Sharp bounds for Toader mean in terms of arithmetic, quadratic, and Neuman means

Abstract

In this paper, we present the best possible parameters $\alpha, \beta \in\mathbb{R}$ and $\lambda, \mu\in(1/2, 1)$ such that the double inequalities $\alpha N_{AQ}(a,b)+(1-\alpha)A(a,b)< T^{\ast}(a,b)<\beta N_{AQ}(a,b)+(1-\beta)A(a,b)$, $Q[\lambda a+(1-\lambda)b, \lambda b+(1-\lambda)a]< T^{\ast}(a,b)< Q[\mu a+(1-\mu)b, \mu b+(1-\mu)a]$ hold for all $a, b>0$ with $a\neq b$, where $T^{\ast}(a,b)$, $A(a,b)$, $Q(a,b)$ and $N_{QA}(a,b)$ are the Toader, arithmetic, quadratic, and Neuman means of a and b, respectively.

1 Introduction

For $a,b>0$ the Toader mean $T^{\ast}(a,b)$ [1] is given by

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

(1.1)

It is well known that the Toader mean satisfies

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

for all $a, b>0$, 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 [2–4]), therefore it cannot be expressed in terms of the elementary transcendental functions.

Recently, the Toader mean $T^{\ast}(a,b)$ has been the subject of intensive research. In particular, many remarkable inequalities for the Toader mean can be found in the literature [5–12].

Let $p\in\mathbb{R}$ and $a, b>0$. Then the pth power mean $M_{p}(a,b)$ is defined by

$$ M_{p}(a,b)= \biggl(\frac{a^{p}+b^{p}}{2} \biggr)^{1/p}\quad (p\neq0),\qquad M_{0}(a,b)=\sqrt{ab}. $$

It is well known that $M_{p}(a,b)$ is continuous and strictly increasing with respect to $p\in\mathbb{R}$ for fixed $a, b>0$ with $a\neq b$.

Vuorinen [13] conjectured that the inequality

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

(1.2)

holds for all $a, b>0$ with $a\neq b$. This conjecture was proved by Qiu and Shen [14], and Barnard et al. [15], respectively.

Alzer and Qiu [16] presented a best possible upper power mean bound for the Toader mean as follows:

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

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

Chu et al. [17] proved that the inequality

$$ T^{\ast}(a,b)< T(a,b) $$

(1.3)

holds for all $a, b>0$ with $a\neq b$, where $T(a,b)=(a-b)/[2\arctan ((a-b)/(a+b))]$ is the second Seiffert mean.

Another important mean of two positive real numbers a and b is the Schwab-Borchardt mean [18, 19]

$$ SB(a,b)= \textstyle\begin{cases} \frac{\sqrt{b^{2}-a^{2}}}{\arccos{(a/b)}}, & a< b, \\ \frac{\sqrt{a^{2}-b^{2}}}{\cosh^{-1}{(a/b)}}, & a>b, \\ a, & a=b, \end{cases} $$

where $\cosh^{-1}(x)=\log(x+\sqrt{x^{2}-1})$ is the inverse hyperbolic cosine function.

It is well known that the Schwab-Borchardt mean $SB(a,b)$ is strictly increasing in both a and b, nonsymmetric and homogeneous of degree 1. Many symmetric bivariate means are special cases of the Schwab-Borchardt mean. For example, $P(a,b)=(a-b)/[2\arcsin((a-b)/(a+b))]=SB[G(a,b), A(a,b)]$ is the first Seiffert mean, $T(a,b)=(a-b)/[2\arctan ((a-b)/(a+b))]=SB[A(a,b), Q(a,b)]$ is the second Seiffert mean, $M(a,b)=(a-b)/[2\sinh^{-1}((a-b)/(a+b))]=SB[Q(a,b), A(a,b)]$ is the Neuman-Sándor mean, $L(a,b)=(a-b)/[2\tanh ^{-1}((a-b)/(a+b))]=SB[A(a,b), G(a,b)]$ is the logarithmic mean, where $G(a,b)=\sqrt{ab}$, $A(a,b)=(a+b)/2$ and $Q(a,b)=\sqrt {(a^{2}+b^{2})/2}$ are the geometric, arithmetic, and quadratic means of a and b, respectively.

Very recently, Neuman [20] introduced the Neuman mean,

$$ N(a,b)=\frac{1}{2} \biggl[a+\frac{b^{2}}{SB(a,b)} \biggr], $$

and presented the explicit formula for $N_{AQ}(a,b)\equiv N[A(a,b), Q(a,b)]$ as follows:

$$ N_{AQ}(a,b)=\frac{1}{2}A(a,b) \biggl[1+ \bigl(1+v^{2} \bigr)\frac{\arctan (v)}{v} \biggr] $$

(1.4)

and proved that the double inequality

$$ T(a,b)< N_{AQ}(a,b)< Q(a,b) $$

(1.5)

holds for all $a, b>0$ with $a\neq b$, where $v=(a-b)/(a+b)$.

Inequalities (1.2), (1.3), and (1.5) lead to

$$ A(a,b)=M_{1}(a,b)< M_{3/2}(a,b)< T^{\ast}(a,b)< N_{AQ}(a,b)< Q(a,b) $$

(1.6)

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

Let $a, b>0$ with $a\neq b$ be fixed and $f(x)=Q[xa+(1-x)b, xb+(1-x)a]$. Then it is not difficult to verify that $f(x)$ is continuous and strictly increasing on $[1/2, 1]$. Note that

$$ f \biggl(\frac{1}{2} \biggr)=A(a,b)< T^{\ast}(a,b)< Q(a,b)=f(1). $$

(1.7)

Motivated by inequalities (1.6) and (1.7), it is natural to ask: what are the best possible parameters $\alpha, \beta\in\mathbb{R}$ and $\lambda, \mu\in(1/2, 1)$ such that the double inequalities

$$\begin{aligned}& \alpha N_{AQ}(a,b)+(1-\alpha)A(a,b)< T^{\ast}(a,b)< \beta N_{AQ}(a,b)+(1-\beta)A(a,b),\\& Q\bigl[\lambda a+(1-\lambda)b, \lambda b+(1-\lambda)a\bigr]< T^{\ast}(a,b)< Q \bigl[\mu a+(1-\mu)b, \mu b+(1-\mu)a\bigr] \end{aligned}$$

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

2 Lemmas

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

For $r\in(0, 1)$ the complete elliptic integrals $\mathcal{K}(r)$ and $\mathcal{E}(r)$ of the first and second kinds are defined by

$$ \mathcal{K}(r)=\int_{0}^{\pi/2} \bigl(1-r^{2}\sin^{2}t \bigr)^{-1/2}\,dt $$

and

$$ \mathcal{E}(r)=\int_{0}^{\pi/2} \bigl(1-r^{2}\sin^{2}t \bigr)^{1/2}\,dt, $$

respectively. We clearly see that

$$ \mathcal{K}\bigl(0^{+}\bigr)=\mathcal{E}\bigl(0^{+}\bigr)= \frac{\pi}{2}, \qquad\mathcal {K}\bigl(1^{-}\bigr)=\infty, \qquad\mathcal{E} \bigl(1^{-}\bigr)=1, $$

and the Toader mean $T^{\ast}(a,b)$ given by (1.1) can be expressed as

$$ T^{\ast}(a,b)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} {2a}{\mathcal{E}} ({\sqrt{1-(b/a)^{2}}} )/{\pi}, & a>b,\\ {2b}{\mathcal{E}} ({\sqrt{1-(a/b)^{2}}} )/{\pi}, & a< b,\\ a, & a=b, \end{array}\displaystyle \right . $$

(2.1)

$\mathcal{K}(r)$ and $\mathcal{E}(r)$ satisfy the formulas (see [21], Appendix E, p.474,475)

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

Lemma 2.1

(See [21], Theorem 1.25)

Let $-\infty< a< b<\infty$, $f, g: [a, b]\rightarrow(-\infty, \infty)$ be continuous on $[a,b]$ and differentiable on $(a, b)$, and $g^{\prime}(x)\neq0$ on $(a,b)$. If $f'(x)/g'(x)$ is increasing (decreasing) on $(a,b)$, then so are

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

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

Lemma 2.2

(See [21], Theorem 3.21)

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

(2) The function $r\mapsto(1-r^{2})^{\lambda}\mathcal{K}(r)$ is strictly decreasing from $(0, 1)$ onto $(0, \pi/2)$ if $\lambda\geq1/4$.

Lemma 2.3

The function $r\mapsto[2(1-r^{2})\mathcal{E}(r)-\pi]/r^{2}$ is strictly increasing from $(0, 1)$ onto $(-5\pi/4, -\pi)$.

Proof

Let $f_{1}(r)=2(1-r^{2})\mathcal{E}(r)-\pi$, $f_{2}(r)=r^{2}$ and $f(r)=[2(1-r^{2})\mathcal{E}(r)-\pi]/r^{2}$. Then

$$\begin{aligned}& f_{1} \bigl(0^{+} \bigr)=f_{2}(0)=0,\qquad f(r)=\frac{f_{1}(r)}{f_{2}(r)}, \end{aligned}$$

(2.2)

$$\begin{aligned}& f\bigl(1^{-}\bigr)=-\pi \end{aligned}$$

(2.3)

and simple computations lead to

$$ \frac{f^{\prime}_{1}(r)}{f^{\prime}_{2}(r)}=-3\mathcal{E}(r)+\frac {\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r)}{r^{2}}. $$

(2.4)

It follows from Lemma 2.2(1), (2.2), and (2.4) that $f^{\prime }_{1}(r)/f^{\prime}_{2}(r)$ is strictly increasing on $(0, 1)$ and

$$ f \bigl(0^{+} \bigr)=\lim_{r\rightarrow0^{+}} \frac{f^{\prime }_{1}(r)}{f^{\prime}_{2}(r)}=-\frac{5\pi}{4}. $$

(2.5)

Therefore, Lemma 2.3 follows from Lemma 2.1, (2.2), (2.3), (2.5), and the monotonicity of $f^{\prime}_{1}(r)/f^{\prime}_{2}(r)$. □

Lemma 2.4

Let $p\in(0, 1)$, $r\in(0, 1)$ and

$$ f(r)=\frac{4 [\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r) ]}{r^{2}}+\frac{2(1-r^{2})\mathcal{E}(r)-\pi}{r^{2}}+\pi(1-p). $$

(2.6)

Then the following statements are true:

• (1) If $p=3/4$, then $f(r)>0$ for all $r\in(0, 1)$;

• (2) If $p=4(4-\pi)/[\pi(\pi-2)]=0.9573\cdots$, then there exists $r_{0}\in(0, 1)$ such that $f(r)<0$ for $r\in(0, r_{0})$ and $f(r)>0$ for $r\in(r_{0}, 1)$.

Proof

For part (1), if $p=3/4$, then (2.6) becomes

$$ f(r)=\frac{4 [\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r) ]}{r^{2}}+\frac{2(1-r^{2})\mathcal{E}(r)-\pi}{r^{2}}+\frac{\pi}{4}, $$

and Lemma 2.2(1) and Lemma 2.3 lead to

$$ f(r)>4\times\frac{\pi}{4}-\frac{5\pi}{4}+\frac{\pi}{4}=0 $$

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

For part (2), if $p=4(4-\pi)/[\pi(\pi-2)]$, then it follows from Lemma 2.2(1), Lemma 2.3, and (2.6) that

$$\begin{aligned}& f \bigl(0^{+} \bigr)=-\frac{64-3\pi^{2}-10\pi}{4(\pi-2)}=-0.6515 \cdots< 0, \end{aligned}$$

(2.7)

$$\begin{aligned}& f \bigl(1^{-} \bigr)=\frac{8(\pi-3)}{\pi-2}=0.9922\cdots>0 \end{aligned}$$

(2.8)

and $f(r)$ is strictly increasing on $(0, 1)$.

Therefore, part (2) follows from (2.7) and (2.8) together with the monotonicity of $f(r)$. □

3 Main results

Theorem 3.1

The double inequality

$$ \alpha N_{AQ}(a,b)+(1-\alpha)A(a,b)< T^{\ast}(a,b)< \beta N_{AQ}(a,b)+(1-\beta)A(a,b) $$

(3.1)

holds for all $a, b>0$ with $a\neq b$ if and only if $\alpha\leq3/4$ and $\beta\geq4(4-\pi)/[\pi(\pi-2)]=0.9573\cdots$.

Proof

Since $A(a,b)$, $T^{\ast}(a,b)$ and $N_{AQ}(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $a>b$. Let $r=(a-b)/(a+b)\in(0, 1)$ and $p\in(0, 1)$. Then (2.1) leads to

$$ T^{\ast}(a,b)=\frac{2}{\pi}A(a,b) \bigl[2 \mathcal{E}(r)-\bigl(1-r^{2}\bigr)\mathcal {K}(r) \bigr]. $$

(3.2)

It follows from (1.4), Lemma 2.2(2), and (3.2) that

$$\begin{aligned}& \frac{T^{\ast}(a,b)-A(a,b)}{N_{AQ}(a,b)-A(a,b)}=\frac{\frac{2}{\pi} [2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r) ]-1}{ \frac{(1+r^{2})\arctan(r)}{2r}-\frac{1}{2}}, \end{aligned}$$

(3.3)

$$\begin{aligned}& T^{\ast}(a,b)-\bigl[pN_{AQ}(a,b)+(1-p)A(a,b) \bigr]=\frac{p(1+r^{2})}{2r}A(a,b)F(r), \end{aligned}$$

(3.4)

where

$$\begin{aligned}& F(r)=\frac{4r [2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r) ]+\pi (p-2)r}{p\pi(1+r^{2})}-\arctan(r), \\& F\bigl(0^{+}\bigr)=0, \end{aligned}$$

(3.5)

$$\begin{aligned}& F\bigl(1^{-}\bigr)=\frac{4(4-\pi)+p\pi(2-\pi)}{4p\pi}, \end{aligned}$$

(3.6)

$$\begin{aligned}& F^{\prime}(r)=\frac{2r^{2}}{p\pi(1+r^{2})^{2}}f(r), \end{aligned}$$

(3.7)

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

We divide the proof into two cases.

Case 1 $p=3/4$. Then Lemma 2.4(1) and (3.7) lead to the conclusion that $F(r)$ is strictly increasing on $(0, 1)$. Therefore,

$$ T^{\ast}(a,b)>\frac{3}{4}N_{AQ}(a,b)+ \frac{1}{4}A(a,b) $$

follows from (3.4) and (3.5) together with the monotonicity of $F(r)$.

Case 2 $p=4(4-\pi)/[\pi(\pi-2)]$. Then (3.6) becomes

$$ F\bigl(1^{-}\bigr)=0, $$

(3.8)

and Lemma 2.4(2) and (3.7) imply that there exists $r_{0}\in(0, 1)$ such that $F(r)$ is strictly decreasing on $(0, r_{0}]$ and strictly increasing on $[r_{0}, 1)$. Therefore,

$$ T^{\ast}(a,b)< \frac{4(4-\pi)}{\pi(\pi-2)}N_{AQ}(a,b)+ \biggl[1- \frac {4(4-\pi)}{\pi(\pi-2)} \biggr]A(a,b) $$

follows from (3.4), (3.5), (3.8), and the piecewise monotonicity of $F(r)$.

Next, we prove that $\alpha=3/4$ and $\beta=4(4-\pi)/[\pi(\pi-2)]$ are the best possible parameters such that the double inequality (3.1) holds for all $a, b>0$ with $a\neq b$. It is not difficult to verify that

$$\begin{aligned}& \lim_{r\rightarrow0^{+}}\frac{\frac{2}{\pi} [2\mathcal {E}(r)-(1-r^{2})\mathcal{K}(r) ]-1}{ \frac{(1+r^{2})\arctan(r)}{2r}-\frac{1}{2}}=\frac{3}{4}, \end{aligned}$$

(3.9)

$$\begin{aligned}& \lim_{r\rightarrow1^{-}}\frac{\frac{2}{\pi} [2\mathcal {E}(r)-(1-r^{2})\mathcal{K}(r) ]-1}{ \frac{(1+r^{2})\arctan(r)}{2r}-\frac{1}{2}}=\frac{4(4-\pi)}{\pi(\pi-2)}. \end{aligned}$$

(3.10)

If $\alpha>3/4$, then (3.3) and (3.9) imply that there exists $0<\delta _{1}<1$ such that

$$ T^{\ast}(a,b)< \alpha N_{AQ}(a,b)+(1-\alpha)A(a,b) $$

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

If $\beta<4(4-\pi)/[\pi(\pi-2)]$, then (3.3) and (3.10) imply that there exists $0<\delta_{2}<1$ such that

$$ T^{\ast}(a,b)>\beta N_{AQ}(a,b)+(1-\beta)A(a,b) $$

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

Theorem 3.2

Let $\lambda, \mu\in(1/2, 1)$. Then the double inequality

$$ Q\bigl[\lambda a+(1-\lambda)b, \lambda b+(1-\lambda)a \bigr]< T^{\ast}(a,b)< Q\bigl[\mu a+(1-\mu)b, \mu b+(1-\mu)a\bigr] $$

(3.11)

holds for all $a, b>0$ with $a\neq b$ if and only if $\lambda\leq 1/2+\sqrt{2}/4=0.8535\cdots$ and $\mu\geq1/2+\sqrt{16/\pi ^{2}-1}/2=0.8940\cdots$.

Proof

Without loss of generality, we assume that $a>b>0$. Let $r=(a-b)/(a+b)\in(0, 1)$ and $p\in(0, 1)$. Then from (3.2) and

$$ Q\bigl[pa+(1-p)b, pb+(1-p)a\bigr]=A(a,b)\sqrt{(2p-1)^{2}r^{2}+1} $$

we get

$$\begin{aligned} &T^{\ast}(a,b)-Q\bigl[pa+(1-p)b, pb+(1-p)a\bigr] \\ &\quad= \biggl[\frac{2}{\pi} \bigl(2\mathcal{E}(r)-\bigl(1-r^{2}\bigr) \mathcal{K}(r) \bigr)-\sqrt{(2p-1)^{2}r^{2}+1} \biggr]A(a,b) \\ &\quad=\frac{g(r)}{\frac{2}{\pi} (2\mathcal{E}(r)-(1-r^{2})\mathcal {K}(r) )+\sqrt{(2p-1)^{2}r^{2}+1}}A(a,b), \end{aligned}$$

(3.12)

where

$$\begin{aligned}& g(r)=\frac{4}{\pi^{2}} \bigl[2\mathcal{E}(r)- \bigl(1-r^{2}\bigr)\mathcal {K}(r) \bigr]^{2}-(2p-1)^{2}r^{2}-1, \end{aligned}$$

(3.13)

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

(3.14)

$$\begin{aligned}& g\bigl(1^{-}\bigr)=\frac{16}{\pi^{2}}-(2p-1)^{2}-1. \end{aligned}$$

(3.15)

Let

$$ g_{1}(r)=g^{\prime}(r)/r. $$

(3.16)

Then (3.13) and Lemma 2.2 lead to

$$\begin{aligned}& g_{1}(r)=\frac{8}{\pi^{2}} \bigl[2\mathcal{E}(r)- \bigl(1-r^{2}\bigr)\mathcal {K}(r) \bigr] \frac{\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r)}{r^{2}}-2(2p-1)^{2}, \end{aligned}$$

(3.17)

$$\begin{aligned}& g_{1}\bigl(0^{+}\bigr)=1-2(2p-1)^{2}, \end{aligned}$$

(3.18)

$$\begin{aligned}& g_{1}\bigl(1^{-}\bigr)=\frac{16}{\pi^{2}}-2(2p-1)^{2}. \end{aligned}$$

(3.19)

We divide the proof into two cases.

Case 1 $p=1/2+\sqrt{2}/4$. Then (3.18) becomes

$$ g_{1}\bigl(0^{+}\bigr)=0. $$

(3.20)

From Lemma 2.2(1) and $d[2\mathcal{E}(r)-(1-r^{2})\mathcal {K}(r)]/dr=[\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r)]/r$ we know that the function $r\mapsto2\mathcal{E}(r)-(1-r^{2})\mathcal{K}(r)$ is strictly increasing on $(0, 1)$. Then from Lemma 2.2(1) and (3.17) together with (3.20) we know that $g_{1}(r)$ is strictly increasing on $(0, 1)$ and

$$ g_{1}(r)>g_{1}\bigl(0^{+}\bigr)=0 $$

(3.21)

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

$$ T^{\ast}(a, b)>Q \biggl[ \biggl(\frac{1}{2}+\frac{\sqrt{2}}{4} \biggr)a+ \biggl(\frac{1}{2}-\frac{\sqrt{2}}{4} \biggr)b, \biggl( \frac{1}{2}+\frac{\sqrt {2}}{4} \biggr)b+ \biggl(\frac{1}{2}- \frac{\sqrt{2}}{4} \biggr)a \biggr] $$

follows from (3.12), (3.14), (3.16), and (3.21).

Case 2 $p=1/2+\sqrt{16/\pi^{2}-1}/2$. Then (3.15), (3.18), and (3.19) lead to

$$\begin{aligned}& g\bigl(1^{-}\bigr)=0, \end{aligned}$$

(3.22)

$$\begin{aligned}& g_{1}\bigl(0^{+}\bigr)=- \frac{32-3\pi^{2}}{\pi^{2}}< 0, \end{aligned}$$

(3.23)

$$\begin{aligned}& g_{1}\bigl(1^{-}\bigr)=\frac{2\pi^{2}-16}{\pi^{2}}>0. \end{aligned}$$

(3.24)

It follows from (3.16), (3.23), and (3.24) together with the monotonicity of $g_{1}(r)$ that there exists $r^{\ast}\in(0, 1)$ such that $g(r)$ is strictly decreasing on $(0, r^{\ast}]$ and strictly increasing on $[r^{\ast}, 1)$. Therefore,

$$ T^{\ast}(a, b)< Q\bigl[pa+(1-p)b, pb+(1-p)a\bigr] $$

follows from (3.12), (3.14), (3.22), and the piecewise monotonicity of $g(r)$.

Next, we prove that $\lambda=1/2+\sqrt{2}/4$ and $\mu=1/2+\sqrt{16/\pi ^{2}-1}/2$ are the best possible parameters in $(1/2, 1)$ such that the double inequality (3.11) holds for all $a, b>0$ with $a\neq b$.

If $1/2+\sqrt{2}/4< p<1$, then (3.18) leads to

$$ g_{1}\bigl(0^{+}\bigr)< 0. $$

(3.25)

Equations (3.12), (3.14), and (3.16) and inequality (3.25) imply that there exists $\delta_{3}\in(0, 1)$ such that

$$ T^{\ast}(a,b)< Q\bigl[pa+(1-p)b, pb+(1-p)a\bigr] $$

for all $a>b>0$ with $(a-b)/(a+b)\in(0, \delta_{3})$.

If $1/2< p<1/2+\sqrt{16/\pi^{2}-1}/2$, then (3.15) leads to

$$ g\bigl(1^{-}\bigr)>0. $$

(3.26)

Equation (3.12) and inequality (3.26) imply that there exists $\delta _{4}\in(0, 1)$ such that

$$ T^{\ast}(a,b)>Q\bigl[pa+(1-p)b, pb+(1-p)a\bigr] $$

for all $a>b>0$ with $(a-b)/(a+b)\in(1-\delta_{4}, 1)$. □

Let $r\in(0, 1)$, $r^{\ast}=r^{2}/(1+\sqrt{1-r^{2}})^{2}$, $a=1$, $b=\sqrt{1-r^{2}}$, $\alpha=3/4$, $\beta=4(4-\pi)/[\pi(\pi-2)]$, $\lambda=1/2+\sqrt{2}/4$, and $\mu=1/2+\sqrt{16/\pi^{2}-1}/2$. Then Theorems 3.1 and 3.2 lead to Corollary 3.3 as follows.

Corollary 3.3

The double inequalities

$$\begin{aligned} &\frac{\pi(1+\sqrt{1-r^{2}})}{32} \biggl[5+3 \biggl(r^{\ast}+\frac{1}{r^{\ast }} \biggr) \arctan \bigl(r^{\ast} \bigr) \biggr]\\ &\quad< \mathcal{E}(r)< \frac{1+\sqrt{1-r^{2}}}{4(\pi-2)} \biggl[\pi^{2}-8+2(4-\pi) \biggl(r^{\ast }+ \frac{1}{r^{\ast}} \biggr)\arctan \bigl(r^{\ast} \bigr) \biggr] \end{aligned}$$

and

$$ \frac{\pi\sqrt{6+2\sqrt{1-r^{2}}-3r^{2}}}{4\sqrt{2}}< \mathcal {E}(r)< \frac{\sqrt{8+(\pi^{2}-8)\sqrt{1-r^{2}}-4r^{2}}}{2} $$

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