## 1 Introduction

For $$a,b>0$$ with $$a\neq b$$, the first and second Seiffert means $$P(a,b)$$ [1] and $$T(a,b)$$ [2] are defined by

$$P(a,b)=\frac{a-b}{4\arctan(\sqrt{{a}/{b}})-\pi}$$

and

$$T(a,b)=\frac{a-b}{2\arctan[(a-b)/(a+b)]},$$
(1.1)

respectively.

Recently, both means P and T have been the subject of intensive research. In particular, many remarkable inequalities for P and T can be found in the literature [39]. The first Seiffert mean $$P(a,b)$$ can be rewritten as (see [10, Eq. (2.4)])

$$P(a,b)=\frac{a-b}{2\arcsin[(a-b)/(a+b)]}.$$
(1.2)

Let $$H(a,b)={2ab}/({a+b})$$, $$G(a,b)=\sqrt{ab}$$, $$L(a,b)=(b-a)/(\log b-\log a)$$, $$I(a,b)={1}/{e}({b^{b}}/{a^{a}})^{{1}/({b-a})}$$, $$A(a,b)=(a+b)/2$$, $$Q(a,b)=\sqrt{(a^{2}+b^{2})/2}$$, $$C(a,b)=(a^{2}+b^{2})/(a+b)$$, $$L_{r}(a,b)=(a^{r+1}+b^{r+1})/(a^{r}+b^{r})$$, and $$M_{r}(a,b)=[(a^{r}+b^{r})/2]^{1/r}$$ ($$r\neq0$$) and $$M_{0}(a,b)=G(a,b)$$ be the harmonic, geometric, logarithmic, identric, arithmetic, quadratic, contraharmonic, rth Lehmer and rth power means of two distinct positive real numbers a and b, respectively. Then both $$L_{r}(a,b)$$ and $$M_{r}(a,b)$$ are strictly increasing with respect to $$r\in\mathbb{R}$$ for fixed $$a, b>0$$ with $$a\neq b$$, and the inequalities

\begin{aligned} H(a,b)&=L_{-1}(a,b)=M_{-1}(a,b)< G(a,b)=L_{-1/2}(a,b) \\ &=M_{0}(a,b)<L(a,b)<P(a,b)<I(a,b)<A(a,b)=L_{0}(a,b) \\ &=M_{1}(a,b)<T(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 $$a\neq b$$.

Jagers [11] and Seiffert [2] proved that the inequalities

$$M_{1/2}(a,b)< P(a,b)<M_{2/3}(a,b),\qquad M_{1}(a,b)<T(a,b)<M_{2}(a,b)$$

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

Costin and Toader [12] proved that the double inequality

$$M_{4/3}(a,b)< T(a,b)<M_{5/3}(a,b)$$

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

In [1317], the authors proved that the inequalities

\begin{aligned}& M_{p}(a,b)< P(a,b)<M_{q}(a,b), \qquad M_{r}(a,b)<T(a,b)<M_{s}(a,b), \\& L_{\alpha}(a,b)<P(a,b)<L_{\beta}(a,b),\qquad L_{\sigma}(a,b)<T(a,b)<L_{\tau}(a,b), \\& P(a,b)> \biggl[\frac{b^{\lambda}-a^{\lambda}}{\lambda(\log b-\log a)} \biggr]^{1/\lambda},\qquad T(a,b)> \biggl[ \frac{b^{\mu}-a^{\mu}}{\mu(\log b-\log a)} \biggr]^{1/\mu} \end{aligned}

hold for $$a,b>0$$ with $$a\neq b$$ if and only if $$p\leq\log\pi/\log 2$$, $$q\geq2/3$$, $$r\leq\log2/(\log\pi-\log2)$$, $$s\geq5/3$$, $$\alpha\leq-1/6$$, $$\beta\geq0$$, $$\sigma\leq0$$, $$\tau\geq1/3$$, $$\lambda\geq2$$ and $$\mu\geq5$$.

Gao [18] proved that $$\alpha=e/\pi$$, $$\beta=1$$, $$\lambda=1$$ and $$\mu=2e/\pi$$ are the best possible constants such that the double inequalities

$$\alpha I(a,b)< P(a,b)<\beta I(a,b),\qquad \lambda I(a,b)<T(a,b)<\mu I(a,b)$$

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

In [19, 20], the authors proved that the double inequalities

\begin{aligned}& \alpha_{1}C(a,b)+(1-\alpha_{1})G(a,b)< P(a,b)< \beta_{1}C(a,b)+(1-\beta _{1})G(a,b), \\& \alpha_{2}A(a,b)+(1-\alpha_{2})G(a,b)<P(a,b)< \beta_{2}A(a,b)+(1-\beta _{2})G(a,b), \\& \alpha_{3}C(a,b)+(1-\alpha_{3})H(a,b)<T(a,b)< \beta_{3}C(a,b)+(1-\beta_{3})H(a,b) \end{aligned}

hold for $$a,b>0$$ with $$a\neq b$$ if and only if $$\alpha_{1}\leq2/9$$, $$\beta_{1}\geq1/\pi$$, $$\alpha_{2}\leq2/\pi$$, $$\beta_{2}\geq2/3$$, $$\alpha_{3}\leq2/\pi$$ and $$\beta_{3}\geq2/3$$.

Let $$p\geq1/2$$, $$q\geq1$$, $$t_{1}, t_{2}\in(1/2, 1)$$ and $$t_{3}, t_{4}\in(0, 1/2)$$. Then the authors in [21, 22] proved that the double inequalities

\begin{aligned}& C^{p}\bigl[t_{1}a+(1-t_{1})b, t_{1}b+(1-t_{1})a\bigr]A^{1-p}(a,b) \\& \quad < T(a,b)<C^{p}\bigl[t_{2}a+(1-t_{2})b, t_{2}b+(1-t_{2})a\bigr]A^{1-p}(a,b), \\& G^{q}\bigl[t_{3}a+(1-t_{3})b, t_{3}b+(1-t_{3})a\bigr]A^{1-q}(a,b) \\& \quad <P(a,b)<G^{q}\bigl[t_{4}a+(1-t_{4})b, t_{4}b+(1-t_{4})a\bigr]A^{1-q}(a,b) \end{aligned}

hold for $$a,b>0$$ with $$a\neq b$$ if and only if $$t_{1}\leq [1+\sqrt{(4/\pi)^{1/p}-1}]/2$$, $$t_{2}\geq1/2+\sqrt{3p}/(6p)$$, $$t_{3}\leq[1-\sqrt{1-(2/\pi)^{2/q}}]/2$$ and $$t_{4}\geq (1-1/\sqrt{3q})/2$$.

Yang et al. [23] proved that the double inequality

$$\frac{Q^{2}(a,b)}{L_{p-1}(a,b)}< T(a,b)<\frac{Q^{2}(a,b)}{L_{q-1}(a,b)}$$

holds for $$a,b>0$$ with $$a\neq b$$ if and only if $$p\geq5/3$$ and $$q\leq1$$.

Sándor [24] and Jiang et al. [25] proved that the inequalities

\begin{aligned}& G^{1/3}(a,b)A^{2/3}(a,b)< P(a,b)<\frac{1}{3}G(a,b)+ \frac{2}{3}A(a,b), \end{aligned}
(1.3)
\begin{aligned}& T(a,b)< \frac{1}{3}C(a,b)+\frac{2}{3}A(a,b) \end{aligned}
(1.4)

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

In [26], Sándor found that $$T(a,b)$$ is the common limit of the sequences $$\{u_{n}\}$$ and $$\{v_{n}\}$$ given by

$$u_{0}=A(a,b),\qquad v_{0}=Q(a,b),\qquad u_{n+1}= \frac{u_{n}+v_{n}}{2},\qquad v_{n+1}=\sqrt{u_{n+1}v_{n}} \quad (n\geq0)$$

and established a more general inequality

$$\sqrt[3]{u_{n}v_{n}^{2}}< T(a,b)< \frac{u_{n}+2v_{n}}{3}$$
(1.5)

for all $$n\geq0$$ and $$a, b>0$$ with $$a\neq b$$. In particular, let $$n=0$$, then (1.4) and (1.5) together with the identity $$Q^{2/3}(a,b)A^{1/3}(a,b)=C^{1/3}(a,b)A^{2/3}(a,b)$$ lead to

$$C^{1/3}(a,b)A^{2/3}(a,b)< T(a,b)< \frac{1}{3}C(a,b)+\frac{2}{3}A(a,b)$$
(1.6)

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

Motivated by inequalities (1.3) and (1.6), we naturally ask: what are the best possible parameters α, β, λ and μ such that the double inequalities

\begin{aligned}& \alpha \biggl[\frac{1}{3}G(a,b)+\frac{2}{3}A(a,b) \biggr]+(1-\alpha )G^{1/3}(a,b)A^{2/3}(a,b) \\& \quad < P(a,b)<\beta \biggl[\frac{1}{3}G(a,b)+\frac{2}{3}A(a,b) \biggr]+(1-\beta )G^{1/3}(a,b)A^{2/3}(a,b), \\& \lambda \biggl[\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b) \biggr]+(1- \lambda )C^{1/3}(a,b)A^{2/3}(a,b) \\& \quad <T(a,b)<\mu \biggl[\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b) \biggr]+(1-\mu )C^{1/3}(a,b)A^{2/3}(a,b) \end{aligned}

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

## 2 Lemmas

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

### Lemma 2.1

Let $$g(t)=-p^{2}t^{6}-2p^{2}t^{5}+3(p^{2}-4p+2)t^{4}+2(2p^{2}-9p+6)t^{3}-(4p^{2}+6p-9)t^{2}+6(1-p)t+3(1-p)$$. Then the following statements are true:

1. (1)

If $$p=4/5$$, then $$g(t)>0$$ for all $$t\in(0,1)$$.

2. (2)

If $$p=3/\pi$$, then there exists $$\lambda_{0}\in(0,1)$$ such that $$g(t)>0$$ for $$t\in(0,\lambda_{0})$$ and $$g(t)<0$$ for $$t\in(\lambda_{0},1)$$.

### Proof

Part (1) follows easily from

$$g(t)=\frac{1}{25}(1-t) \bigl(16t^{5}+48t^{4}+90t^{3}+86t^{2}+45t+15 \bigr)>0$$

for all $$t\in(0,1)$$ if $$p=4/5$$.

For part (2), if $$p=3/\pi$$, then numerical computations lead to

\begin{aligned}& p^{2}-4p+2=\frac{2\pi^{2}-12\pi+9}{\pi^{2}}< 0, \end{aligned}
(2.1)
\begin{aligned}& 2p^{2}-9p+6=\frac{6\pi^{2}-27\pi+18}{\pi^{2}}< 0, \end{aligned}
(2.2)
\begin{aligned}& 4p^{2}+6p-9=\frac{-9\pi^{2}+18\pi+36}{\pi^{2}}>0, \end{aligned}
(2.3)
\begin{aligned}& g(0)=\frac{3(\pi-3)}{\pi}>0, \end{aligned}
(2.4)
\begin{aligned}& g(1)=\frac{9(4\pi-15)}{\pi}< 0, \end{aligned}
(2.5)
\begin{aligned}& g^{\prime}(t)=-6p^{2}t^{5}-10p^{2}t^{4}+12 \bigl(p^{2}-4p+2\bigr)t^{3} \\& \hphantom{g^{\prime}(t)=}{}+6\bigl(2p^{2}-9p+6\bigr)t^{2}-2 \bigl(4p^{2}+6p-9\bigr)t+6(1-p), \\& g^{\prime}(0)=\frac{6(\pi-3)}{\pi}>0, \end{aligned}
(2.6)
\begin{aligned}& g^{\prime}(1)=\frac{84\pi-360}{\pi}< 0, \end{aligned}
(2.7)
\begin{aligned}& g^{\prime\prime}(t)=-30p^{2}t^{4}-40p^{2}t^{3}+36 \bigl(p^{2}-4p+2\bigr)t^{2} \\& \hphantom{g^{\prime\prime}(t)=}{}+12\bigl(2p^{2}-9p+6\bigr)t-2\bigl(4p^{2}+6p-9 \bigr). \end{aligned}
(2.8)

It follows from (2.1)-(2.3) and (2.8) that $$g^{\prime}$$ is strictly decreasing on $$(0, 1)$$. Then (2.6) and (2.7) lead to the conclusion that there exists $$\lambda_{1}\in(0, 1)$$ such that g is strictly increasing on $$(0, \lambda_{1}]$$ and strictly decreasing on $$[\lambda_{1}, 1)$$.

Therefore, part (2) follows from (2.4) and (2.5) together with the piecewise monotonicity of $$g^{\prime}$$. □

### Lemma 2.2

Let $$h(t)=q(q+3)t^{4}+2q(q+3)t^{3}-3(q^{2}-6q+1)t^{2}-2(2q^{2}-9q+3)t+4q^{2}$$. Then the following statements are true:

1. (1)

If $$q=1/5$$, then $$h(t)>0$$ for $$t\in(1,\sqrt[3]{2})$$.

2. (2)

If $$q=[3(\sqrt[3]{2}\pi-4)]/[(3\sqrt[3]{2}-4)\pi]=0.1814\ldots$$ , then there exists $$\mu_{0}\in(1, \sqrt[3]{2})$$ such that $$h(t)<0$$ for $$t\in(1, \mu_{0})$$ and $$h(t)>0$$ for $$t\in(\mu_{0}, \sqrt[3]{2})$$.

### Proof

Part (1) follows easily from

$$h(t)=\frac{4(t-1)}{25}\bigl(4t^{3}+12t^{2}+15t-1\bigr)>0$$

for all $$t\in(1,\sqrt[3]{2})$$ if $$q=1/5$$.

For part (2), if $$q=[3(\sqrt[3]{2}\pi-4)]/[(3\sqrt[3]{2}-4)\pi]$$, then numerical computations lead to

\begin{aligned}& q^{2}-6q+1=-0.0556\ldots< 0, \end{aligned}
(2.9)
\begin{aligned}& h(1)=9(5q-1)=-0.836\ldots< 0, \end{aligned}
(2.10)
\begin{aligned}& h\bigl(\sqrt[3]{2}\bigr)=0.548\ldots>0, \end{aligned}
(2.11)
\begin{aligned}& h^{\prime}(t)=4q(q+3)t^{3}+6q(q+3)t^{2}-6 \bigl(q^{2}-6q+1\bigr)t-2\bigl(2q^{2}-9q+3 \bigr). \end{aligned}
(2.12)

It follows from (2.9) and (2.12) that

\begin{aligned} h^{\prime}(t)&>4q(q+3)+6q(q+3)-6\bigl(q^{2}-6q+1\bigr)-2 \bigl(2q^{2}-9q+3\bigr) \\ &=12(7q-1)=3.239\ldots>0 \end{aligned}
(2.13)

for $$t\in(1,\sqrt[3]{2})$$.

Therefore, part (2) follows easily from (2.10) and (2.11) together with (2.13). □

## 3 Main results

### Theorem 3.1

The double inequality

\begin{aligned}& \alpha \biggl[\frac{1}{3}G(a,b)+\frac{2}{3}A(a,b) \biggr]+(1-\alpha )G^{1/3}(a,b)A^{2/3}(a,b) \\& \quad < P(a,b)<\beta \biggl[\frac{1}{3}G(a,b)+\frac{2}{3}A(a,b) \biggr]+(1-\beta )G^{1/3}(a,b)A^{2/3}(a,b) \end{aligned}

holds for all $$a,b>0$$ with $$a\neq b$$ if and only if $$\alpha\leq 4/5$$ and $$\beta\geq3/\pi$$.

### Proof

Firstly, we prove that the inequalities

\begin{aligned}& P(a,b)>\frac{4}{5} \biggl[\frac{1}{3}G(a,b)+\frac{2}{3}A(a,b) \biggr]+\frac {1}{5}G^{1/3}(a,b)A^{2/3}(a,b), \end{aligned}
(3.1)
\begin{aligned}& P(a,b)< \frac{3}{\pi} \biggl[\frac{1}{3}G(a,b)+\frac{2}{3}A(a,b) \biggr]+ \biggl(1-\frac{3}{\pi} \biggr)G^{1/3}(a,b)A^{2/3}(a,b) \end{aligned}
(3.2)

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

Since $$P(a,b)$$, $$A(a,b)$$ and $$G(a,b)$$ are symmetric and homogenous of degree 1, without loss of generality, we assume that $$a>b$$. Let $$x=(a-b)/(a+b)\in(0, 1)$$ and $$p\in(0, 1)$$. Then (1.2) leads to

\begin{aligned}& \frac{P(a,b)}{A(a,b)}=\frac{x}{\arcsin(x)},\qquad \frac{G(a,b)}{A(a,b)}= \sqrt{1-x^{2}}, \\& \frac {P(a,b)-G^{1/3}(a,b)A^{2/3}(a,b)}{G(a,b)/3+2A(a,b)/3-G^{1/3}(a,b)A^{2/3}(a,b)} \\& \quad =\frac{x/\arcsin(x)-\sqrt[6]{1-x^{2}}}{2/3+\sqrt{1-x^{2}}/3-\sqrt[6]{1-x^{2}}}, \end{aligned}
(3.3)
\begin{aligned}& \lim_{x\rightarrow 0^{+}}\frac{x/\arcsin(x)-\sqrt[6]{1-x^{2}}}{2/3+\sqrt{1-x^{2}}/3-\sqrt [6]{1-x^{2}}}=\frac{4}{5}, \end{aligned}
(3.4)
\begin{aligned}& \lim_{x\rightarrow 1^{-}}\frac{x/\arcsin(x)-\sqrt[6]{1-x^{2}}}{2/3+\sqrt{1-x^{2}}/3-\sqrt [6]{1-x^{2}}}=\frac{3}{\pi}, \end{aligned}
(3.5)
\begin{aligned}& P(a, b)-p \biggl[\frac{1}{3}G(a,b)+\frac{2}{3}A(a,b) \biggr]-(1-p)G^{1/3}(a,b)A^{2/3}(a,b) \\& \quad =A(a,b) \biggl[\frac{x}{\arcsin(x)}-p \biggl(\frac{1}{3} \sqrt{1-x^{2}}+\frac {2}{3} \biggr)-(1-p)\sqrt[6]{1-x^{2}} \biggr] \\& \quad =\frac{A(a,b)[p(2+\sqrt{1-x^{2}})+3(1-p)\sqrt[6]{1-x^{2}}]}{3\arcsin(x)}G(x), \end{aligned}
(3.6)

where

\begin{aligned}& G(x)=\frac{3x}{p(2+\sqrt{1-x^{2}})+3(1-p)\sqrt[6]{1-x^{2}}}-\arcsin(x), \\& G(0)=0, \end{aligned}
(3.7)
\begin{aligned}& G(1)=\frac{3}{2p}-\frac{\pi}{2}, \end{aligned}
(3.8)
\begin{aligned}& G^{\prime}(x)=\frac{(1-\sqrt[6]{1-x^{2}})^{2}}{\sqrt [6]{(1-x^{2})^{5}}[p(2+\sqrt{1-x^{2}})+3(1-p)\sqrt [6]{1-x^{2}}]^{2}}g\bigl(\sqrt[6]{1-x^{2}} \bigr), \end{aligned}
(3.9)

where the function $$g(\cdot)$$ is defined as in Lemma 2.1.

We divide the proof into two cases.

Case 1. $$p=4/5$$. Then (3.1) follows easily from (3.6), (3.7), (3.9) and Lemma 2.1(1).

Case 2. $$p=3/\pi$$. Then Lemma 2.1(2) and (3.9) lead to the conclusion that there exists $$x_{0}\in(0, 1)$$ such that G is strictly decreasing on $$(0, x_{0}]$$ and strictly increasing on $$[x_{0}, 1)$$.

Note that (3.8) becomes

$$G(1)=0.$$
(3.10)

It follows from (3.7) and (3.10) together with the piecewise monotonicity of G that

$$G(x)< 0$$
(3.11)

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

Therefore, (3.2) follows from (3.6) and (3.11), and Theorem 3.1 follows from (3.1) and (3.2) in conjunction with the following statements.

• If $$\alpha>4/5$$, then equations (3.3) and (3.4) lead to the conclusion that there exists $$0<\delta_{1}<1$$ such that $$P(a,b)<\alpha [G(a,b)/3+2A(a,b)/3 ]+(1-\alpha )G^{1/3}(a,b)A^{2/3}(a,b)$$ for all $$a,b>0$$ with $$(a-b)/(a+b)\in(0,\delta_{1})$$.

• If $$\beta<3/\pi$$, then equations (3.3) and (3.5) imply that there exists $$0<\delta_{2}<1$$ such that $$P(a,b)>\beta [G(a,b)/3+2A(a,b)/3 ]+(1-\beta)G^{1/3}(a,b)A^{2/3}(a,b)$$ for all $$a,b>0$$ with $$(a-b)/(a+b)\in(1-\delta_{2},1)$$.

□

### Theorem 3.2

The double inequality

\begin{aligned}& \lambda \biggl[\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b) \biggr]+(1- \lambda )C^{1/3}(a,b)A^{2/3}(a,b) \\& \quad < T(a,b)<\mu \biggl[\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b) \biggr]+(1-\mu )C^{1/3}(a,b)A^{2/3}(a,b) \end{aligned}

holds for all $$a,b>0$$ with $$a\neq b$$ if and only if $$\lambda\leq [3(\sqrt[3]{2}\pi-4)]/[(3\sqrt[3]{2}-4)\pi]=0.1814\ldots$$ and $$\mu \geq1/5$$.

### Proof

Let $$\lambda^{\ast}=[3(\sqrt[3]{2}\pi-4)]/[(3\sqrt [3]{2}-4)\pi]$$. Firstly, we prove that the inequalities

\begin{aligned}& T(a,b)< \frac{1}{5} \biggl[\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b) \biggr]+\frac {4}{5}C^{1/3}(a,b)A^{2/3}(a,b), \end{aligned}
(3.12)
\begin{aligned}& T(a,b)>\lambda^{\ast} \biggl[\frac{1}{3}C(a,b)+ \frac{2}{3}A(a,b) \biggr]+\bigl(1-\lambda^{\ast} \bigr)C^{1/3}(a,b)A^{2/3}(a,b) \end{aligned}
(3.13)

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

Since $$T(a,b)$$, $$A(a,b)$$ and $$C(a,b)$$ are symmetric and homogenous of degree 1, without loss of generality, we assume that $$a>b$$. Let $$x=(a-b)/(a+b)\in(0, 1)$$ and $$q\in(0, 1)$$. Then (1.1) leads to

\begin{aligned}& \frac{T(a,b)}{A(a,b)}=\frac{x}{\arctan(x)},\qquad \frac{C(a,b)}{A(a,b)}=1+x^{2}, \\& \frac {T(a,b)-C^{1/3}(a,b)A^{2/3}(a,b)}{C(a,b)/3+2A(a,b)/3-C^{1/3}(a,b)A^{2/3}(a,b)} \\& \quad =\frac{x/\arctan{x}-\sqrt[3]{1+x^{2}}}{2/3+(1+x^{2})/3-\sqrt[3]{1+x^{2}}}, \end{aligned}
(3.14)
\begin{aligned}& \lim_{x\rightarrow 0^{+}}\frac{x/\arctan{x}-\sqrt[3]{1+x^{2}}}{2/3+(1+x^{2})/3-\sqrt [3]{1+x^{2}}}=\frac{1}{5}, \end{aligned}
(3.15)
\begin{aligned}& \lim_{x\rightarrow 1^{-}}\frac{x/\arctan{x}-\sqrt[3]{1+x^{2}}}{2/3+(1+x^{2})/3-\sqrt [3]{1+x^{2}}}=\lambda^{\ast}, \end{aligned}
(3.16)
\begin{aligned}& T(a,b)-q \biggl[\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b) \biggr]-(1-q)C^{1/3}(a,b)A^{2/3}(a,b) \\& \quad =A(a,b) \biggl[\frac{x}{\arctan(x)}-\frac{q}{3}\bigl(3+x^{2} \bigr)-(1-q)\sqrt [3]{1+x^{2}} \biggr] \\& \quad =\frac{A(a,b)[q(3+x^{2})+3(1-q)\sqrt[3]{1+x^{2}}]}{3\arctan(x)}H(x), \end{aligned}
(3.17)

where

\begin{aligned}& H(x)=\frac{3x}{q(3+x^{2})+3(1-q)\sqrt[3]{1+x^{2}}}-\arctan(x), \\& H(0)=0, \end{aligned}
(3.18)
\begin{aligned}& H(1)=\frac{3}{4q+3\sqrt[3]{2}(1-q)}-\frac{\pi}{4}, \end{aligned}
(3.19)
\begin{aligned}& H^{\prime}(x)=-\frac{(\sqrt [3]{1+x^{2}}-1)^{2}}{(1+x^{2})[q(3+x^{2})+3(1-q)\sqrt [3]{1+x^{2}}]^{2}}h\bigl(\sqrt[3]{1+x^{2}} \bigr), \end{aligned}
(3.20)

where the function $$h(\cdot)$$ is defined as in Lemma 2.2.

We divide the proof into two cases.

Case 1. $$q=1/5$$. Then (3.12) follows easily from Lemma 2.2(1), (3.17), (3.18) and (3.20).

Case 2. $$q=\lambda^{\ast}$$. Then Lemma 2.2(2) and (3.20) lead to the conclusion that there exists $$x^{\ast}\in(0, 1)$$ such that H is strictly increasing on $$(0, x^{\ast}]$$ and strictly decreasing on $$[x^{\ast}, 1)$$.

Note that (3.19) becomes

$$H(1)=0.$$
(3.21)

Therefore, (3.13) follows from (3.17), (3.18), (3.21) and the piecewise monotonicity of H, and Theorem 3.2 follows from (3.12) and (3.13) in conjunction with the following statements.

• If $$\mu<1/5$$, then equations (3.14) and (3.15) lead to the conclusion that there exists $$0<\delta_{3}<1$$ such that $$T(a,b)>\mu [C(a,b)/3+2A(a,b)/3 ]+(1-\mu)C^{1/3}(a,b)A^{2/3}(a,b)$$ for all $$a,b>0$$ with $$(a-b)/(a+b)\in(0,\delta_{3})$$.

• If $$\lambda>\lambda^{\ast}$$, then equations (3.14) and (3.16) imply that there exists $$0<\delta_{4}<1$$ such that $$T(a,b)<\lambda [C(a,b)/3+2A(a,b)/3 ]+(1-\lambda )C^{1/3}(a,b)A^{2/3}(a,b)$$ for all $$a,b>0$$ with $$(a-b)/(a+b)\in(1-\delta_{4},1)$$.

□