Open Access

Optimal bounds for the first and second Seiffert means in terms of geometric, arithmetic and contraharmonic means

Journal of Inequalities and Applications20152015:44

https://doi.org/10.1186/s13660-015-0570-2

Received: 30 October 2014

Accepted: 20 January 2015

Published: 4 February 2015

Abstract

In this paper, we find the greatest values α, λ and the least values β, μ such that the double inequalities \(\alpha[G(a,b)/3+2A(a,b)/3]+(1-\alpha)G^{1/3}(a,b)A^{2/3}(a,b)< P(a,b) <\beta[G(a,b)/3+2A(a,b)/3]+(1-\beta)G^{1/3}(a,b)A^{2/3}(a,b)\) and \(\lambda [C(a,b)/3+2A(a,b)/3 ]+ (1-\lambda)C^{1/3}(a,b) A^{2/3}(a,b)< T(a,b)<\mu [C(a,b)/3+2A(a,b)/3 ]+(1-\mu)C^{1/3}(a,b) A^{2/3}(a,b)\) hold for all \(a,b>0\) with \(a\neq b\). Here \(G(a,b)\), \(A(a,b)\), \(C(a,b)\), \(P(a,b)\) and \(T(a,b)\) denote the geometric, arithmetic, contraharmonic, first Seiffert and second Seiffert means of two positive numbers a and b, respectively.

Keywords

first Seiffert meansecond Seiffert meangeometric meanarithmetic meancontraharmonic mean

MSC

26E60

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 \((a-b)/(a+b)\in(1-\delta_{4},1)\).

 □

Declarations

Acknowledgements

The research was supported by the Natural Science Foundation of China under Grants 11401191, 11171307 and 61374086, and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
School of Mathematics and Computation Sciences, Hunan City University
(2)
School of Distance Education, Huzhou Broadcast and TV University
(3)
Department of Mathematics, Huzhou University

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© Chu et al.; licensee Springer. 2015