Optimal evaluation of a Toader-type mean by power mean

Abstract

In this paper, we present the best possible parameters $p, q\in\mathbb {R}$ such that the double inequality $M_{p}(a,b)< T[A(a,b), Q(a,b)]< M_{q}(a,b)$ holds for all $a, b>0$ with $a\neq b$, and we get sharp bounds for the complete elliptic integral $\mathcal{E}(t)=\int _{0}^{\pi/2}(1-t^{2}\sin^{2}\theta)^{1/2}\,d\theta$ of the second kind on the interval $(0, \sqrt{2}/2)$, where $T(a,b)=\frac{2}{\pi }\int _{0}^{\pi/2}\sqrt{a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}\,d\theta$, $A(a,b)=(a+b)/2$, $Q(a,b)=\sqrt{(a^{2}+b^{2})/2}$, $M_{r}(a,b)=[(a^{r}+b^{r})/2]^{1/r}$ ($r\neq0$), and $M_{0}(a,b)=\sqrt {ab}$ are the Toader, arithmetic, quadratic, and rth power means of a and b, respectively.

1 Introduction

For $r\in\mathbb{R}$ and $a,b>0$, the Toader mean $T(a,b)$ (see [1]) and rth power mean $M_{r}(a, b)$ are defined by

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

(1.1)

and

$$ M_{r}(a,b)= \textstyle\begin{cases} (\frac{a^{r}+b^{r}}{2} )^{1/r}, & r\neq0 ,\\ \sqrt{ab}, & r=0, \end{cases} $$

(1.2)

respectively.

It is well known that $M_{r}(a, b)$ is continuous and strictly increasing with respect to $r\in\mathbb{R}$ for fixed $a, b>0$ with $a\neq b$. Many classical bivariate means are a special case of the power mean, for example, $H(a,b)=2ab/(a+b)=M_{-1}(a,b)$ is the harmonic mean, $G(a,b)=\sqrt{ab}=M_{0}(a,b)$ is the geometric mean,

$$ A(a,b)=(a+b)/2=M_{1}(a,b) $$

(1.3)

is the arithmetic mean, and

$$ Q(a,b)=\sqrt{\bigl(a^{2}+b^{2}\bigr)/2}=M_{2}(a,b) $$

(1.4)

is the quadratic mean. The main properties of the power mean are given in [2]. The Toader mean $T(a,b)$ has been well known in the mathematical literature for many years, it satisfies

$$ T(a,b)=R_{E} \bigl(a^{2}, b^{2} \bigr), $$

where

$$ R_{E}(a,b)=\frac{1}{\pi} \int_{0}^{\infty}\frac {[a(t+b)+b(t+a)]t}{(t+a)^{3/2}(t+b)^{3/2}}\,dt $$

stands for the symmetric complete elliptic integral of the second kind (see [3–5]), therefore it cannot be expressed in terms of the elementary transcendental functions.

Let $r\in(0, 1)$, $\mathcal{K}(r)=\int_{0}^{\pi /2}(1-r^{2}\sin ^{2}\theta)^{-1/2}\,d\theta$, and $\mathcal{E}(r)=\int _{0}^{\pi /2}(1-r^{2}\sin^{2}\theta)^{1/2}\,d\theta$ be, respectively, the complete elliptic integrals of the first and second kind. Then $\mathcal{K}(0^{+})=\mathcal{E}(0^{+})=\pi/2$, the Toader mean $T(a,b)$ given in (1.1) can be expressed as

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

(1.5)

and $\mathcal{K}(r)$ and $\mathcal{E}(r)$ satisfy the derivatives formulas (see [6], Appendix E, p. 474-475)

$$ \frac{d\mathcal{K}(r)}{dr}=\frac{\mathcal{E}(r)-(1-r^{2})\mathcal {K}(r)}{r(1-r^{2})}, \qquad \frac{d\mathcal{E}(r)}{dr}=\frac{\mathcal {E}(r)-\mathcal{K}(r)}{r}, \qquad \frac{d(\mathcal{K}(r)-\mathcal {E}(r))}{dr}=\frac{r\mathcal{E}(r)}{1-r^{2}}. $$

Numerical computations show that

$$\begin{aligned}& \mathcal{E} \biggl(\frac{\sqrt{2}}{2} \biggr)=1.3506\ldots,\qquad \mathcal {K} \biggl(\frac{3}{5} \biggr)=1.7507\ldots, \qquad \mathcal{E} \biggl(\frac{3}{5} \biggr)=1.4180\ldots, \\& \mathcal{K} \biggl(\frac{17}{25} \biggr)=1.8234\ldots, \qquad \mathcal {E} \biggl( \frac{17}{25} \biggr)=1.3693\ldots. \end{aligned}$$

Recently, the power mean $M_{r}(a,b)$ and Toader mean $T(a,b)$ have been the subject of intensive research. In particular, many remarkable inequalities for both means can be found in the literature [7–18].

Vuorinen [19] conjectured that the inequality

$$ M_{3/2}(a,b)< T(a,b) $$

holds for all $a, b>0$ with $a\neq b$. This conjecture was proved by Qiu and Shen [20], and Barnard et al. [21], respectively.

Alzer and Qiu [22] presented a best possible upper power mean bound for the Toader mean as follows:

$$ T(a,b)< M_{\log2/(\log\pi-\log2)}(a,b) $$

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

Neuman [3], and Kazi and Neuman [4] proved that the inequalities

$$\begin{aligned}& \frac{(a+b)\sqrt{ab}-ab}{AGM(a,b)}< T(a,b)< \frac{4(a+b)\sqrt {ab}+(a-b)^{2}}{8AGM(a,b)}, \\& T(a,b)< \frac{1}{4} \bigl(\sqrt{(2+\sqrt{2})a^{2}+(2-\sqrt {2})b^{2}}+\sqrt {(2+\sqrt{2})b^{2}+(2-\sqrt{2})a^{2}} \bigr) \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$, where $AGM(a,b)$ is the arithmetic-geometric mean of a and b.

Let $\lambda, \mu, \alpha, \beta\in(1/2, 1)$. Then Chu et al. [23], and Hua and Qi [24] proved that the double inequalities

$$\begin{aligned}& C\bigl[\lambda a+(1-\lambda)b, \lambda b+(1-\lambda)a\bigr]< T(a,b)< C\bigl[\mu a+(1-\mu )b, \mu b+(1-\mu)a\bigr], \\& \overline{C}\bigl[\alpha a+(1-\alpha)b, \alpha b+(1-\alpha )a\bigr]< T(a,b)< \overline{C}\bigl[\beta a+(1-\beta)b, \beta b+(1-\beta)a\bigr] \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$ if and only if $\lambda\leq3/4$, $\mu\geq1/2+\sqrt{\pi(4-\pi)}/(2\pi)$, $\alpha\leq1/2+\sqrt{3}/4$, and $\beta\geq1/2+\sqrt{12/\pi-3}/2$, where $C(a,b)=(a^{2}+b^{2})/(a+b)$ and $\overline {C}(a,b)=2(a^{2}+ab+b^{2})/[3(a+b)]$ are, respectively, the contraharmonic and centroidal means of a and b.

In [25–29], the authors proved that the double inequalities

$$\begin{aligned}& \alpha_{1}Q(a,b)+(1-\alpha_{1})A(a,b)< T(a,b)< \beta _{1}Q(a,b)+(1-\beta _{1})A(a,b), \\& Q^{\alpha_{2}}(a,b)A^{(1-\alpha_{2})}(a,b)< T(a,b)< Q^{\beta _{2}}(a,b)A^{(1-\beta_{2})}(a,b), \\& \alpha_{3}C(a,b)+(1-\alpha_{3})A(a,b)< T(a,b)< \beta _{3}C(a,b)+(1-\beta _{3})A(a,b), \\& \frac{\alpha_{4}}{A(a,b)}+\frac{1-\alpha_{4}}{C(a,b)}< \frac {1}{T(a,b)}< \frac{\beta_{4}}{A(a,b)}+ \frac{1-\beta_{4}}{C(a,b)}, \\& \alpha_{5}C(a,b)+(1-\alpha_{5})H(a,b)< T(a,b)< \beta _{5}C(a,b)+(1-\beta _{5})H(a,b), \\& \alpha_{6}\bigl[C(a,b)-H(a,b)\bigr]+A(a,b)< T(a,b)< \beta_{6}\bigl[C(a,b)-H(a,b)\bigr]+A(a,b), \\& \alpha_{7}\overline{C}(a,b)+(1-\alpha_{7})A(a,b)< T(a,b)< \beta _{7}\overline{C}(a,b)+(1-\beta_{7})A(a,b), \\& \frac{\alpha_{8}}{A(a,b)}+\frac{1-\alpha_{8}}{\overline {C}(a,b)}< \frac {1}{T(a,b)}< \frac{\beta_{8}}{A(a,b)}+ \frac{1-\beta_{8}}{\overline{C}(a,b)}, \\& \alpha_{9}Q(a,b)+(1-\alpha_{9})H(a,b)< T(a,b)< \beta _{9}Q(a,b)+(1-\beta _{9})H(a,b), \\& \frac{\alpha_{10}}{H(a,b)}+\frac{1-\alpha_{10}}{Q(a,b)}< \frac {1}{T(a,b)}< \frac{\beta_{10}}{H(a,b)}+ \frac{1-\beta_{10}}{Q(a,b)} \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$ if and only if $\alpha_{1}\leq 1/2$, $\beta_{1}\geq(4-\pi)/[(\sqrt{2}-1)\pi]$, $\alpha_{2}\leq1/2$, $\beta_{2}\geq4-2\log\pi/\log2$, $\alpha _{3}\leq1/4$, $\beta_{3}\geq4/\pi-1$, $\alpha_{4}\leq\pi/2-1$, $\beta_{4}\geq3/4$, $\alpha_{5}\leq5/8$, $\beta_{5}\geq2/\pi$, $\alpha_{6}\leq1/8$, $\beta_{6}\geq2/\pi-1/2$, $\alpha_{7}\leq3/4$, $\beta_{7}\geq12/\pi-3$, $\alpha_{8}\leq\pi-3$, $\beta_{8}\geq1/4$, $\alpha_{9}\leq5/6$, $\beta_{9}\geq2\sqrt {2}/\pi$, $\alpha_{10}\leq0$, and $\beta_{10}\geq1/6$.

The main purpose of this paper is to present the best possible parameters $p, q\in\mathbb{R}$ such that the double inequality

$$ M_{p}(a,b)< T\bigl[A(a,b), Q(a,b)\bigr]< M_{q}(a,b) $$

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

2 Lemmas

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

Lemma 2.1

(See [30], Theorem 1.1)

The inequality $\mathcal {E} [M_{p}(x, y) ]>M_{q}[\mathcal{E}(x), \mathcal{E}(y)]$ holds for all $x, y\in(0, 1)$ if and only if

$$ p\leq C(q):=\inf_{r\in(0, 1)} \biggl\{ \frac{r^{2}\mathcal {E}(r)}{(1-r^{2})[\mathcal{K}(r)-\mathcal{E}(r)]}+ \frac {(1-q)[\mathcal {K}(r)-\mathcal{E}(r)]}{\mathcal{E}(r)} \biggr\} , $$

where $q\rightarrow C(q)$ is a continuous function which satisfies $C(q)=2$ for all $q\leq5/2$ and $C(q)<2$ for all $q>5/2$.

Lemma 2.2

The double inequality

$$ \frac{(1-t^{2})^{5/8}+1}{(1-t^{2})^{1/8}+1}< 1-\frac{t^{2}}{4}< \biggl[\frac{(\sqrt{1-t^{2}}+t)^{3/2}+(\sqrt{1-t^{2}}-t)^{3/2}}{2} \biggr]^{2/3} $$

(2.1)

holds for all $t\in(0, \sqrt{2}/2)$.

Proof

Let $u=(1-t^{2})^{1/8}$. Then $u\in(1/\sqrt[8]{2}, 1)$, $t^{2}=1-u^{8}$, and the first inequality of (2.1) is equivalent to

$$ \frac{u^{5}+1}{u+1}< \frac{u^{8}+3}{4} $$

(2.2)

for all $u\in(1/\sqrt[8]{2}, 1)$.

We clearly see that (2.2) follows from

$$ (u+1) \bigl(u^{8}+3\bigr)-4\bigl(u^{5}+1\bigr)=(u+1) \bigl(u^{2}+1\bigr) (1-u)^{2}\bigl[(2u-1)+2u^{2}+2u^{3}+u^{4} \bigr]>0 $$

for all $u\in(1/\sqrt[8]{2}, 1)$.

For the second inequality of (2.1), let $v=\sqrt{1-t^{2}}\in(\sqrt {2}/2, 1)$, then it suffices to prove that

$$\begin{aligned} \rho(v)&:=\frac{ [(v+\sqrt{1-v^{2}})^{3/2}+(v-\sqrt {1-v^{2}})^{3/2} ]^{2}}{4}-\frac{(v^{2}+3)^{3}}{64} \\ &=\frac{1}{2} \biggl[3v-2v^{3}+\bigl(2v^{2}-1 \bigr)^{3/2}-\frac {(v^{2}+3)^{3}}{32} \biggr]>0 \end{aligned}$$

(2.3)

for all $v\in(\sqrt{2}/2, 1)$.

We claim that

$$ \bigl(2v^{2}-1\bigr)^{3/2}>2-6v+3v^{2}+2v^{3} $$

(2.4)

for all $v\in(\sqrt{2}/2, 1)$.

Indeed, if $v\in(\sqrt{2}/2, (\sqrt{6}-1)/2]$, then we clearly see that the function $2-6v+3v^{2}+2v^{3}$ is strictly increasing on $(\sqrt{2}/2, (\sqrt{6}-1)/2]$, and (2.4) follows from

$$\begin{aligned} 2-6v+3v^{2}+2v^{3}&\leq2-6\times\frac{\sqrt{6}-1}{2}+3\times \biggl(\frac {\sqrt{6}-1}{2} \biggr)^{2}+2\times \biggl(\frac{\sqrt {6}-1}{2} \biggr)^{3} \\ &=\frac{22-9\sqrt{6}}{4}< 0. \end{aligned} $$

If $v\in((\sqrt{6}-1)/2, 1)$, then (2.4) follows easily from

$$\begin{aligned} \bigl(2v^{2}-1 \bigr)^{3}- \bigl(2-6v+3v^{2}+2v^{3} \bigr)^{2}&=\bigl(1-v^{4}\bigr) \bigl(-5+4v+4v^{2} \bigr) \\ &>\bigl(1-v^{4}\bigr) \biggl[-5+4\times\frac{\sqrt{6}-1}{2}+4 \biggl( \frac {\sqrt {6}-1}{2} \biggr)^{2} \biggr]=0. \end{aligned}$$

Therefore, inequality (2.3) follows from (2.4) and

$$\begin{aligned} 3v-2v^{3}+\bigl(2v^{2}-1\bigr)^{3/2}- \frac{(v^{2}+3)^{3}}{32} & >3v-2v^{3}+\bigl(2-6v+3v^{2}+2v^{3}\bigr)- \frac {(v^{2}+3)^{3}}{32} \\ & =\frac{(1-v^{3})(37+15v+3v^{2}+v^{3})}{32}>0 \end{aligned}$$

for all $v\in(\sqrt{2}/2, 1)$. □

Lemma 2.3

The inequality

$$ \mathcal{E}(t)>\frac{\pi}{2} \biggl(1-\frac{5t^{2}}{18} \biggr) $$

holds for all $t\in(0, 3/5)$.

Proof

Let

$$ f(t)=\mathcal{E}(t)-\frac{\pi}{2} \biggl(1-\frac{5t^{2}}{18} \biggr). $$

(2.5)

Then simple computations lead to

$$\begin{aligned}& f\bigl(0^{+}\bigr) =0, \qquad f \biggl(\frac{3}{5} \biggr)=0.00436 \ldots>0, \end{aligned}$$

(2.6)

$$\begin{aligned}& f'(t) =tf_{1}(t), \end{aligned}$$

(2.7)

where

$$\begin{aligned}& f_{1}(t) =\frac{\mathcal{E}(t)-\mathcal{K}(t)}{t^{2}}+\frac{5\pi }{18}, \\& f_{1}\bigl(0^{+}\bigr) =\frac{\pi}{36}>0,\qquad f_{1} \biggl(\frac{3}{5} \biggr)=-0.0514\ldots< 0, \end{aligned}$$

(2.8)

$$\begin{aligned}& f'_{1}(t) =\frac{f_{2}(t)}{t^{3}(1-t^{2})}, \end{aligned}$$

(2.9)

where

$$\begin{aligned}& f_{2}(t) =2\bigl(1-t^{2}\bigr)\mathcal{K}(t)- \bigl(2-t^{2}\bigr)\mathcal{E}(t), \\& f_{2}\bigl(0^{+}\bigr) =0, \end{aligned}$$

(2.10)

$$\begin{aligned}& f'_{2}(t) =-3t\bigl[\mathcal{K}(t)-\mathcal{E}(t) \bigr]< 0 \end{aligned}$$

(2.11)

for $t\in(0, 3/5)$.

From (2.9)-(2.11) we clearly see that $f_{1}(t)$ is strictly decreasing on $(0, 3/5)$. Then (2.7) and (2.8) lead to the conclusion that there exists $t_{0}\in(0, 3/5)$ such that $f(t)$ is strictly increasing on $(0, t_{0}]$ and strictly decreasing on $[t_{0}, 3/5)$.

Therefore, Lemma 2.3 follows easily from (2.5) and (2.6) together with the piecewise monotonicity of $f(t)$. □

Lemma 2.4

The inequality

$$ \biggl(\frac{18+13t^{2}}{18\sqrt{1+t^{2}}} \biggr)^{7/5}>1+\frac{14t^{2}}{45} $$

holds for all $t\in(0, 3/4)$.

Proof

It suffices to prove that the inequalities

$$ \frac{18+13t^{2}}{18\sqrt{1+t^{2}}}>1+\frac{2t^{2}}{9} $$

(2.12)

and

$$ \biggl(1+\frac{2t^{2}}{9} \biggr)^{7/5}>1+\frac{14t^{2}}{45} $$

(2.13)

hold for all $t\in(0, 3/4)$.

Indeed, inequalities (2.12) and (2.13) follow easily from the identities

$$ \bigl(18+13t^{2} \bigr)^{2}-4\bigl(1+t^{2}\bigr) \bigl(9+2t^{2}\bigr)^{2}=t^{4}(3-4t) (3+4t) $$

and

$$\begin{aligned} & \biggl(1+\frac{2t^{2}}{9} \biggr)^{7}- \biggl(1+ \frac {14t^{2}}{45} \biggr)^{5} \\ &\quad =4t^{2} \biggl(\frac{7}{405}+\frac{14t^{2}}{675}+ \frac {2\mbox{,}632t^{4}}{273\mbox{,}375}+\frac{390\mbox{,}544t^{6}}{184\mbox{,}528\mbox{,}125}+\frac {112t^{8}}{531\mbox{,}441}+\frac{32t^{10}}{4\mbox{,}782\mbox{,}969} \biggr). \end{aligned}$$

 □

Lemma 2.5

Let $\lambda=2\log2/[2\log\pi-\log2-2\log\mathcal {E}(\sqrt{2}/2)]=1.3930\ldots$ and

$$ g(t)=\frac{2}{\pi}\sqrt{1+t^{2}}\mathcal{E} \biggl( \frac{t}{\sqrt {1+t^{2}}} \biggr)- \biggl[\frac{(1+t)^{\lambda}+(1-t)^{\lambda }}{2} \biggr]^{1/\lambda}. $$

Then $g(t)>0$ for all $t\in(0, 3/4)$.

Proof

It follows from $t/\sqrt{1+t^{2}}\in(0, 3/5)$, $\lambda<7/5$, Lemma 2.3, Lemma 2.4 and the monotonicity of $M_{r}(1+t, 1-t)$ with respect to $r\in\mathbb{R}$ that

$$\begin{aligned} \begin{aligned}[b] g(t)&>\frac{2}{\pi}\sqrt{1+t^{2}}\times \frac{\pi}{2} \biggl[1-\frac {5t^{2}}{18(1+t^{2})} \biggr]- \biggl[\frac {(1+t)^{7/5}+(1-t)^{7/5}}{2} \biggr]^{5/7} \\ &=\frac{18+13t^{2}}{18\sqrt{1+t^{2}}}- \biggl[\frac {(1+t)^{7/5}+(1-t)^{7/5}}{2} \biggr]^{5/7} \\ &> \biggl(1+\frac{14t^{2}}{45} \biggr)^{5/7}- \biggl[\frac {(1+t)^{7/5}+(1-t)^{7/5}}{2} \biggr]^{5/7} \end{aligned} \end{aligned}$$

(2.14)

for $t\in(0, 3/4)$. Let

$$ g_{1}(t)=2 \biggl(1+\frac{14t^{2}}{45} \biggr)- \bigl[(1+t)^{7/5}+(1-t)^{7/5} \bigr]. $$

(2.15)

Then simple computations lead to

$$\begin{aligned}& g_{1}(0) =0, \qquad g_{1} \biggl(\frac{3}{4} \biggr)=0.0173\ldots>0, \end{aligned}$$

(2.16)

$$\begin{aligned}& g'_{1}(t) =\frac{7}{45} \bigl[8t-9(1+t)^{2/5}+9(1-t)^{2/5} \bigr], \\& g'_{1}(0) =0,\qquad g'_{1} \biggl( \frac{3}{4} \biggr)=-0.0138\ldots < 0, \end{aligned}$$

(2.17)

$$\begin{aligned}& g''_{1}(t) =\frac{14}{225} \biggl[20-\frac{9}{(1+t)^{3/5}}-\frac {9}{(1-t)^{3/5}} \biggr], \\& g''_{1}(0) =\frac{28}{225}>0,\qquad g''_{1} \biggl(\frac{3}{4} \biggr)=-0.4423\ldots< 0, \end{aligned}$$

(2.18)

$$\begin{aligned}& g'''_{1}(t) = \frac{42}{125} \biggl[\frac{1}{(1+t)^{8/5}}-\frac {1}{(1-t)^{8/5}} \biggr]< 0 \end{aligned}$$

(2.19)

for $t\in(0, 3/4)$.

From (2.18) and (2.19) we know that there exists $t_{1}\in(0, 3/4)$ such that $g^{\prime}_{1}(t)$ is strictly increasing on $(0, t_{1}]$ and strictly decreasing on $[t_{1}, 3/4)$. Then (2.17) leads to the conclusion that there exists $t_{2}\in(0, 3/4)$ such that $g_{1}(t)$ is strictly increasing on $(0, t_{2}]$ and strictly decreasing on $[t_{2}, 3/4)$.

Therefore, Lemma 2.5 follows from (2.14)-(2.16) and the piecewise monotonicity of $g_{1}(t)$. □

Lemma 2.6

Let $\lambda=2\log2/[2\log\pi-\log2-2\log\mathcal {E}(\sqrt{2}/2)]=1.3930\ldots$ . Then the function $t^{-1}\mathcal {E}^{\lambda-1}(t)[\mathcal{E}(t)-\mathcal{K}(t)]$ is strictly decreasing on $(0, 1)$.

Proof

From Lemma 2.1 we clearly see that the inequality

$$ \mathcal{E}\bigl(M_{\lambda}(x,y)\bigr)>M_{\lambda}\bigl( \mathcal{E}(x),\mathcal {E}(y)\bigr)= \biggl(\frac{\mathcal{E}^{\lambda}(x)+\mathcal{E}^{\lambda }(y)}{2} \biggr)^{1/\lambda} $$

(2.20)

holds for all $x, y\in(0, 1)$ with $x\neq y$.

It follows from the monotonicity of the function $\mathcal{E}(t)$ and the power mean $M_{p}(x, y)$ with respect to $p\in\mathbb{R}$ together with $\lambda>1$ that

$$ \mathcal{E} \biggl(\frac{x+y}{2} \biggr)=\mathcal {E}\bigl(M_{1}(x,y) \bigr)>\mathcal {E}\bigl(M_{\lambda}(x,y)\bigr) $$

(2.21)

for all $x, y\in(0, 1)$ with $x\neq y$.

Inequalities (2.20) and (2.21) lead to

$$ \mathcal{E}^{\lambda} \biggl(\frac{x+y}{2} \biggr)>\frac{\mathcal {E}^{\lambda}(x)+\mathcal{E}^{\lambda}(y)}{2} $$

for all $x, y\in(0, 1)$ with $x\neq y$, which implies that the function $\mathcal{E}^{\lambda}(t)$ is strictly concave on $(0, 1)$.

Note that

$$ t^{-1}\mathcal{E}^{\lambda-1}(t)\bigl[\mathcal{E}(t)-\mathcal {K}(t)\bigr]=\frac {1}{\lambda}\frac{d\mathcal{E}^{\lambda}(t)}{dt}. $$

(2.22)

 □

Therefore, Lemma 2.6 follows easily from (2.22) and the concavity of $\mathcal{E}^{\lambda}(t)$ on $(0, 1)$.

Lemma 2.7

Let $\lambda=2\log2/[2\log\pi-\log2-2\log\mathcal {E}(\sqrt{2}/2)]=1.3930\ldots$

$$ h_{1}(t)=\frac{2^{1+\lambda}}{\pi^{\lambda}}\mathcal{E}^{\lambda }(t)- \frac{(5+t)^{\lambda}+(5-7t)^{\lambda}}{4^{\lambda}} $$

and

$$ h_{2}(t)=\frac{2^{1+\lambda}}{\pi^{\lambda}}\mathcal{E}^{\lambda }(t)-( \sqrt{2}-2t)^{\lambda}-2^{\lambda/2}. $$

Then $h_{1}(t)>0$ for $t\in[3/5, 17/25)$ and $h_{2}(t)>0$ for $t\in [17/25, \sqrt{2}/2)$.

Proof

Simple computations lead to

$$\begin{aligned}& h_{1} \biggl(\frac{17}{25} \biggr) =0.0022\ldots>0,\qquad h_{2} \biggl(\frac {\sqrt{2}}{2} \biggr)=0, \end{aligned}$$

(2.23)

$$\begin{aligned}& h'_{1}(t) =\frac{\lambda}{4^{\lambda}} \biggl[ \frac{2^{3\lambda +1}}{\pi ^{\lambda}}t^{-1}\mathcal{E}^{\lambda-1}(t) \bigl( \mathcal{E}(t)-\mathcal{K}(t) \bigr)+7(5-7t)^{\lambda -1}-(5+t)^{\lambda-1} \biggr], \end{aligned}$$

(2.24)

$$\begin{aligned}& h'_{2}(t) =2\lambda \biggl[ \biggl(\frac{2}{\pi} \biggr)^{\lambda }t^{-1}\mathcal{E}^{\lambda-1}(t) \bigl( \mathcal{E}(t)-\mathcal{K}(t) \bigr)+(\sqrt{2}-2t)^{\lambda -1} \biggr], \end{aligned}$$

(2.25)

$$\begin{aligned}& h'_{1} \biggl(\frac{3}{5} \biggr) =-0.0471 \ldots< 0,\qquad h'_{2} \biggl(\frac {17}{25} \biggr)=-0.236 \ldots< 0. \end{aligned}$$

(2.26)

From (2.24) and (2.25) together with Lemma 2.6 we clearly see that both $h'_{1}(t)$ and $h'_{2}(t)$ are strictly decreasing on $(0, \sqrt {2}/2)$. Then (2.26) leads to the conclusion that $h_{1}(t)$ is strictly decreasing on $[3/5, 17/25]$ and $h_{2}(t)$ is strictly decreasing on $[17/25, \sqrt{2}/2)$.

Therefore, Lemma 2.7 follows from (2.23) and the monotonicity of $h_{1}(t)$ on $[3/5, 17/25]$ and $h_{2}(t)$ on $[17/25, \sqrt{2}/2)$. □

Lemma 2.8

(See [18], Corollary 3.2)

The inequality

$$ \frac{2}{\pi}\mathcal{E}(t)< \frac{(1-t^{2})^{5/8}+1}{(1-t^{2})^{1/8}+1} $$

(2.27)

holds for all $t\in(0, 1)$.

3 Main results

Theorem 3.1

Let $\lambda=2\log2/[2\log\pi-\log2-2\log\mathcal {E}(\sqrt{2}/2)]=1.3930\ldots$ . Then the double inequality

$$ M_{p}(a,b)< T\bigl[A(a,b), Q(a,b)\bigr]< M_{q}(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $p\leq\lambda$ and $q\geq3/2$.

Proof

Since the arithmetic mean $A(a,b)$, quadratic mean $Q(a,b)$, Toader mean $T(a,b)$, and rth power mean $M_{r}(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $a>b$. Let $t=(a-b)/\sqrt{2(a^{2}+b^{2})}$. Then $t\in(0, \sqrt{2}/2)$ and equations (1.2)-(1.5) lead to

$$\begin{aligned}& M_{r}(a,b)=\frac{A(a,b)}{\sqrt{1-t^{2}}} \biggl[\frac{(\sqrt {1-t^{2}}+t)^{r}+(\sqrt{1-t^{2}}-t)^{r}}{2} \biggr]^{1/r}, \end{aligned}$$

(3.1)

$$\begin{aligned}& T\bigl[A(a,b), Q(a,b)\bigr]=\frac{2A(a,b)\mathcal{E}(t)}{\pi\sqrt{1-t^{2}}}. \end{aligned}$$

(3.2)

We divide the proof into three cases.

Case 1 $r\geq3/2$. Then it follows from (3.1) and (3.2) together with the monotonicity of $M_{r}(a,b)$ with respect to r that

$$\begin{aligned} \begin{aligned}[b] &T\bigl[A(a,b), Q(a,b)\bigr]-M_{r}(a,b)\\ &\quad \leq T\bigl[A(a,b), Q(a,b)\bigr]-M_{3/2}(a,b) \\ &\quad =\frac{A(a,b)}{\sqrt{1-t^{2}}} \biggl[\frac{2}{\pi}\mathcal {E}(t)-\frac {(1-t^{2})^{5/8}+1}{(1-t^{2})^{1/8}+1} \biggr] \\ &\qquad {}+\frac{A(a,b)}{\sqrt{1-t^{2}}} \biggl[\frac {(1-t^{2})^{5/8}+1}{(1-t^{2})^{1/8}+1}- \biggl(\frac{(\sqrt{1-t^{2}}+t)^{3/2} +(\sqrt{1-t^{2}}-t)^{3/2}}{2} \biggr)^{2/3} \biggr]. \end{aligned} \end{aligned}$$

(3.3)

Therefore,

$$ T\bigl[A(a,b), Q(a,b)\bigr]< M_{r}(a,b) $$

for all $a, b>0$ with $a\neq b$ follows from Lemmas 2.2 and 2.8 together with (3.3).

Case 2 $r\leq\lambda$. Then equations (3.1) and (3.2) together with the monotonicity of $M_{r}(a,b)$ with respect to r lead to

$$\begin{aligned} \begin{aligned}[b] &T\bigl[A(a,b), Q(a,b)\bigr]-M_{r}(a,b) \\ &\quad \geq T\bigl[A(a,b), Q(a,b)\bigr]-M_{\lambda}(a,b) \\ &\quad =\frac{A(a,b)}{\sqrt{1-t^{2}}} \biggl[\frac{2}{\pi}\mathcal {E}(t)- \biggl( \frac{(\sqrt{1-t^{2}}+t)^{\lambda} +(\sqrt{1-t^{2}}-t)^{\lambda}}{2} \biggr)^{1/\lambda} \biggr]. \end{aligned} \end{aligned}$$

(3.4)

We divide the proof into two subcases.

Subcase 2.1 $t\in(0, 3/5)$. Let $u=t/\sqrt{1-t^{2}}$. Then $u\in(0, 3/4)$ and (3.4) leads to

$$\begin{aligned} \begin{aligned}[b] &T\bigl[A(a,b), Q(a,b)\bigr]-M_{r}(a,b) \\ &\quad >A(a,b) \biggl[\frac{2}{\pi}\sqrt{1+u^{2}}\mathcal{E} \biggl( \frac {u}{\sqrt {1+u^{2}}} \biggr) - \biggl(\frac{(1+u)^{\lambda}+(1-u)^{\lambda}}{2} \biggr)^{1/\lambda } \biggr]. \end{aligned} \end{aligned}$$

(3.5)

Therefore,

$$ T\bigl[A(a,b), Q(a,b)\bigr]>M_{r}(a,b) $$

for $0<\vert a-b\vert /\sqrt{2(a^{2}+b^{2})}<3/5$ with $a\neq b$ follows from Lemma 2.5 and (3.5).

Subcase 2.2 $t\in[3/5, \sqrt{2}/2)$. Let

$$ h(t)=\frac{2^{1+\lambda}}{\pi^{\lambda}}\mathcal{E}^{\lambda }(t)-\bigl(\sqrt {1-t^{2}}+t\bigr)^{\lambda}-\bigl(\sqrt{1-t^{2}}-t \bigr)^{\lambda}. $$

(3.6)

It is easy to verify that

$$ \sqrt{1-t^{2}}\leq\frac{5-3t}{4}\quad \text{and}\quad \sqrt {1-t^{2}}< \sqrt{2}-t $$

(3.7)

for all $t\in(0, \sqrt{2}/2)$.

Equation (3.6) and inequality (3.7) lead to

$$ h(t)>\frac{2^{1+\lambda}}{\pi^{\lambda}}\mathcal{E}^{\lambda }(t)-\frac {(5+t)^{\lambda}+(5-7t)^{\lambda}}{4^{\lambda}} $$

(3.8)

and

$$ h(t)>\frac{2^{1+\lambda}}{\pi^{\lambda}}\mathcal{E}^{\lambda }(t)-(\sqrt {2}-2t)^{\lambda}-2^{\lambda/2}. $$

(3.9)

Therefore,

$$ T\bigl[A(a,b), Q(a,b)\bigr]>M_{r}(a,b) $$

for $3/5\leq \vert a-b\vert /\sqrt{2(a^{2}+b^{2})}$ with $a\neq b$ follows from Lemma 2.7, (3.4), (3.6), (3.8), and (3.9).

Case 3 $\lambda< r<3/2$. On the one hand, equations (1.2) and (1.5) lead to

$$\begin{aligned} \begin{aligned}[b] &\lim_{x\rightarrow0^{+}}\bigl[\log T\bigl[A(1,x), Q(1, x)\bigr]-\log M_{r}(1,x)\bigr] \\ &\quad =\log \biggl[\frac{\sqrt{2}\mathcal{E} (\frac{\sqrt {2}}{2} )}{\pi} \biggr]+\frac{\log2}{r} \\ &\quad =-\frac{(r-\lambda)\log2}{\lambda r}< 0. \end{aligned} \end{aligned}$$

(3.10)

Inequality (3.10) implies that there exists $\delta_{1}>0$ such that

$$ T\bigl[A(a,b), Q(a,b)\bigr]< M_{r}(a,b) $$

for all $a, b>0$ with $a/b\in(0, \delta_{1})$.

On the other hand, by the Taylor expansion and let $x>0$ and $x\rightarrow0$, then equations (1.2) and (1.5) lead to

$$\begin{aligned} \begin{aligned}[b] &T\bigl[A(1,1-x), Q(1, 1-x)\bigr]- M_{r}(1,1-x) \\ &\quad =\frac{2}{\pi}\sqrt{1-x+\frac{x^{2}}{2}}\mathcal{E} \biggl( \frac {x}{2\sqrt{1-x+\frac{x^{2}}{2}}} \biggr)- \biggl[\frac {1+(1-x)^{r}}{2} \biggr]^{1/r} \\ &\quad =1-\frac{x}{2}+\frac{x^{2}}{16}- \biggl[1-\frac{x}{2}+ \biggl(\frac {1}{16}-\frac{3-2r}{16} \biggr)x^{2} \biggr]+o \bigl(x^{2}\bigr) \\ &\quad =\frac{3-2r}{16}x^{2}+o\bigl(x^{2}\bigr). \end{aligned} \end{aligned}$$

(3.11)

Equation (3.11) implies there exists $\delta_{2}\in(0, 1)$ such that

$$ T\bigl[A(a,b), Q(a,b)\bigr]> M_{r}(a,b) $$

for all $a, b>0$ with $a/b\in(1-\delta_{2}, 1)$. □

From Theorem 3.1 we get Corollary 3.2 immediately.

Corollary 3.2

Let $\lambda=2\log2/[2\log\pi-\log2-2\log \mathcal {E}(\sqrt{2}/2)]=1.3930\ldots$ . Then the double inequality

$$ \frac{\pi}{2} \biggl[\frac{(\sqrt{1-t^{2}}+t)^{p}+(\sqrt {1-t^{2}}-t)^{p}}{2} \biggr]^{1/p}< \mathcal{E}(t)< \frac{\pi}{2} \biggl[\frac{(\sqrt{1-t^{2}}+t)^{q}+(\sqrt{1-t^{2}}-t)^{q}}{2} \biggr]^{1/q} $$

holds for all $t\in(0, \sqrt{2}/2)$ if and only if $p\leq\lambda$ and $q\geq3/2$.