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

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.

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

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

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

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

• (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 $(a-b)/(a+b)\in(1-\delta^{\prime}, 1)$. □