# Optimal inequalities for bounding Toader mean by arithmetic and quadratic means

## Abstract

In this paper, we present the best possible parameters $$\alpha(r)$$ and $$\beta(r)$$ such that the double inequality

\begin{aligned} \bigl[\alpha(r)A^{r}(a,b)+ \bigl(1-\alpha(r) \bigr)Q^{r}(a,b) \bigr]^{1/r} < & TD \bigl[A(a,b), Q(a,b) \bigr] \\ < & \bigl[\beta(r)A^{r}(a,b)+ \bigl(1-\beta(r) \bigr)Q^{r}(a,b) \bigr]^{1/r} \end{aligned}

holds for all $$r\leq 1$$ and $$a, b>0$$ with $$a\neq b$$, and we provide new bounds for the complete elliptic integral $$\mathcal{E}(r)=\int_{0}^{\pi/2}(1-r^{2}\sin^{2}\theta)^{1/2}\,d\theta$$ $$(r\in (0, \sqrt{2}/2))$$ of the second kind, where $$TD(a,b)=\frac{2}{\pi}\int_{0}^{\pi/2}\sqrt{a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}\,d\theta$$, $$A(a,b)=(a+b)/2$$ and $$Q(a,b)=\sqrt{(a^{2}+b^{2})/2}$$ are the Toader, arithmetic, and quadratic means of a and b, respectively.

## 1 Introduction

For $$p\in [0, 1]$$, $$q\in \mathbb{R}$$ and $$a,b>0$$ with $$a\neq b$$, the pth generalized Seiffert mean $$S_{p}(a, b)$$, qth Gini mean $$G_{q}(a, b)$$, qth power mean $$M_{q}(a, b)$$, qth Lehmer mean $$L_{q}(a,b)$$, harmonic mean $$H(a,b)$$, geometric mean $$G(a,b)$$, arithmetic mean $$A(a,b)$$, quadratic mean $$Q(a,b)$$, Toader mean $$TD(a,b)$$ [1], centroidal mean $$\overline{C}(a,b)$$, contraharmonic mean $$C(a,b)$$ are, respectively, defined by

\begin{aligned}& S_{p}(a,b)= \textstyle\begin{cases} \displaystyle\frac{p(a-b)}{\arctan [2p(a-b)/(a+b) ]}, &0< p\leq 1,\\ (a+b)/2, &p=0, \end{cases}\displaystyle \\& G_{q}(a,b)= \textstyle\begin{cases} [(a^{q-1}+b^{q-1})/(a+b) ]^{1/(q-2)}, &q\neq 2,\\ (a^{a}b^{b} )^{1/(a+b)}, &q=2, \end{cases}\displaystyle \\& M_{q}(a,b)= \textstyle\begin{cases} [(a^{q}+b^{q})/2 ]^{1/q}, &q\neq 0,\\ \sqrt{ab}, &q=0, \end{cases}\displaystyle \\& L_{q}(a,b)=\frac{a^{q+1}+b^{q+1}}{a^{q}+b^{q}}, \qquad H(a,b)=\frac{2ab}{a+b},\qquad G(a,b)= \sqrt{ab}, \\& A(a,b)=\frac{a+b}{2},\qquad Q(a,b)=\sqrt{\frac{a^{2}+b^{2}}{2}}, \\& TD(a,b)=\frac{2}{\pi} \int_{0}^{{\pi}/{2}}\sqrt{a^{2} \cos^{2}{\theta}+b^{2}\sin^{2}{\theta}}\,d\theta, \\& \overline{C}(a,b)=\frac{2(a^{2}+ab+b^{2})}{3(a+b)}, \qquad C(a,b)=\frac{a^{2}+b^{2}}{a+b}. \end{aligned}
(1.1)

It is well known that $$S_{p}(a, b)$$, $$G_{q}(a, b)$$, $$M_{q}(a, b)$$, and $$L_{q}(a,b)$$ are continuous and strictly increasing with respect to $$p\in [0, 1]$$ and $$q\in \mathbb{R}$$ for fixed $$a, b>0$$ with $$a\neq b$$, and the inequalities

\begin{aligned} H(a,b)&=M_{-1}(a,b)=L_{-1}(a,b)< G(a,b)=M_{0}(a,b)=L_{-1/2}(a,b) \\ & < A(a,b)=M_{1}(a,b)=L_{0}(a,b)< TD(a,b)< \overline{C}(a,b) \\ &< Q(a,b)=M_{2}(a,b)< C(a,b)=L_{1}(a,b) \end{aligned}

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

The Toader mean $$TD(a,b)$$ has been well known in the mathematical literature for many years, it satisfies

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

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 [2â€“4]), 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 $$\mathcal{K}(0^{+})=\mathcal{E}(0^{+})=\pi/2$$, $$\mathcal{K}(r)$$, and $$\mathcal{E}(r)$$ satisfy the derivatives formulas (see [5], Appendix E, p.474-475)

\begin{aligned}& \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}, \\& \frac{d[\mathcal{K}(r)-\mathcal{E}(r)]}{dr}=\frac{r\mathcal{E}(r)}{1-r^{2}}, \end{aligned}

the values $$\mathcal{K}(\sqrt{2}/2)$$ and $$\mathcal{E}(\sqrt{2}/2)$$ can be expressed as (see [6], TheoremÂ 1.7)

$$\mathcal{K} \biggl(\frac{\sqrt{2}}{2} \biggr)=\frac{\Gamma^{2} ({1}/{4} )}{4\sqrt{\pi}}=1.854\ldots, \qquad \mathcal{E} \biggl(\frac{\sqrt{2}}{2} \biggr)=\frac{4\Gamma^{2} ({3}/{4} )+\Gamma^{2} ({1}/{4} )}{8\sqrt{\pi}}=1.350 \ldots,$$

where $$\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}\,dt(\operatorname{Re}{x}>0)$$ is the Euler gamma function, and the Toader mean $$TD(a,b)$$ can be rewritten as

$$TD(a,b)= \textstyle\begin{cases} {2a}\mathcal{E} (\sqrt{1- ({b}/{a} )^{2}} )/\pi, \quad a\geq b,\\ {2b}\mathcal{E} (\sqrt{1- ({a}/{b} )^{2}} )/\pi, \quad a< b. \end{cases}$$
(1.2)

Recently, the Toader mean $$TD(a,b)$$ has been the subject of intensive research. Vuorinen [7] conjectured that the inequality

$$TD(a,b)>M_{3/2}(a,b)$$

holds for all $$a, b>0$$ with $$a\neq b$$. This conjecture was proved by Qiu and Shen [8], and Barnard, Pearce and Richards [9], respectively.

Alzer and Qiu [10] presented a best possible upper power mean bound for the Toader mean as follows:

$$TD(a,b)< M_{\log 2/(\log\pi-\log 2)}(a,b)$$

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

Neuman [2], and Kazi and Neuman [3] proved that the inequalities

\begin{aligned}& \frac{(a+b)\sqrt{ab}-ab}{AGM(a,b)}< TD(a,b)< \frac{4(a+b)\sqrt{ab}+(a-b)^{2}}{8AGM(a,b)}, \\& TD(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.

In [11â€“13], the authors presented the best possible parameters $$\lambda_{1}, \mu_{1}\in [0, 1]$$ and $$\lambda_{2}, \mu_{2}, \lambda_{3}, \mu_{3}\in \mathbb{R}$$ such that the double inequalities $$S_{\lambda_{1}}(a,b)< TD(a,b)< S_{\mu_{1}}(a,b)$$, $$G_{\lambda_{2}}(a,b)< TD(a,b)< G_{\mu_{2}}(a,b)$$ and $$L_{\lambda_{3}}(a,b)< TD(a,b)< L_{\mu_{3}}(a,b)$$ hold for all $$a, b>0$$ with $$a\neq b$$.

Let $$\lambda, \mu, \alpha, \beta\in (1/2, 1)$$. Then Chu, Wang and Ma [14], and Hua and Qi [15] proved that the double inequalities

\begin{aligned}& C \bigl[\lambda a+(1-\lambda)b, \lambda b+(1-\lambda)a \bigr]< TD(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]< TD(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 $$\lambda\leq 3/4$$, $$\mu\geq 1/2+\sqrt{\pi(4-\pi)}/(2\pi)$$, $$\alpha\leq 1/2+\sqrt{3}/4$$ and $$\beta\geq 1/2+\sqrt{12/\pi-3}/2$$.

In [16â€“20], the authors proved that the double inequalities

\begin{aligned}& \alpha_{1}Q(a,b)+(1-\alpha_{1})A(a,b)< TD(a,b)< \beta_{1}Q(a,b)+(1-\beta_{1})A(a,b), \\& Q^{\alpha_{2}}(a,b)A^{(1-\alpha_{2})}(a,b)< TD(a,b)< Q^{\beta_{2}}(a,b)A^{(1-\beta_{2})}(a,b), \\& \alpha_{3}C(a,b)+(1-\alpha_{3})A(a,b)< TD(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}{TD(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)< TD(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)< TD(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)< TD(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}{TD(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)< TD(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}{TD(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 $$\alpha_{1}\leq 1/2$$, $$\beta_{1}\geq (4-\pi)/[(\sqrt{2}-1)\pi]$$, $$\alpha_{2}\leq 1/2$$, $$\beta_{2}\geq 4-2\log\pi/\log 2$$, $$\alpha_{3}\leq 1/4$$, $$\beta_{3}\geq 4/\pi-1$$, $$\alpha_{4}\leq \pi/2-1$$, $$\beta_{4}\geq 3/4$$, $$\alpha_{5}\leq 5/8$$, $$\beta_{5}\geq 2/\pi$$, $$\alpha_{6}\leq 1/8$$, $$\beta_{6}\geq 2/\pi-1/2$$, $$\alpha_{7}\leq 3/4$$, $$\beta_{7}\geq 12/\pi-3$$, $$\alpha_{8}\leq \pi-3$$, $$\beta_{8}\geq 1/4$$, $$\alpha_{9}\leq 5/6$$, $$\beta_{9}\geq 2\sqrt{2}/\pi$$, $$\alpha_{10}\leq 0$$, and $$\beta_{10}\geq 1/6$$.

The main purpose of this paper is to present the best possible parameters $$\alpha(r)$$ and $$\beta(r)$$ such that the double inequality

\begin{aligned} \bigl[\alpha(r)A^{r}(a,b)+ \bigl(1-\alpha(r) \bigr)Q^{r}(a,b) \bigr]^{1/r} & < TD \bigl[A(a,b), Q(a,b) \bigr] \\ &< \bigl[\beta(r)A^{r}(a,b)+ \bigl(1-\beta(r) \bigr)Q^{r}(a,b) \bigr]^{1/r} \end{aligned}

holds for all $$r\leq 1$$ and $$a, b>0$$ with $$a\neq b$$.

## 2 Lemmas

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

### Lemma 2.1

Let $$p\in (0, 1)$$, $$t\in (0, \sqrt{2}/2)$$, $$\lambda=(2+\sqrt{2})[1-2\mathcal{E}(\sqrt{2}/2)/\pi]=0.478\ldots$$ and

$$f(t)=\frac{\pi p}{2}\sqrt{1-t^{2}}+\frac{\pi}{2}(1-p)- \mathcal{E}(t).$$
(2.1)

Then $$f(t)<0$$ for all $$t\in (0, \sqrt{2}/2)$$ if and only if $$p\geq 1/2$$ and $$f(t)>0$$ for all $$t\in (0, \sqrt{2}/2)$$ if and only if $$p\leq \lambda$$.

### Proof

It follows from (2.1) that

\begin{aligned}& f \bigl(0^{+} \bigr)=0, \end{aligned}
(2.2)
\begin{aligned}& f \biggl(\frac{\sqrt{2}}{2} \biggr)=\frac{\pi}{2} \biggl(1- \frac{\sqrt{2}}{2} \biggr) (\lambda-p), \end{aligned}
(2.3)
\begin{aligned}& f^{\prime}(t)=\frac{f_{1}(t)}{t\sqrt{1-t^{2}}}, \end{aligned}
(2.4)

where

\begin{aligned}& f_{1}(t)=\sqrt{1-t^{2}} \bigl[\mathcal{K}(t)-\mathcal{E}(t) \bigr]-\frac{\pi p}{2}t^{2},\qquad f_{1} \bigl(0^{+} \bigr)=0, \end{aligned}
(2.5)
\begin{aligned}& f_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)=\frac{\sqrt{2}}{2} \biggl[ \mathcal{K} \biggl(\frac{\sqrt{2}}{2} \biggr)-\mathcal{E} \biggl( \frac{\sqrt{2}}{2} \biggr) \biggr] -\frac{\pi p}{4}, \end{aligned}
(2.6)
\begin{aligned}& f^{\prime}_{1}(t)=\frac{t[2\mathcal{E}(t)-\mathcal{K}(t)]}{\sqrt{1-t^{2}}}-\pi pt, \end{aligned}
(2.7)
\begin{aligned}& f^{\prime}_{1} \bigl(0^{+} \bigr)=0, \end{aligned}
(2.8)
\begin{aligned}& f^{\prime}_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)=2\mathcal{E} \biggl(\frac{\sqrt{2}}{2} \biggr)-\mathcal{K} \biggl(\frac{\sqrt{2}}{2} \biggr)- \frac{\sqrt{2}\pi p}{2}, \end{aligned}
(2.9)
\begin{aligned}& f^{\prime\prime}_{1}(t)=\frac{ (3-2t^{2} )\mathcal{E}(t)- (2-t^{2} )\mathcal{K}(t)}{ (1-t^{2} )^{3/2}}-\pi p, \end{aligned}
(2.10)
\begin{aligned}& f^{\prime\prime}_{1} \bigl(0^{+} \bigr)=\pi \biggl( \frac{1}{2}-p \biggr), \end{aligned}
(2.11)
\begin{aligned}& f^{\prime\prime}_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)=\sqrt{2} \biggl[4\mathcal{E} \biggl(\frac{\sqrt{2}}{2} \biggr)-3\mathcal{K} \biggl( \frac{\sqrt{2}}{2} \biggr) \biggr] -\pi p, \end{aligned}
(2.12)
\begin{aligned}& f^{\prime\prime\prime}_{1}(t)=-\frac{(1+t^{2})[\mathcal{K}(t)-\mathcal{E}(t)]+t^{2}\mathcal{K}(t)}{t (1-t^{2} )^{5/2}}< 0 \end{aligned}
(2.13)

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

It follows from (2.13) that $$f_{1}^{\prime\prime}(t)$$ is strictly decreasing on $$(0, \sqrt{2}/2)$$.

We divide the proof into three cases.

Case 1 $$p\geq 1/2$$. Then (2.11) leads to

$$f^{\prime\prime}_{1} \bigl(0^{+} \bigr)\leq 0.$$
(2.14)

From (2.14) and the monotonicity of $$f_{1}^{\prime\prime}(t)$$ we clearly see that $$f_{1}^{\prime}(t)$$ is strictly decreasing on $$(0, \sqrt{2}/2)$$. Therefore, $$f(t)<0$$ for all $$t\in (0, \sqrt{2}/2)$$ follows easily from (2.2), (2.4), (2.5), (2.8), and the monotonicity of $$f_{1}^{\prime}(t)$$.

Case 2 $$0< p\leq \lambda$$. Then from (2.11) and (2.12) together with $$4\mathcal{E}(\sqrt{2}/2)-3\mathcal{K}(\sqrt{2}/2)=-0.159\ldots$$ we clearly see that

$$f^{\prime\prime}_{1} \bigl(0^{+} \bigr)> 0, \qquad f^{\prime\prime}_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)< 0.$$
(2.15)

It follows from (2.15) and the monotonicity of $$f_{1}^{\prime\prime}(t)$$ that there exists $$t_{0}\in (0, \sqrt{2}/2)$$ such that $$f_{1}^{\prime}(t)$$ is strictly increasing on $$(0, t_{0}]$$ and strictly decreasing on $$[t_{0}, \sqrt{2}/2)$$.

Let $$\lambda^{\ast}=\frac{\sqrt{2}}{\pi} [2\mathcal{E}(\frac{\sqrt{2}}{2})-\mathcal{K}(\frac{\sqrt{2}}{2}) ]=0.381\ldots$$ and $$\lambda^{\ast\ast}=\frac{2\sqrt{2}}{\pi} [\mathcal{K}(\frac{\sqrt{2}}{2})-\mathcal{E}(\frac{\sqrt{2}}{2}) ]=0.453\ldots$$â€‰. We divide the proof into three subcases.

Subcase 2.1 $$0< p\leq\lambda^{\ast}$$. Then (2.9) leads to

$$f^{\prime}_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)\geq 0.$$
(2.16)

It follows from (2.8) and (2.16) together with the piecewise monotonicity of $$f_{1}^{\prime}(t)$$ that

$$f^{\prime}_{1}(t)>0$$
(2.17)

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

Therefore, $$f(t)>0$$ for all $$t\in (0, \sqrt{2}/2)$$ follows easily from (2.2), (2.4), (2.5), and (2.17).

Subcase 2.2 $$\lambda^{\ast}< p\leq\lambda^{\ast\ast}$$. Then (2.6) and (2.9) lead to

\begin{aligned}& f_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)\geq 0, \end{aligned}
(2.18)
\begin{aligned}& f^{\prime}_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)< 0. \end{aligned}
(2.19)

It follows from (2.8) and (2.19) together with the piecewise monotonicity of $$f_{1}^{\prime}(t)$$ that there exists $$t_{1}\in (0, \sqrt{2}/2)$$ such that $$f_{1}(t)$$ is strictly increasing on $$(0, t_{1}]$$ and strictly decreasing on $$[t_{1}, \sqrt{2}/2)$$.

Equation (2.5) and inequality (2.18) together with the piecewise monotonicity of $$f_{1}(t)$$ lead to the conclusion that

$$f_{1}(t)>0$$
(2.20)

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

Therefore, $$f(t)>0$$ for all $$t\in (0, \sqrt{2}/2)$$ follows easily from (2.2) and (2.4) together withÂ (2.20).

Subcase 2.3 $$\lambda^{\ast\ast}< p\leq \lambda$$. Then (2.3), (2.6), and (2.9) lead to

\begin{aligned}& f \biggl(\frac{\sqrt{2}}{2} \biggr)\geq 0, \end{aligned}
(2.21)
\begin{aligned}& f_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)< 0, \end{aligned}
(2.22)
\begin{aligned}& f_{1}^{\prime} \biggl(\frac{\sqrt{2}}{2} \biggr)< 2\mathcal{E} \biggl(\frac{\sqrt{2}}{2} \biggr)-\mathcal{K} \biggl(\frac{\sqrt{2}}{2} \biggr) - \frac{\sqrt{2}\pi}{2}\lambda^{\ast\ast} \\& \phantom{f_{1}^{\prime} \biggl(\frac{\sqrt{2}}{2} \biggr)} < 2\mathcal{E} \biggl(\frac{\sqrt{2}}{2} \biggr)-\mathcal{K} \biggl( \frac{\sqrt{2}}{2} \biggr) -\frac{\sqrt{2}\pi}{2}\lambda^{\ast}=0. \end{aligned}
(2.23)

It follows from (2.8) and (2.23) together with the piecewise monotonicity of $$f_{1}^{\prime}(t)$$ that there exists $$t_{2}\in (0, \sqrt{2}/2)$$ such that $$f_{1}(t)$$ is strictly increasing on $$(0, t_{2}]$$ and strictly decreasing on $$[t_{2}, \sqrt{2}/2)$$.

From (2.4), (2.5), and (2.22) together with the piecewise monotonicity of $$f_{1}(t)$$ we clearly see that there exists $$t_{3}\in (0, \sqrt{2}/2)$$ such that $$f(t)$$ is strictly increasing on $$(0, t_{3}]$$ and strictly decreasing on $$[t_{3}, \sqrt{2}/2)$$.

Therefore, $$f(t)>0$$ for all $$t\in (0, \sqrt{2}/2)$$ follows easily from (2.2) and (2.21) together with the piecewise monotonicity of $$f(t)$$.

Case 3 $$\lambda< p<1/2$$. Then (2.3), (2.6), (2.9), (2.11), and (2.12) lead to

\begin{aligned}& f \biggl(\frac{\sqrt{2}}{2} \biggr)< 0, \end{aligned}
(2.24)
\begin{aligned}& f_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)< \frac{\sqrt{2}}{2} \biggl[ \mathcal{K} \biggl(\frac{\sqrt{2}}{2} \biggr)-\mathcal{E} \biggl( \frac{\sqrt{2}}{2} \biggr) \biggr] -\frac{\pi \lambda^{\ast\ast}}{4}=0, \end{aligned}
(2.25)
\begin{aligned}& f^{\prime}_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)< 2\mathcal{E} \biggl(\frac{\sqrt{2}}{2} \biggr)-\mathcal{K} \biggl(\frac{\sqrt{2}}{2} \biggr)- \frac{\sqrt{2}\pi \lambda^{\ast}}{2}=0, \end{aligned}
(2.26)
\begin{aligned}& f^{\prime\prime}_{1} \bigl(0^{+} \bigr)>0, \end{aligned}
(2.27)
\begin{aligned}& f^{\prime\prime}_{1} \biggl(\frac{\sqrt{2}}{2} \biggr) < \sqrt{2} \biggl[4\mathcal{E} \biggl(\frac{\sqrt{2}}{2} \biggr)-3\mathcal{K} \biggl( \frac{\sqrt{2}}{2} \biggr) \biggr] -\pi \lambda^{\ast} \\& \phantom{f^{\prime\prime}_{1} \biggl(\frac{\sqrt{2}}{2} \biggr)} =-2\sqrt{2} \biggl[\mathcal{K} \biggl(\frac{\sqrt{2}}{2} \biggr)-\mathcal{E} \biggl(\frac{\sqrt{2}}{2} \biggr) \biggr]< 0. \end{aligned}
(2.28)

It follows from (2.27) and (2.28) together with the monotonicity of $$f_{1}^{\prime\prime}(t)$$ that there exists $$t_{4}\in (0, \sqrt{2}/2)$$ such that $$f_{1}^{\prime}(t)$$ is strictly increasing on $$(0, t_{4}]$$ and strictly decreasing on $$[t_{4}, \sqrt{2}/2)$$.

Equation (2.8) and inequality (2.26) together with the piecewise monotonicity of $$f_{1}^{\prime}(t)$$ lead to the conclusion that there exists $$t_{5}\in (0, \sqrt{2}/2)$$ such that $$f_{1}(t)$$ is strictly increasing on $$(0, t_{5}]$$ and strictly decreasing on $$[t_{5}, \sqrt{2}/2)$$.

From (2.4), (2.5), (2.25), and the piecewise monotonicity of $$f_{1}(t)$$ we clearly see that there exists $$t_{6}\in (0, \sqrt{2}/2)$$ such that $$f(t)$$ is strictly increasing on $$(0, t_{6}]$$ and strictly decreasing on $$[t_{6}, \sqrt{2}/2)$$.

Therefore, there exists $$t_{7}\in (0, \sqrt{2}/2)$$ such that $$f(t)>0$$ for $$t\in (0, t_{7})$$ and $$f(t)<0$$ for $$t\in (t_{7}, \sqrt{2}/2)$$ follows from (2.2) and (2.24) together with the piecewise monotonicity of $$f(t)$$.â€ƒâ–¡

### Lemma 2.2

Let $$r\in \mathbb{R}$$, $$a, b>0$$ with $$1< b/a<\sqrt{2}$$, $$c_{0}=2\mathcal{E}(\sqrt{2}/2)/\pi=0.859\ldots$$â€‰, $$c_{1}=\sqrt{2}/2$$, $$\lambda(r)$$ and $$U(r; a,b)$$ be defined by

$$\lambda(r)=\frac{1-c^{r}_{0}}{1-c^{r}_{1}} \quad (r\neq 0),\qquad \lambda_{0}= \frac{\log c_{0}}{\log c_{1}} ,$$
(2.29)

and

$$U(r; a,b)= \bigl[\lambda(r)a^{r}+ \bigl(1-\lambda(r) \bigr)b^{r} \bigr]^{1/r} \quad (r\neq 0),\qquad U(0; a,b)=a^{\lambda_{0}}b^{1-\lambda_{0}},$$
(2.30)

respectively. Then the function $$r\mapsto U(r; a,b)$$ is strictly decreasing on $$(-\infty, \infty)$$.

### Proof

Let $$x=b/a\in (1, \sqrt{2})$$, $$r\neq 0$$, and

$$V(r, x)= \bigl(1-\lambda(r) \bigr)\log x- \bigl(\log \lambda(r) \bigr)^{\prime}.$$
(2.31)

Then from (2.29)-(2.31) one has

\begin{aligned}& \log U(r; a,b)=\log a+\frac{1}{r}\log \bigl(\lambda(r)+ \bigl(1- \lambda(r) \bigr)x^{r} \bigr), \\& \frac{\partial\log U(r; a,b)}{\partial r}=\frac{\lambda^{\prime}(r)(1-x^{r})+(1-\lambda(r))x^{r}\log x}{r (\lambda(r)+(1-\lambda(r))x^{r} )}-\frac{\log (\lambda(r)+(1-\lambda(r))x^{r} )}{r^{2}}, \\& \frac{\partial\log U(r; a,b)}{\partial r}\bigg|_{x=1}=0, \end{aligned}
(2.32)
\begin{aligned}& \lambda^{\prime}(r)=\frac{ (c^{r}_{1}-1 )c^{r}_{0}\log c_{0}- (c^{r}_{0}-1 )c^{r}_{1}\log c_{1}}{ (c^{r}_{1}-1 )^{2}}, \\& \bigl(\lambda(r)+ \bigl(1-\lambda(r) \bigr)x^{r} \bigr)|_{x=\sqrt{2}}= \frac{1-c^{r}_{0}}{1-c^{r}_{1}} + \biggl(1-\frac{1-c^{r}_{0}}{1-c^{r}_{1}} \biggr)\frac{1}{c^{r}_{1}}= \frac{c^{r}_{0}}{c^{r}_{1}}, \\& \frac{\lambda^{\prime}(r)(1-x^{r})+(1-\lambda(r))x^{r}\log x}{r (\lambda(r)+(1-\lambda(r))x^{r} )}\bigg|_{x=\sqrt{2}}=\frac{1}{r}\log\frac{c_{0}}{c_{1}}, \\& \frac{\partial\log U(r; a, b)}{\partial r}\bigg|_{x=\sqrt{2}}=0, \end{aligned}
(2.33)
\begin{aligned}& \frac{\partial^{2}\log U(r; a, b)}{\partial x\,\partial r}=\frac{\lambda(r)x^{r-1}}{ (\lambda(r)+(1-\lambda(r))x^{r} )^{2}}V(r, x), \end{aligned}
(2.34)
\begin{aligned}& V(r, 1)=\frac{\log\frac{1}{c_{1}}}{ (\frac{1}{c_{1}} )^{r}-1}-\frac{\log\frac{1}{c_{0}}}{ (\frac{1}{c_{0}} )^{r}-1}< 0, \end{aligned}
(2.35)
\begin{aligned}& V(r, \sqrt{2})=c^{r}_{0} \biggl(\frac{\log c_{1}}{c^{r}_{1}-1}- \frac{\log c_{0}}{c^{r}_{0}-1} \biggr)>0, \end{aligned}
(2.36)

where inequalities (2.35) and (2.36) hold due to $$c_{0}>c_{1}$$ and the function $$t\mapsto \log t/(t^{r}-1)$$ is strictly decreasing on $$(0, \infty)$$.

Note that $$\lambda(r)\in (0, 1)$$ and the function $$x\rightarrow V(r, x)$$ is strictly increasing on $$(1, \sqrt{2})$$. Then (2.34)-(2.36) lead to the conclusion that there exists $$x_{0}\in (1, \sqrt{2})$$ such that the function $$x\mapsto \partial \log U(r; a, b)/\partial r$$ is strictly decreasing on $$(1, x_{0})$$ and strictly increasing on $$(x_{0}, \sqrt{2})$$.

It follows from (2.32) and (2.33) together with the piecewise monotonicity of the function $$x\mapsto \partial \log U(r; a, b)/\partial r$$ on the interval $$(1, \sqrt{2})$$ that

$$\frac{\partial\log U(r; a, b)}{\partial r}< 0$$
(2.37)

for all $$a, b>0$$ with $$1< b/a<\sqrt{2}$$.

Therefore, LemmaÂ 2.2 follows from (2.37).â€ƒâ–¡

## 3 Main result

### Theorem 3.1

Let $$c_{0}=2\mathcal{E}(\sqrt{2}/2)/\pi=0.859\ldots$$â€‰, $$c_{1}=\sqrt{2}/2$$ and $$\lambda(r)$$ be defined by (2.29). Then the double inequality

\begin{aligned} \bigl[\alpha(r)A^{r}(a,b)+ \bigl(1-\alpha(r) \bigr)Q^{r}(a,b) \bigr]^{1/r}&< TD \bigl[A(a,b), Q(a,b) \bigr] \\ &< \bigl[\beta(r)A^{r}(a,b)+ \bigl(1-\beta(r) \bigr)Q^{r}(a,b) \bigr]^{1/r} \end{aligned}

holds for all $$r\leq 1$$ and $$a, b>0$$ with $$a\neq b$$ if and only if $$\alpha(r)\geq 1/2$$ and $$\beta(r)\leq \lambda(r)$$, where $$r=0$$ is the limit value of $$r\rightarrow 0$$.

### Proof

We first prove that TheoremÂ 3.1 holds for $$r=1$$.

Since $$A(a,b)< TD[A(a,b), Q(a,b)]< Q(a,b)$$ for all $$a, b>0$$ with $$a\neq b$$, and $$A(a,b)$$, $$TD(a,b)$$ and $$Q(a,b)$$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $$\alpha(1)$$, $$\beta(1)\in (0, 1)$$ and $$a>b$$. Let $$t=(a-b)/\sqrt{2(a^{2}+b^{2})}\in (0, \sqrt{2}/2)$$ and $$p\in (0, 1)$$. Then (1.1) and (1.2) lead to

\begin{aligned}& A(a,b)=Q(a,b)\sqrt{1-t^{2}}, \qquad TD \bigl[A(a,b), Q(a,b) \bigr]= \frac{2}{\pi}Q(a,b)\mathcal{E}(t), \\ & \\ & pA(a,b)+(1-p)Q(a,b)-TD \bigl[A(a,b), Q(a,b) \bigr]=\frac{2}{\pi}Q(a,b)f(t), \end{aligned}
(3.1)

where $$f(t)$$ is defined as in LemmaÂ 2.1.

Therefore, TheoremÂ 3.1 for $$r=1$$ follows easily from LemmaÂ 2.1 and (3.1).

Next, let $$r<1$$ and $$a, b>0$$ with $$a\neq b$$, then it follows from TheoremÂ 3.1 for $$r=1$$ that

$$\frac{A(a,b)+Q(a,b)}{2}< TD \bigl[A(a,b), Q(a,b) \bigr]< \lambda(1)A(a,b)+ \bigl(1- \lambda(1) \bigr)Q(a,b).$$
(3.2)

Note that

\begin{aligned}& 1< \frac{Q(a,b)}{A(a,b)}< \sqrt{2}, \end{aligned}
(3.3)
\begin{aligned}& \frac{TD[A(a,b), Q(a,b)]}{ [\frac{A^{r}(a,b)+Q^{r}(a,b)}{2} ]^{1/r}}=\frac{2^{1+1/r}}{\pi}\frac{\mathcal{E}(t)}{ [1+(1-t^{2})^{r/2} ]^{1/r}}, \end{aligned}
(3.4)
\begin{aligned}& \frac{TD[A(a,b), Q(a,b)]}{ [\lambda(r)A^{r}(a,b)+(1-\lambda(r))Q^{r}(a,b) ]^{1/r}} =\frac{2}{\pi}\frac{\mathcal{E}(t)}{ [\lambda(r)(1-t^{2})^{r/2}+1-\lambda(r) ]^{1/r}}, \end{aligned}
(3.5)
\begin{aligned}& \lim_{t\rightarrow 0^{+}}\frac{2^{1+1/r}}{\pi}\frac{\mathcal{E}(t)}{ [1+(1-t^{2})^{r/2} ]^{1/r}} =\lim _{t\rightarrow \sqrt{2}/2}\frac{2}{\pi}\frac{\mathcal{E}(t)}{ [\lambda(r)(1-t^{2})^{r/2}+1-\lambda(r) ]^{1/r}}=1. \end{aligned}
(3.6)

Therefore, TheoremÂ 3.1 for $$r<1$$ follows from (3.2)-(3.6) and LemmaÂ 2.2 together with the monotonicity of the function $$r\mapsto [(a^{r}+b^{r})/2]^{1/r}$$.â€ƒâ–¡

Let $$r=1$$. Then TheoremsÂ 3.1 leads to CorollaryÂ 3.2 immediately.

### Corollary 3.2

Let $$\lambda=(2+\sqrt{2})[1-2\mathcal{E}(\sqrt{2}/2)/\pi]$$. Then the double inequality

$$\frac{\pi}{4}\sqrt{1-t^{2}}+\frac{\pi}{4}< \mathcal{E}(t)< \frac{\pi}{2}\lambda\sqrt{1-t^{2}}+\frac{\pi}{2}(1-\lambda)$$

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

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## Acknowledgements

The research was supported by the Natural Science Foundation of China under Grants 61673169, 11371125, 11401191, and 61374086.

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Correspondence to Yu-Ming Chu.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Zhao, TH., Chu, YM. & Zhang, W. Optimal inequalities for bounding Toader mean by arithmetic and quadratic means. J Inequal Appl 2017, 26 (2017). https://doi.org/10.1186/s13660-017-1300-8