Open Access

Sharp bounds for Sándor mean in terms of arithmetic, geometric and harmonic means

Journal of Inequalities and Applications20152015:221

https://doi.org/10.1186/s13660-015-0741-1

Received: 23 February 2015

Accepted: 23 June 2015

Published: 9 July 2015

Abstract

In the article, we present the best possible parameters \(\alpha_{1}, \alpha_{2}, \beta_{1}, \beta_{2}\in(0, 1)\) and \(\alpha_{3}, \alpha_{4}, \beta_{3}, \beta_{4}\in(0, 1/2)\) such that the double inequalities
$$\begin{aligned}& \alpha_{1}A(a,b)+(1-\alpha_{1})H(a,b)< X(a,b)< \beta _{1}A(a,b)+(1-\beta_{1})H(a,b), \\ & \alpha_{2}A(a,b)+(1-\alpha_{2})G(a,b)< X(a,b)< \beta _{2}A(a,b)+(1-\beta_{2})G(a,b), \\ & H\bigl[\alpha_{3}a+(1-\alpha_{3})b, \alpha_{3}b+(1- \alpha _{3})a\bigr]< X(a,b)< H\bigl[\beta_{3}a+(1- \beta_{3})b, \beta_{3}b+(1-\beta_{3})a\bigr], \\ & G\bigl[\alpha_{4}a+(1-\alpha_{4})b, \alpha_{4}b+(1- \alpha _{4})a\bigr]< X(a,b)< G\bigl[\beta_{4}a+(1- \beta_{4})b, \beta_{4}b+(1-\beta_{4})a\bigr] \end{aligned}$$
hold for all \(a, b>0\) with \(a\neq b\). Here, \(X(a,b)\), \(A(a,b)\), \(G(a,b)\) and \(H(a,b)\) are the Sándor, arithmetic, geometric and harmonic means of a and b, respectively.

Keywords

Sándor mean arithmetic mean geometric mean harmonic mean

MSC

26E60

1 Introduction

Let \(r\in\mathbb{R}\) and \(a, b>0\) with \(a\neq b\). Then the harmonic mean \(H(a,b)\), geometric mean \(G(a,b)\), logarithmic mean \(L(a,b)\), Seiffert mean \(P(a,b)\), arithmetic mean \(A(a,b)\), Sándor mean \(X(a,b)\) [1] and rth power mean \(M_{r}(a,b)\) of a and b are, respectively, defined by
$$\begin{aligned}& H(a,b)=\frac{2ab}{a+b}, \qquad G(a,b)=\sqrt{ab}, \qquad L(a,b)= \frac{a-b}{\log a-\log b}, \end{aligned}$$
(1.1)
$$\begin{aligned}& P(a,b)=\frac{a-b}{2\arcsin (\frac{a-b}{a+b} )},\qquad A(a,b)=\frac{a+b}{2}, \qquad X(a,b)=A(a,b)e^{\frac{G(a,b)}{P(a,b)}-1} \end{aligned}$$
(1.2)
and
$$ M_{r}(a,b)= \biggl(\frac{a^{r}+b^{r}}{2} \biggr)^{1/r} \quad (r\neq0),\qquad M_{0}(a,b)=\sqrt{ab}. $$
(1.3)
It is well known that \(M_{r}(a, b)\) is continuous and strictly increasing with respect to \(r\in\mathbb{R}\) for fixed \(a, b>0\) with \(a\neq b\), and the inequalities
$$ H(a,b)< G(a,b)< L(a,b)< P(a,b)< A(a,b) $$
(1.4)
hold for all \(a, b>0\) with \(a\neq b\).
Recently, the Sándor mean has attracted the attention of several researchers. In [2], Sándor established the inequalities
$$\begin{aligned}& X(a,b)< \frac{P^{2}(a,b)}{A(a,b)}, \qquad \frac {A(a,b)G(a,b)}{P(a,b)}< X(a,b)< \frac{A(a,b)P(a,b)}{2P(a,b)-G(a,b)}, \\& X(a,b)>\frac{A(a,b)L(a,b)}{P(a,b)}e^{\frac{G(a,b)}{L(a,b)}-1},\qquad X(a,b)>\frac{A(a,b)[P(a,b)+G(a,b)]}{3P(a,b)-G(a,b)}, \\& \frac{A^{2}(a,b)G(a,b)}{P(a,b)L(a,b)}e^{\frac {L(a,b)}{A(a,b)}-1}< X(a,b)< A(a,b) \biggl[\frac{1}{e}+ \biggl(1-\frac {1}{e} \biggr)\frac{G(a,b)}{P(a,b)} \biggr], \\& A(a,b)+G(a,b)-P(a,b)< X(a,b)< A^{-1/3}(a,b) \biggl[\frac {A(a,b)+G(a,b)}{2} \biggr]^{4/3}, \\& P^{1/(\log\pi-\log2)}(a,b)A^{1-1/(\log\pi-\log 2)}(a,b) \\& \quad < X(a,b)< P^{-1}(a,b) \biggl[ \frac{A(a,b)+G(a,b)}{2} \biggr]^{2} \end{aligned}$$
for all \(a, b>0\) with \(a\neq b\).
Yang et al. [3] proved that the double inequality
$$ M_{p}(a,b)< X(a,b)< M_{q}(a,b) $$
(1.5)
holds for all \(a, b>0\) with \(a\neq b\) if and only if \(p\leq1/3\) and \(q\geq\log2/(1+\log2)=0.4903\ldots\) .
In [4], Zhou et al. proved that the double inequality
$$ H_{\alpha}(a,b)< X(a,b)< H_{\beta}(a,b) $$
(1.6)
holds for all \(a, b>0\) with \(a\neq b\) if and only if \(\alpha\leq1/2\) and \(\beta\geq\log3/(1+\log2)=0.6488\ldots\) , where \(H_{p}(a,b)=[(a^{p}+(ab)^{p/2}+b^{p})/3]^{1/p}\) (\(p\neq0\)) and \(H_{0}(p)=\sqrt{ab}\) is the pth power-type Heronian mean of a and b.
Inequalities (1.4) and (1.5) together with the identities \(H(a, b)=M_{-1}(a, b)\), \(G(a, b)=M_{0}(a, b)\) and \(A(a, b)=M_{1}(a, b)\) lead to the inequalities
$$ H(a,b)< G(a,b)< X(a,b)< A(a,b) $$
(1.7)
for all \(a, b>0\) with \(a\neq b\).
Let \(a, b>0\) with \(a\neq b\), \(x\in[0, 1/2]\), \(f(x)=H[xa+(1-x)b, xb+(1-x)a]\) and \(g(x)=G[xa+(1-x)b, xb+(1-x)a]\). Then both functions f and g are continuous and strictly increasing on \([0, 1/2]\). Note that
$$ f(0)=H(a,b)< X(a,b)< f(1/2)=A(a,b) $$
(1.8)
and
$$ g(0)=G(a,b)< X(a,b)< g(1/2)=A(a,b). $$
(1.9)
Motivated by inequalities (1.7)-(1.9), we naturally ask: what are the best possible parameters \(\alpha_{1}, \alpha_{2}, \beta_{1}, \beta _{2}\in(0, 1)\) and \(\alpha_{3}, \alpha_{4}, \beta_{3}, \beta _{4}\in(0, 1/2)\) such that the double inequalities
$$\begin{aligned}& \alpha_{1}A(a,b)+(1-\alpha_{1})H(a,b)< X(a,b)< \beta _{1}A(a,b)+(1-\beta_{1})H(a,b), \\& \alpha_{2}A(a,b)+(1-\alpha_{2})G(a,b)< X(a,b)< \beta _{2}A(a,b)+(1-\beta_{2})G(a,b), \\& H\bigl[\alpha_{3}a+(1-\alpha_{3})b, \alpha_{3}b+(1- \alpha _{3})a\bigr]< X(a,b)< H\bigl[\beta_{3}a+(1- \beta_{3})b, \beta_{3}b+(1-\beta_{3})a\bigr], \\& G\bigl[\alpha_{4}a+(1-\alpha_{4})b, \alpha_{4}b+(1- \alpha _{4})a\bigr]< X(a,b)< G\bigl[\beta_{4}a+(1- \beta_{4})b, \beta_{4}b+(1-\beta_{4})a\bigr] \end{aligned}$$
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 prove our main results, we need four lemmas, which we present in this section.

Lemma 2.1

Let \(p\in(0, 1)\) and
$$ f(x)=\frac{x\sqrt{1-x^{2}}[(1-p)x^{2}+1]}{p+(1-p)(1-x^{2})}-\arcsin(x). $$
(2.1)
Then the following statements are true:
  1. (1)

    If \(p=2/3\), then \(f(x)<0\) for all \(x\in(0, 1)\).

     
  2. (2)

    If \(p=1/e\), then there exists \(\lambda_{1}\in(0, 1)\) such that \(f(x)>0\) for \(x\in(0, \lambda_{1})\) and \(f(x)<0\) for \(x\in(\lambda _{1}, 1)\).

     

Proof

Simple computations lead to
$$\begin{aligned}& f(0)=0, \qquad f(1)=-\frac{\pi}{2}, \end{aligned}$$
(2.2)
$$\begin{aligned}& f^{\prime}(x)=\frac{2x^{2}}{\sqrt{1-x^{2}}[p+(1-p)(1-x^{2})]^{2}}f_{1}(x), \end{aligned}$$
(2.3)
where
$$ f_{1}(x)=(1-p)^{2}x^{4}-(1-p) (3-p)x^{2}+2-3p. $$
(2.4)
(1) If \(p=2/3\), then (2.4) leads to
$$ f_{1}(x)=-\frac{x^{2}}{9}\bigl(7-x^{2} \bigr)< 0 $$
(2.5)
for \(x\in(0, 1)\).

Therefore, \(f(x)<0\) for \(x\in(0, 1)\) follows easily from (2.2), (2.3) and (2.5).

(2) If \(p=1/e\), then (2.4) leads to
$$\begin{aligned}& f_{1}(0)=\frac{2e-3}{e}>0, \qquad f_{1}(1)=- \frac{1}{e}< 0, \end{aligned}$$
(2.6)
$$\begin{aligned}& f_{1}^{\prime}(x)=2(1-p)\bigl[2(1-p)x^{2}-(3-p) \bigr]x< -2\bigl(1-p^{2}\bigr)x< 0 \end{aligned}$$
(2.7)
for \(x\in(0, 1)\).

From (2.6) and (2.7) we clearly see that there exists \(\lambda_{0}\in (0, 1)\) such that \(f_{1}(x)>0\) for \(x\in(0, \lambda_{0})\) and \(f_{1}(x)<0\) for \(x\in(\lambda_{0}, 1)\).

We divide the proof into two cases.

Case 1. \(x\in(0, \lambda_{0}]\). Then \(f(x)>0\) follows easily from (2.2) and (2.3) together with \(f_{1}(x)>0\) on the interval \((0, \lambda_{0})\).

Case 2. \(x\in(\lambda_{0}, 1)\). Then (2.3) and \(f_{1}(x)<0\) on the interval \((\lambda_{0}, 1)\) lead to the conclusion that \(f(x)\) is strictly decreasing on \([\lambda_{0}, 1)\).

From (2.2) and \(f(\lambda_{0})>0\) together with the monotonicity of \(f(x)\) on \([\lambda_{0}, 1)\) we clearly see that there exists \(\lambda_{1}\in(\lambda_{0}, 1)\subset(0, 1)\) such that \(f(x)>0\) for \(x\in(\lambda_{0}, \lambda_{1})\) and \(f(x)<0\) for \(x\in(\lambda_{1}, 1)\). □

Lemma 2.2

Let \(p\in(0, 1)\) and
$$ g(x)=\frac{px\sqrt{1-x^{2}}+(1-p)x}{(1-p)\sqrt{1-x^{2}}+p}-\arcsin(x). $$
(2.8)
Then the following statements are true:
  1. (1)

    If \(p=1/3\), then \(g(x)>0\) for all \(x\in(0, 1)\).

     
  2. (2)

    If \(p=1/e\), then there exists \(\mu_{1}\in(0, 1)\) such that \(g(x)<0\) for \(x\in(0, \mu_{1})\) and \(g(x)>0\) for \(x\in(\mu_{1}, 1)\).

     

Proof

Simple computations lead to
$$\begin{aligned}& g(0)=0, \qquad g(1)=\frac{1}{p}-1-\frac{\pi}{2}, \end{aligned}$$
(2.9)
$$\begin{aligned}& g^{\prime}(x)=\frac{x^{2}}{\sqrt{1-x^{2}}[p+(1-p)\sqrt {1-x^{2}}]^{2}}g_{1}(x), \end{aligned}$$
(2.10)
where
$$ g_{1}(x)=p(p-1)\sqrt{1-x^{2}}+1-2p-p^{2}. $$
(2.11)
(1) If \(p=1/3\), then (2.11) leads to
$$ g_{1}(x)=\frac{2}{9}\bigl(1- \sqrt{1-x^{2}}\bigr)>0 $$
(2.12)
for \(x\in(0, 1)\).

Therefore, \(g(x)>0\) for all \(x\in(0, 1)\) follows easily from (2.9), (2.10) and (2.12).

(2) If \(p=1/e\), then (2.11) leads to
$$\begin{aligned}& g_{1}(0)=\frac{e-3}{e}< 0, \qquad g_{1}(1)=\frac{e^{2}-2e-1}{e^{2}}>0, \end{aligned}$$
(2.13)
$$\begin{aligned}& g_{1}^{\prime}(x)=\frac{p(1-p)x}{\sqrt{1-x^{2}}}>0 \end{aligned}$$
(2.14)
for all \(x\in(0, 1)\).

From (2.13) and (2.14) we clearly see that there exists \(\mu_{0}\in (0, 1)\) such that \(g_{1}(x)<0\) for \(x\in(0, \mu_{0})\) and \(g_{1}(x)>0\) for \(x\in(\mu_{0}, 1)\).

We divide the proof into two cases.

Case 1. \(x\in(0, \mu_{0}]\). Then \(g(x)<0\) for \(x\in(0, \mu_{0}]\) follows easily from (2.9) and (2.10) together with \(g_{1}(x)<0\) on the interval \((0, \mu_{0})\).

Case 2. \(x\in(\mu_{0}, 1)\). Then (2.10) and \(g_{1}(x)>0\) on the interval \((\mu_{0}, 1)\) lead to the conclusion that \(g(x)\) is strictly increasing on \([\mu_{0}, 1)\). Note that
$$ g(\mu_{0})< 0, \qquad g(1)=e-1-\frac{\pi}{2}>0. $$
(2.15)

From (2.15) and the monotonicity of \(g(x)\) on the interval \([\mu_{0}, 1)\) we clearly see that there exists \(\mu_{1}\in(\mu_{0}, 1)\subset (0, 1)\) such that \(g(x)<0\) for \(x\in(\mu_{0}, \mu_{1})\) and \(g(x)>0\) for \(x\in(\mu_{1}, 1)\). □

Lemma 2.3

Let \(p\in(0, 1/2)\) and
$$ h(x)=\arcsin(x)-\frac{x\sqrt{1-x^{2}}[1+(1-2p)^{2}x^{2}]}{1-(1-2p)^{2}x^{2}}. $$
(2.16)
Then the following statements are true:
  1. (1)

    If \(p=1/2-\sqrt{3}/6=0.2113\ldots\) , then \(h(x)>0\) for all \(x\in (0, 1)\).

     
  2. (2)

    If \(p=1/2-\sqrt{1-1/e}/2=0.1024\ldots\) , then there exists \(\sigma _{1}\in(0, 1)\) such that \(h(x)<0\) for \(x\in(0, \sigma_{1})\) and \(h(x)>0\) for \(x\in(\sigma_{1}, 1)\).

     

Proof

Simple computations lead to
$$\begin{aligned}& h(0)=0, \qquad h(1)=\frac{\pi}{2}, \end{aligned}$$
(2.17)
$$\begin{aligned}& h^{\prime}(x)=-\frac{x^{2}}{\sqrt{1-x^{2}}[1-(1-2p)^{2}x^{2}]^{2}}h_{1}(x), \end{aligned}$$
(2.18)
where
$$\begin{aligned} h_{1}(x) =&\bigl(16p^{4}-32p^{3}+24p^{2}-8p+1 \bigr)x^{4} \\ &{}+\bigl(-16p^{4}+32p^{3}-32p^{2}+16p-3 \bigr)x^{2}+2\bigl(6p^{2}-6p+1\bigr). \end{aligned}$$
(2.19)
(1) If \(p=1/2-\sqrt{3}/6\), then (2.19) leads to
$$ h_{1}(x)=-\frac{4}{9}x^{2} \bigl(7-x^{2}\bigr)< 0 $$
(2.20)
for \(x\in(0, 1)\).

Therefore, \(h(x)>0\) for all \(x\in(0, 1)\) follows easily from (2.17) and (2.18) together with (2.10).

(2) If \(p=1/2-\sqrt{1-1/e}/2\), then
$$\begin{aligned}& h_{1}(0)=2\bigl(6p^{2}-6p+1\bigr)>0, \qquad h_{1}(1)=-4p(1-p)< 0, \end{aligned}$$
(2.21)
$$\begin{aligned}& h_{1}^{\prime }(x)=4\bigl(16p^{4}-32p^{3}+24p^{2}-8p+1 \bigr)x^{3} \\& \hphantom{h_{1}^{\prime }(x)=}{}+2\bigl(-16p^{4}+32p^{3}-32p^{2}+16p-3 \bigr)x. \end{aligned}$$
(2.22)
Note that
$$\begin{aligned}& 16p^{4}-32p^{3}+24p^{2}-8p+1=0.3995 \ldots>0, \end{aligned}$$
(2.23)
$$\begin{aligned}& 16p^{4}-32p^{3}+16p^{2}-1=-0.8646 \ldots< 0. \end{aligned}$$
(2.24)
It follows from (2.22)-(2.24) that
$$\begin{aligned} h_{1}^{\prime }(x) < &4\bigl(16p^{4}-32p^{3}+24p^{2}-8p+1 \bigr)x+2\bigl(-16p^{4}+32p^{3}-32p^{2}+16p-3 \bigr)x \\ =&2\bigl(16p^{4}-32p^{3}+16p^{2}-1\bigr)x< 0 \end{aligned}$$
(2.25)
for \(x\in(0, 1)\).

From (2.21) and (2.25) we clearly see that there exists \(\sigma_{0}\in (0, 1)\) such that \(h_{1}(x)>0\) for \(x\in(0, \sigma_{0})\) and \(h_{1}(x)<0\) for \(x\in(\sigma_{0}, 1)\).

We divide the proof into two cases.

Case 1. \(x\in(0, \sigma_{0}]\). Then \(h(x)<0\) for \(x\in(0, \sigma _{0}]\) follows easily from (2.17) and (2.18) together with \(h_{1}(x)>0\) on the interval \((0, \sigma_{0})\).

Case 2. \(x\in(\sigma_{0}, 1)\). Then (2.18) and \(h_{1}(x)<0\) on the interval \((\sigma_{0}, 1)\) lead to the conclusion that \(h(x)\) is strictly increasing on \((\sigma_{0}, 1)\). Therefore, there exists \(\sigma_{1}\in(\sigma_{0}, 1)\subset(0, 1)\) such that \(h(x)<0\) for \(x\in(\sigma_{0}, \sigma_{1})\) and \(h(x)>0\) for \(x\in (\sigma_{1}, 1)\) follows from (2.17) and \(h(\sigma_{0})<0\) together with the monotonicity of \(h(x)\) on the interval \((\sigma_{0}, 1)\). □

Lemma 2.4

Let \(p\in(0, 1/2)\) and
$$ J(x)=\arcsin(x)-\frac{x\sqrt{1-x^{2}}}{1-(1-2p)^{2}x^{2}}. $$
(2.26)
Then the following statements are true:
  1. (1)

    If \(p=1/2-\sqrt{6}/6=0.0917\ldots\) , then \(J(x)>0\) for all \(x\in (0, 1)\).

     
  2. (2)

    If \(p=1/2-\sqrt{1-1/e^{2}}/2=0.0350\ldots\) , then there exists \(\tau_{1}\in(0, 1)\) such that \(J(x)<0\) for \(x\in(0, \tau_{1})\) and \(h(x)>0\) for \(x\in(\tau_{1}, 1)\).

     

Proof

Simple computations lead to
$$\begin{aligned}& J(0)=0, \qquad J(1)=\frac{\pi}{2}, \end{aligned}$$
(2.27)
$$\begin{aligned}& J^{\prime}(x)=\frac{x^{2}}{\sqrt{1-x^{2}}[1-(1-2p)^{2}x^{2}]^{2}}J_{1}(x), \end{aligned}$$
(2.28)
where
$$ J_{1}(x)=\bigl(16p^{4}-32p^{3}+24p^{2}-8p+1 \bigr)x^{2}-\bigl(12p^{2}-12p+1\bigr). $$
(2.29)
(1) If \(p=1/2-\sqrt{6}/6\), then (2.29) leads to
$$ J_{1}(x)=\frac{4}{9}x^{2}>0 $$
(2.30)
for \(x\in(0, 1)\).

Therefore, \(J(x)>0\) for all \(x\in(0, 1)\) follows easily from (2.27) and (2.28) together with (2.30).

(2) If \(p=1/2-\sqrt{1-1/e^{2}}/2\), then (2.29) leads to
$$\begin{aligned}& J_{1}(0)=-\bigl(12p^{2}-12p+1\bigr)< 0, \qquad J_{1}(1)=4p\bigl(4p^{3}-8p^{2}+3p+1 \bigr)>0, \end{aligned}$$
(2.31)
$$\begin{aligned}& J_{1}^{\prime}(x)=2\bigl(16p^{4}-32p^{3}+24p^{2}-8p+1 \bigr)x>0 \end{aligned}$$
(2.32)
for \(x\in(0, 1)\).

It follows from (2.31) and (2.32) that there exists \(\tau_{0}\in(0, 1)\) such that \(J_{1}(x)<0\) for \(x\in(0, \tau_{0})\) and \(J_{1}(x)>0\) for \(x\in(\tau_{0}, 1)\).

We divide the proof into two cases.

Case 1. \(x\in(0, \tau_{0}]\). Then \(J(x)<0\) for \(x\in(0, \tau_{0}]\) follows easily from (2.27) and (2.28) together with \(J_{1}(x)<0\) on the interval \((0, \tau_{0})\).

Case 2. \(x\in(\tau_{0}, 1)\). Then (2.28) and \(J_{1}(x)>0\) on the interval \((\tau_{0}, 1)\) lead to the conclusion that \(J(x)\) is strictly increasing on \((\tau_{0}, 1)\).

Therefore, there exists \(\tau_{1}\in(\tau_{0}, 1)\subset(0, 1)\) such that \(J(x)<0\) for \(x\in(\tau_{0}, \tau_{1})\) and \(J(x)>0\) for \(x\in(\tau_{1}, 1)\) follows from (2.27) and \(J(\tau _{0})<0\) together with the monotonicity of \(J(x)\) on the interval \((\tau_{0}, 1)\). □

3 Main results

Theorem 3.1

The double inequality
$$ \alpha_{1}A(a,b)+(1-\alpha_{1})H(a,b)< X(a,b)< \beta _{1}A(a,b)+(1-\beta_{1})H(a,b) $$
holds for all \(a, b>0\) with \(a\neq b\) if and only if \(\alpha_{1}\leq 1/e=0.3678\ldots\) and \(\beta_{1}\geq2/3\).

Proof

Since \(H(a,b)\), \(X(a,b)\) and \(A(a,b)\) are symmetric and homogenous of degree one, we assume that \(a>b>0\). Let \(x=(a-b)/(a+b)\in (0, 1)\) and \(p\in(0, 1)\). Then (1.1) and (1.2) lead to
$$\begin{aligned}& \frac{X(a,b)-H(a,b)}{A(a,b)-H(a,b)}=\frac{e^{\frac{\sqrt {1-x^{2}}\arcsin(x)}{x}-1}-(1-x^{2})}{x^{2}}, \end{aligned}$$
(3.1)
$$\begin{aligned}& \log\frac{X(a,b)}{pA(a,b)+(1-p)H(a,b)}=\frac{\sqrt{1-x^{2}}\arcsin (x)}{x}-1-\log\bigl[p+(1-p) \bigl(1-x^{2}\bigr)\bigr]. \end{aligned}$$
(3.2)
Let
$$ F(x)=\frac{\sqrt{1-x^{2}}\arcsin(x)}{x}-1-\log\bigl[p+(1-p) \bigl(1-x^{2} \bigr)\bigr]. $$
(3.3)
Then simple computations lead to
$$\begin{aligned}& F\bigl(0^{+}\bigr)=0, \end{aligned}$$
(3.4)
$$\begin{aligned}& F(1)=-\log p-1, \end{aligned}$$
(3.5)
$$\begin{aligned}& F^{\prime}(x)=\frac{1}{x^{2}\sqrt{1-x^{2}}}f(x), \end{aligned}$$
(3.6)
where \(f(x)\) is defined by (2.1).

We divide the proof into two cases.

Case 1. \(p=2/3\). Then (3.2)-(3.4) and (3.6) together with Lemma 2.1(1) lead to the conclusion that
$$ X(a,b)< \frac{2}{3}A(a,b)+\frac{1}{3}H(a,b). $$
(3.7)

Case 2. \(p=1/e\). Then (3.6) and Lemma 2.1(2) lead to the conclusion that there exists \(\lambda_{1}\in(0, 1)\) such that \(F(x)\) is strictly increasing on \((0, \lambda_{1}]\) and strictly decreasing on \([\lambda_{1}, 1)\).

Note that (3.5) becomes
$$ F(1)=0. $$
(3.8)
It follows from (3.2)-(3.4) and (3.8) together with the piecewise monotonicity of \(F(x)\) that
$$ X(a,b)>\frac{1}{e}A(a,b)+ \biggl(1-\frac{1}{e} \biggr)H(a,b). $$
(3.9)
Note that
$$\begin{aligned}& \lim_{x\rightarrow0^{+}}\frac{e^{\frac{\sqrt{1-x^{2}}\arcsin (x)}{x}-1}-(1-x^{2})}{x^{2}}=\frac{2}{3}, \end{aligned}$$
(3.10)
$$\begin{aligned}& \lim_{x\rightarrow1^{-}}\frac{e^{\frac{\sqrt{1-x^{2}}\arcsin (x)}{x}-1}-(1-x^{2})}{x^{2}}=\frac{1}{e}. \end{aligned}$$
(3.11)
Therefore, Theorem 3.1 follows from (3.7) and (3.9) in conjunction with the following statements.
•: 

If \(\alpha_{1}>2/3\), then equations (3.1) and (3.10) lead to the conclusion that there exists \(\delta_{1}\in(0, 1)\) such that \(X(a,b)<\alpha_{1}A(a,b)+(1-\alpha _{1})H(a,b)\) for all \(a>b>0\) with \((a-b)/(a+b)\in(0, \delta_{1})\).

••: 

If \(\beta_{1}<1/e\), then equations (3.1) and (3.11) lead to the conclusion that there exists \(\delta_{2}\in(0, 1)\) such that \(X(a,b)>\beta_{1}A(a,b)+(1-\beta _{1})H(a,b)\) for all \(a>b>0\) with \((a-b)/(a+b)\in(1-\delta_{2}, 1)\).

 □

Theorem 3.2

The double inequality
$$ \alpha_{2}A(a,b)+(1-\alpha_{2})G(a,b)< X(a,b)< \beta _{2}A(a,b)+(1-\beta_{2})G(a,b) $$
holds for all \(a, b>0\) with \(a\neq b\) if and only if \(\alpha_{2}\leq 1/3\) and \(\beta_{2}\geq1/e=0.3678\ldots\) .

Proof

Since \(A(a,b)\), \(G(a,b)\) and \(X(a,b)\) are symmetric and homogenous of degree one, we assume that \(a>b>0\). Let \(x=(a-b)/(a+b)\in (0, 1)\) and \(p\in(0, 1)\). Then (1.1) and (1.2) lead to
$$\begin{aligned}& \frac{X(a,b)-G(a,b)}{A(a,b)-G(a,b)}=\frac{e^{\frac{\sqrt {1-x^{2}}\arcsin(x)}{x}-1}-\sqrt{1-x^{2}}}{1-\sqrt{1-x^{2}}}, \end{aligned}$$
(3.12)
$$\begin{aligned}& \log\frac{X(a,b)}{pA(a,b)+(1-p)G(a,b)}=\frac{\sqrt{1-x^{2}}\arcsin (x)}{x}-1-\log\bigl[p+(1-p) \sqrt{1-x^{2}}\bigr]. \end{aligned}$$
(3.13)
Let
$$ G(x)=\frac{\sqrt{1-x^{2}}\arcsin(x)}{x}-1-\log\bigl[p+(1-p)\sqrt{1-x^{2}} \bigr]. $$
(3.14)
Then simple computations lead to
$$\begin{aligned}& G\bigl(0^{+}\bigr)=0, \end{aligned}$$
(3.15)
$$\begin{aligned}& G(1)=-\log p-1, \end{aligned}$$
(3.16)
$$\begin{aligned}& G^{\prime}(x)=\frac{1}{x^{2}\sqrt{1-x^{2}}}g(x), \end{aligned}$$
(3.17)
where \(g(x)\) is defined by (2.8).

We divide the proof into two cases.

Case 1. \(p=1/3\). Then (3.13)-(3.15) and (3.17) together with Lemma 2.2(1) lead to the conclusion that
$$ X(a,b)>\frac{1}{3}A(a,b)+\frac{2}{3}G(a,b). $$
(3.18)
Case 2. \(p=1/e\). Then from Lemma 2.2(2) and (3.17) we know that there exists \(\mu_{1}\in(0, 1)\) such that \(G(x)\) is strictly decreasing on \((0, \mu_{1}]\) and strictly increasing on \([\mu_{1}, 1)\). Note that (3.16) becomes
$$ G(1)=0. $$
(3.19)
It follows from (3.13)-(3.15) and (3.19) together with the piecewise monotonicity of \(G(x)\) that
$$ X(a,b)< \frac{1}{e}A(a,b)+ \biggl(1-\frac{1}{e} \biggr)G(a,b). $$
(3.20)
Note that
$$\begin{aligned}& \lim_{x\rightarrow0^{+}}\frac{e^{\frac{\sqrt{1-x^{2}}\arcsin (x)}{x}-1}-\sqrt{1-x^{2}}}{1-\sqrt{1-x^{2}}}=\frac{1}{3}, \end{aligned}$$
(3.21)
$$\begin{aligned}& \lim_{x\rightarrow1^{-}}\frac{e^{\frac{\sqrt{1-x^{2}}\arcsin (x)}{x}-1}-\sqrt{1-x^{2}}}{1-\sqrt{1-x^{2}}}=\frac{1}{e}. \end{aligned}$$
(3.22)

Therefore, Theorem 3.2 follows easily from (3.12) and (3.18) together with (3.20)-(3.22). □

Theorem 3.3

Let \(\alpha_{3}, \beta_{3}\in(0, 1/2)\). Then the double inequality
$$ H\bigl[\alpha_{3}a+(1-\alpha_{3})b, \alpha_{3}b+(1- \alpha _{3})a\bigr]< X(a,b)< H\bigl[\beta_{3}a+(1- \beta_{3})b, \beta_{3}b+(1-\beta_{3})a\bigr] $$
holds for all \(a, b>0\) with \(a\neq b\) if and only if \(\alpha_{3}\leq 1/2-\sqrt{1-1/e}/2=0.1024\ldots\) and \(\beta_{3}\geq1/2-\sqrt {3}/6=0.2113\ldots\) .

Proof

Since \(H(a,b)\) and \(X(a,b)\) are symmetric and homogenous of degree one, we assume that \(a>b>0\). Let \(x=(a-b)/(a+b)\in(0, 1)\) and \(p\in(0, 1/2)\). Then (1.1) and (1.2) lead to
$$ \log\frac{H[pa+(1-p)b, pb+(1-p)a]}{X(a,b)}=\log \bigl[1-(1-2p)^{2}x^{2} \bigr]-\frac{\sqrt{1-x^{2}}\arcsin(x)}{x}+1. $$
(3.23)
Let
$$ H(x)=\log\bigl[1-(1-2p)^{2}x^{2}\bigr]- \frac{\sqrt{1-x^{2}}\arcsin(x)}{x}+1. $$
(3.24)
Then simple computations lead to
$$\begin{aligned}& H\bigl(0^{+}\bigr)=0, \end{aligned}$$
(3.25)
$$\begin{aligned}& H(1)=1+2\log2+\log\bigl(p-p^{2}\bigr), \end{aligned}$$
(3.26)
$$\begin{aligned}& H^{\prime}(x)=\frac{1}{x^{2}\sqrt{1-x^{2}}}h(x), \end{aligned}$$
(3.27)
where \(h(x)\) is defined by (2.16).

We divide the proof into four cases.

Case 1. \(p=1/2-\sqrt{3}/6\). Then (3.23)-(3.25) and (3.27) together with Lemma 2.3(1) lead to
$$ X(a,b)< H \biggl[ \biggl(\frac{1}{2}-\frac{\sqrt{3}}{6} \biggr)a+ \biggl( \frac{1}{2}+\frac{\sqrt{3}}{6} \biggr)b, \biggl(\frac{1}{2}- \frac{\sqrt{3}}{6} \biggr)b+ \biggl(\frac {1}{2}+\frac{\sqrt{3}}{6} \biggr)a \biggr]. $$
Case 2. \(0< p<1/2-\sqrt{3}/6\). Let \(q=(1-2p)^{2}\) and \(x\rightarrow 0^{+}\), then \(1/3< q<1\) and power series expansion leads to
$$ H(x)=- \biggl(q-\frac{1}{3} \biggr)x^{2}+o \bigl(x^{2}\bigr). $$
(3.28)
Equations (3.23), (3.24) and (3.28) lead to the conclusion that there exists \(0<\delta<1\) such that
$$ X(a,b)>H\bigl[pa+(1-p)b, pb+(1-p)a\bigr] $$
(3.29)
for all \(a>b>0\) with \((a-b)/(a+b)\in(0, \delta)\).

Case 3. \(p=1/2-\sqrt{1-1/e}/2\). Then (3.27) and Lemma 2.3(2) lead to the conclusion that there exists \(\sigma_{1}\in(0, 1)\) such that \(H(x)\) is strictly decreasing on \((0, \sigma_{1}]\) and strictly increasing on \([\sigma_{1}, 1)\).

Note that (3.26) becomes
$$ H(1)=0. $$
(3.30)
Therefore,
$$ X(a,b)> H \biggl[ \biggl(\frac{1}{2}-\frac{\sqrt{1-\frac{1}{e}}}{2} \biggr)a+ \biggl( \frac{1}{2}+\frac{\sqrt{1-\frac{1}{e}}}{2} \biggr)b, \biggl(\frac{1}{2}- \frac{\sqrt{1-\frac{1}{e}}}{2} \biggr)b+ \biggl(\frac{1}{2}+\frac{\sqrt{1-\frac{1}{e}}}{2} \biggr)a \biggr] $$
follows from (3.23)-(3.25) and (3.30) together with the piecewise monotonicity of \(H(x)\).
Case 4. \(1/2-\sqrt{1-1/e}/2< p<1/2\). Then (3.26) leads to
$$ H(1)>0. $$
(3.31)
Equations (3.23) and (3.24) together with inequality (3.31) imply that there exists \(0<\delta^{\prime}<1\) such that
$$ X(a,b)< H\bigl[pa+(1-p)b, pb+(1-p)a\bigr] $$
for \(a>b>0\) with \((a-b)(a+b)\in(1-\delta^{\prime}, 1)\). □

Theorem 3.4

Let \(\alpha_{4}, \beta_{4}\in(0, 1/2)\). Then the double inequality
$$ G\bigl[\alpha_{4}a+(1-\alpha_{4})b, \alpha_{4}b+(1- \alpha _{4})a\bigr]< X(a,b)< G\bigl[\beta_{4}a+(1- \beta_{4})b, \beta_{4}b+(1-\beta_{4})a\bigr] $$
holds for all \(a, b>0\) with \(a\neq b\) if and only if \(\alpha_{4}\leq 1/2-\sqrt{1-1/e^{2}}/2=0.0350\ldots\) and \(\beta_{4}\geq1/2-\sqrt {6}/6=0.0917\ldots\) .

Proof

Since \(G(a,b)\) and \(X(a,b)\) are symmetric and homogenous of degree one, we assume that \(a>b>0\). Let \(x=(a-b)/(a+b)\in(0, 1)\) and \(p\in(0, 1/2)\). Then (1.1) and (1.2) lead to
$$\begin{aligned}& \log\frac{G[pa+(1-p)b, pb+(1-p)a]}{X(a,b)} \\& \quad =\frac{1}{2}\log \bigl[1-(1-2p)^{2}x^{2}\bigr]-\frac{\sqrt{1-x^{2}}\arcsin(x)}{x}+1. \end{aligned}$$
(3.32)
Let
$$ K(x)=\frac{1}{2}\log\bigl[1-(1-2p)^{2}x^{2} \bigr]-\frac{\sqrt{1-x^{2}}\arcsin (x)}{x}+1. $$
(3.33)
Then simple computations lead to
$$\begin{aligned}& K\bigl(0^{+}\bigr)=0, \end{aligned}$$
(3.34)
$$\begin{aligned}& K(1)=1+\log2+\frac{1}{2}\log\bigl(p-p^{2}\bigr), \end{aligned}$$
(3.35)
$$\begin{aligned}& K^{\prime}(x)=\frac{1}{x^{2}\sqrt{1-x^{2}}}J(x), \end{aligned}$$
(3.36)
where \(J(x)\) is defined by (2.26).

We divide the proof into four cases.

Case 1. \(p=p_{0}=1/2-\sqrt{6}/6\). Then
$$ X(a,b)< G\bigl[p_{0}a+(1-p_{0})b, p_{0}b+(1-p_{0})a \bigr] $$
follows from (3.32)-(3.34) and (3.36) together with Lemma 2.4(1).
Case 2. \(0< p<1/2-\sqrt{6}/6\). Let \(q=(1-2p)^{2}\) and \(x\rightarrow 0^{+}\), then \(2/3< q<1\) and power series expansion leads to
$$ K(x)=-\frac{1}{2} \biggl(q-\frac{2}{3} \biggr)x^{2}+o\bigl(x^{2}\bigr). $$
(3.37)
From (3.32), (3.33) and (3.37) we clearly see that there exists \(0<\delta<1\) such that
$$ X(a,b)>G\bigl[pa+(1-p)b, pb+(1-p)a\bigr] $$
for \(a>b>0\) with \((a-b)/(a+b)\in(0, \delta)\).

Case 3. \(p=p_{1}=1/2-\sqrt{1-1/e^{2}}/2\). Then (3.36) and Lemma 2.4(2) lead to the conclusion that there exists \(\tau_{1}\in(0, 1)\) such that \(K(x)\) is strictly decreasing on \((0, \tau_{1}]\) and strictly increasing on \([\tau_{1}, 1)\).

Note that (3.35) becomes
$$ K(1)=0. $$
(3.38)
Therefore,
$$ X(a,b)>G\bigl[p_{1}a+(1-p_{1})b, p_{1}b+(1-p_{1})a \bigr] $$
follows from (3.32)-(3.34) and (3.38) together with the piecewise monotonicity of \(K(x)\).
Case 4. \(1/2-\sqrt{1-1/e^{2}}/2< p<1/2\). Then (3.35) leads to
$$ K(1)>0. $$
(3.39)
Equations (3.32) and (3.33) together with inequality (3.39) imply that there exists \(0<\delta^{\prime}<1\) such that
$$ X(a,b)< G\bigl[pa+(1-p)b, pb+(1-p)a\bigr] $$
(3.40)
for \(a>b>0\) with \((a-b)/(a+b)\in(1-\delta^{\prime}, 1)\). □

Declarations

Acknowledgements

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

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

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© Qian et al. 2015