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Optimal inequalities for bounding Toader mean by arithmetic and quadratic means

Journal of Inequalities and Applications20172017:26

https://doi.org/10.1186/s13660-017-1300-8

  • Received: 28 November 2016
  • Accepted: 17 January 2017
  • Published:

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.

Keywords

  • arithmetic mean
  • Toader mean
  • quadratic mean
  • complete elliptic integral

MSC

  • 26E60

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 [24]), 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], Theorem 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 [1113], 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 [1620], 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 \(\lambda^{\ast\ast}=\frac{2\sqrt{2}}{\pi} [\mathcal{K}(\frac{\sqrt{2}}{2})-\mathcal{E}(\frac{\sqrt{2}}{2}) ]=0.453\ldots\) . 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 with (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 \(f(t)\). □

Lemma 2.2

Let \(r\in \mathbb{R}\), \(a, b>0\) with \(1< b/a<\sqrt{2}\), \(c_{0}=2\mathcal{E}(\sqrt{2}/2)/\pi=0.859\ldots\) , \(c_{1}=\sqrt{2}/2\), \(\lambda(r)\) 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, Lemma 2.2 follows from (2.37). □

3 Main result

Theorem 3.1

Let \(c_{0}=2\mathcal{E}(\sqrt{2}/2)/\pi=0.859\ldots\) , \(c_{1}=\sqrt{2}/2\) 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 Theorem 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 Lemma 2.1.

Therefore, Theorem 3.1 for \(r=1\) follows easily from Lemma 2.1 and (3.1).

Next, let \(r<1\) and \(a, b>0\) with \(a\neq b\), then it follows from Theorem 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, Theorem 3.1 for \(r<1\) follows from (3.2)-(3.6) and Lemma 2.2 together with the monotonicity of the function \(r\mapsto [(a^{r}+b^{r})/2]^{1/r}\). □

Let \(r=1\). Then Theorems 3.1 leads to Corollary 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 \(t\in (0, \sqrt{2}/2)\).

Declarations

Acknowledgements

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

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Computation Sciences, Hunan City University, Yiyang, 413000, China
(2)
Icahn School of Medicine at Mount Sinai, Friedman Brain Institute, New York, 10033, United States

References

  1. Toader, G: Some mean values related to the arithmetic-geometric mean. J. Math. Anal. Appl. 218(2), 358-368 (1998) MathSciNetView ArticleMATHGoogle Scholar
  2. Neuman, E: Bounds for symmetric elliptic integrals. J. Approx. Theory 122(2), 249-259 (2003) MathSciNetView ArticleMATHGoogle Scholar
  3. Kazi, H, Neuman, E: Inequalities and bounds for elliptic integrals. J. Approx. Theory 146(2), 212-226 (2007) MathSciNetView ArticleMATHGoogle Scholar
  4. Kazi, H, Neuman, E: Inequalities and bounds for elliptic integrals II. In: Special Functions and Orthogonal Polynomials. Contemp. Math., vol. 471, pp. 127-138. Amer. Math. Soc., Providence, RI (2008) View ArticleGoogle Scholar
  5. Anderson, GD, Vamanamurthy, MK, Vuorinen, M: Conformal Invariants, Inequalities, and Quasiconformal Maps. Wiley, New York (1997) MATHGoogle Scholar
  6. Borwein, JM, Borwein, PB: Pi and the AGM. Wiley, New York (1987) MATHGoogle Scholar
  7. Vuorinen, M: Hypergeometric functions in geometric function theory. In: Special Functions and Differential Equations, Madras, 1977, pp. 119-126. Allied Publ., New Delhi (1998) Google Scholar
  8. Qiu, S-L, Shen, J-M: On two problems concerning means. J. Hangzhou Inst. Electron. Eng. 17(3), 1-7 (1997) (in Chinese) Google Scholar
  9. Barnard, RW, Pearce, K, Richards, KC: An inequality involving the generalized hypergeometric function and the arc length of an ellipse. SIAM J. Math. Anal. 31(3), 693-699 (2000) MathSciNetView ArticleMATHGoogle Scholar
  10. Alzer, H, Qiu, S-L: Monotonicity theorems and inequalities for the complete elliptic integrals. J. Comput. Appl. Math. 172(2), 289-312 (2004) MathSciNetView ArticleMATHGoogle Scholar
  11. Chu, Y-M, Wang, M-K, Qiu, S-L, Qiu, Y-F: Sharp generalized Seiffert mean bounds for Toader mean. Abstr. Appl. Anal. 2011, Article ID 605259 (2011) MathSciNetMATHGoogle Scholar
  12. Chu, Y-M, Wang, M-K: Inequalities between arithmetic-geometric, Gini, and Toader means. Abstr. Appl. Anal. 2012, Article ID 830585 (2012) MathSciNetMATHGoogle Scholar
  13. Chu, Y-M, Wang, M-K: Optimal Lehmer mean bounds for the Toader mean. Results Math. 61(3-4), 223-229 (2012) MathSciNetView ArticleMATHGoogle Scholar
  14. Chu, Y-M, Wang, M-K, Ma, X-Y: Sharp bounds for Toader mean in terms of contraharmonic mean with applications. J. Math. Inequal. 7(2), 161-166 (2012) MathSciNetMATHGoogle Scholar
  15. Hua, Y, Qi, F: A double inequality for bounding Toader mean by the centroidal mean. Proc. Indian Acad. Sci. Math. Sci. 124(4), 527-531 (2014) MathSciNetView ArticleMATHGoogle Scholar
  16. Chu, Y-M, Wang, M-K, Qiu, S-L: Optimal combination bounds of root-square and arithmetic means for Toader mean. Proc. Indian Acad. Sci. Math. Sci. 122(1), 41-51 (2012) MathSciNetView ArticleMATHGoogle Scholar
  17. Song, Y-Q, Jiang, W-D, Chu, Y-M, Yan, D-D: Optimal bounds for Toader mean in terms of arithmetic and contraharmonic means. J. Math. Inequal. 7(4), 751-757 (2013) MathSciNetView ArticleMATHGoogle Scholar
  18. Li, W-H, Zheng, M-M: Some inequalities for bounding Toader mean. J. Funct. Spaces Appl. 2013, Article ID 394194 (2013) MathSciNetMATHGoogle Scholar
  19. Hua, Y, Qi, F: The best bounds for Toader mean in terms of the centroidal and arithmetic means. Filomat 28(4), 775-780 (2014) MathSciNetView ArticleGoogle Scholar
  20. Sun, H, Chu, Y-M: Bounds for Toader mean by quadratic and harmonic means. Acta Math. Sci. 35A(1), 36-42 (2015) (in Chinese) MATHGoogle Scholar

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