{\frac{A_{n+3}}{B_{n+3}}=\frac{\widetilde{A}_{n}}{\widetilde {B}_{n}}=}{}+192{,}838n^{2}+120{,}672n^{3}+40{,}024n^{4}+ 6720n^{5}+448 n^{6}\bigr), \\ &\frac{\widetilde{A}_{n+1}}{\widetilde{B}_{n+1}}-\frac{\widetilde {A}_{n}}{\widetilde{B}_{n}}=\frac{3\Delta_{1}(n)}{8\Delta_{2}(n)}>0, \end{aligned}$$

(2.10)

for $n\geq0$, where

$$\begin{aligned} \Delta_{1}(n)={}&20{,}417{,}670 +119{,}034{,}009 n + 234{,}552{,}870 n^{2} \\ &{}+ 238{,}084{,}434 n^{3} +144{,}127{,}820 n^{4} \\ &{} + 55{,}145{,}420 n^{5} + 13{,}474{,}832 n^{6} + 2{,}036{,}720 n^{7} + 172{,}928 n^{8} + 6272 n^{9}, \\ \Delta_{2}(n)={}&(n+4) (n+5) (n+6) (2n+1) (2n+3) \bigl(36+30n+7n^{2} \bigr) \bigl(73+44n+7n^{2}\bigr). \end{aligned}$$

It follows from Lemma 2.2 and (2.8)–(2.10) that $g(r)$ is strictly increasing on $(0,1)$. Therefore, Lemma 2.5 follows easily from the monotonicity of $g(r)$, together with the facts that $g(0^{+})=A_{0}/B_{0}=3/16$ and $g(1^{-})=1$. □

3 Main results

Theorem 3.1

The double inequality

$$ \alpha_{1}M_{2}(a,b)+(1-\alpha_{1})M_{1}(a,b)< E(a,b)< \beta _{1}M_{2}(a,b)+(1-\beta_{1})M_{1}(a,b) $$

holds for $a,b>0$ with $a\neq b$ if and only if $\alpha_{1}\leq3/16$ and $\beta_{1}\geq64/\pi^{2}-6$.

Proof

Since $M_{1}(a,b), M_{2}(a,b)$ and $E(a,b)$ are symmetric and homogeneous of degree one, without loss of generality, we assume that $a>b>0$. Let $r=(1-\sqrt{b/a})/(1+\sqrt{b/a})\in(0,1)$, then (1.6) and (1.10), together with Landen identities, lead to

$$\begin{aligned} & E(a,b)=A(a,b)\frac{4(1+r)^{2}}{\pi^{2}(1+r^{2})}\mathcal{E}^{2} \biggl( \frac {2\sqrt{r}}{1+r} \biggr)=A(a,b)\frac{4}{\pi^{2}}\frac{(2\mathcal {E}-r^{\prime 2}\mathcal{K})^{2}}{1+r^{2}}, \end{aligned}$$

(3.1)

$$\begin{aligned} & M_{1}(a,b)=A(a,b)\frac{r^{2}+2}{2(1+r^{2})},\qquad M_{2}(a,b)=A(a,b)\frac {2+3r^{2}+2r^{4}}{2(1+r^{2})^{2}} \end{aligned}$$

(3.2)

and

$$\begin{aligned} &E(a,b)-pM_{2}(a,b)-(1-p)M_{1}(a,b) \\ &\quad=A(a,b) \biggl[\frac{4}{\pi^{2}}\frac{(2\mathcal{E}-r^{\prime 2}\mathcal {K})^{2}}{1+r^{2}}-p \frac{2+3r^{2}+2r^{4}}{2(1+r^{2})^{2}}-(1-p)\frac {r^{2}+2}{2(1+r^{2})} \biggr] \\ &\quad =\frac{A(a,b)r^{4}}{2(1+r^{2})^{2}}\bigl[f(r)-p\bigr], \end{aligned}$$

(3.3)

where $f(r)$ is defined as in Lemma 2.4.

Therefore, Theorem 3.1 follows from Lemma 2.4 and (3.3) immediately. □

Theorem 3.2

The double inequality

$$ M_{2}(a,b)^{\alpha_{2}}M_{1}(a,b)^{1-\alpha_{2}}< E(a,b)< M_{2}(a,b)^{\beta _{2}}M_{1}(a,b)^{1-\beta_{2}} $$

holds for $a,b>0$ with $a\neq b$ if and only if $\alpha_{2}\leq3/16$ and $\beta_{2}\geq\log[32/(3\pi^{2})]/\log(7/6)$.

Proof

Without loss of generality, we may assume that $a>b>0$. Let $r=(1-\sqrt {b/a})/(1+\sqrt{b/a})\in(0,1)$, then (3.1) and (3.2) lead to

$$\begin{aligned} &\log E(a,b)-\lambda\log M_{2}(a,b)-(1-\lambda)\log M_{1}(a,b) \\ &\quad=\log\frac{8}{\pi^{2}}+\log\frac{(2\mathcal{E}-r^{\prime 2}\mathcal {K})^{2}}{r^{2}+2}-\lambda\log \frac{2r^{4}+3r^{2}+2}{(r^{2}+1)(r^{2}+2)} \\ &\quad\triangleq\varphi(r). \end{aligned}$$

(3.4)

Elaborated computations lead to

$$\begin{aligned} & \varphi(0)=0,\qquad \varphi(1)=\log\frac{32}{3\pi^{2}}-\lambda\log \frac {7}{6}, \end{aligned}$$

(3.5)

$$\begin{aligned} & \varphi'(r)=\frac {2r(3r^{2}+4)}{(r^{2}+1)(r^{2}+2)(2r^{4}+3r^{2}+2)}\bigl[g(r)-\lambda \bigr], \end{aligned}$$

(3.6)

where $g(r)$ is defined as in Lemma 2.5.

We divide the proof into three cases.

Case 1. $\lambda_{1}=3/16$. We clearly see from Lemma 2.5 that

$$ g(r)>\lambda_{1} $$

(3.7)

for $r\in(0,1)$. It follows from (3.5)–(3.7) that $\varphi(r)>0$ for $r\in(0,1)$. This, in conjunction with (3.4), yields

$$ E(a,b)>M_{2}(a,b)^{\lambda_{1}}M_{1}(a,b)^{1-\lambda_{1}} $$

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

Case 2. $\lambda_{2}=\log[32/(3\pi^{2})]/\log(7/6)$. It follows from Lemma 2.5 that there exists $\delta\in(0,1)$ such that $g(r)<\lambda_{2}$ for $r\in(0,\delta)$ and $g(r)>\lambda_{2}$ for $r\in(\delta,1)$. This, in conjunction with (3.6), implies that $\varphi(r)$ is strictly decreasing on $(0,\delta)$ and is strictly increasing on $(\delta,1)$. Moreover, we clearly see from (3.5) that

$$ \varphi(0)=\varphi(1)=0. $$

(3.8)

The piecewise monotonicity property of $g(r)$ and (3.8) lead to the conclusion that $\varphi(r)<0$ for $r\in(0,1)$. Therefore,

$$ E(a,b)< M_{2}(a,b)^{\lambda_{2}}M_{1}(a,b)^{1-\lambda_{2}} $$

for all $a,b>0$ with $a\neq b$ follows from (3.4).

Case 3. $3/16<\lambda_{3}<\log[32/(3\pi^{2})]/\log(7/6)$. By the locally sign-preserving property of limit, Lemma 2.5 and (3.6) enable us to know that there exists $\tau_{1}\in(0,1)$ such that $\varphi(r)$ is strictly decreasing on $(0,\tau_{1})$. This, in conjunction with (3.5), implies that $\varphi(r)<0$ for $0< r<\tau_{1}$. Therefore,

$$ E(a,b)< M_{2}(a,b)^{\lambda_{3}}M_{1}(a,b)^{1-\lambda_{3}} $$

for $b< a<[(1+\tau_{1})/(1-\tau_{1})]^{2}b$ follows from (3.4).

On the other hand, we clearly see from (3.5) that $\varphi (1)>0$. This, in conjunction with the continuity of $\varphi(r)$, implies that there exists $\tau_{2}\in(0,1)$ such that $\varphi(r)>0$ for $\tau_{2}< r<1$. Therefore, it follows from (3.4) that

$$ E(a,b)>M_{2}(a,b)^{\lambda_{3}}M_{1}(a,b)^{1-\lambda_{3}} $$

for $a>[(1+\tau_{2})/(1-\tau_{2})]^{2}b$. □

Let $a=1$ and $b=1-r^{2}=r^{\prime 2}$, then (1.6), and Theorems 3.1 and 3.2 give rise to Corollary 3.3 immediately.

Corollary 3.3

The double inequalities

$$\begin{aligned} &\frac{3(7+18r^{\prime 2}+7r^{\prime 4})}{256(1+r^{\prime 2})}+\frac{13(1+6 r' + r^{\prime 2})}{128}\\ &\quad < \mathcal{E}(r) \\ &\quad < \frac{(64-6\pi^{2})(7+18r^{\prime 2}+7r^{\prime 4})}{16\pi^{2}(1+r^{\prime 2})}+\frac{(7\pi ^{2}-64)(1+6 r' + r^{\prime 2})}{8\pi^{2}}, \\ &\biggl[\frac{7 + 18 r^{\prime 2} + 7 r^{\prime 4}}{16(1 + r^{\prime 2})} \biggr]^{3/16} \biggl(\frac{1+6r'+r^{\prime 2}}{8} \biggr)^{13/16}\\ &\quad< \mathcal{E}(r) \\ &\quad< \biggl[\frac{7 + 18 r^{\prime 2} + 7 r^{\prime 4}}{16(1 + r^{\prime 2})} \biggr]^{\frac {\log32/(3\pi^{2})}{\log(7/6)}} \biggl(\frac{1+6r'+r^{\prime 2}}{8} \biggr)^{\frac{\log(7\pi^{2}/64)}{\log(7/6)}} \end{aligned}$$

hold for all $r\in(0,1)$.

4 Results and discussion

In this article, we find the best possible parameters $\alpha_{1}$, $\beta_{1}$, $\alpha_{2}$ and $\beta_{2}$ on the interval $(0, 1)$ such that the double inequalities

$$\begin{aligned} &\alpha_{1} \biggl[\frac{7C(a,b)}{16}+\frac{9H(a,b)}{16} \biggr]+(1- \alpha_{1}) \biggl[\frac{3A(a,b)}{4}+\frac{G(a, b)}{4} \biggr]\\ &\quad< E(a,b) \\ &\quad< \beta_{1} \biggl[\frac{7C(a,b)}{16}+\frac{9H(a,b)}{16} \biggr]+(1- \beta_{1}) \biggl[\frac{3A(a,b)}{4}+\frac{G(a, b)}{4} \biggr], \\ &\biggl[\frac{7C(a,b)}{16}+\frac{9H(a,b)}{16} \biggr]^{\alpha _{2}} \biggl[ \frac{3A(a,b)}{4}+\frac{G(a, b)}{4} \biggr]^{1-\alpha_{2}}\\ &\quad< E(a,b) \\ &\quad < \biggl[\frac{7C(a,b)}{16}+\frac{9H(a,b)}{16} \biggr]^{\beta _{2}} \biggl[ \frac{3A(a,b)}{4}+\frac{G(a, b)}{4} \biggr]^{1-\beta_{2}} \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$. Our results improve and refine the results given in [67, 68].

5 Conclusion

We present several sharp bounds for the quasi-arithmetic mean in terms of the combination of harmonic, geometric, arithmetic and contra-harmonic means. Our approach may have further applications in the theory of bivariate means.