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 [3–9]. 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 [13–17], 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 \((a-b)/(a+b)\in(1-\delta_{4},1)\).

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