# 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.

## 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 $$X(a,b)$$  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 , 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.  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 , 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.

## 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

\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

\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

\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

\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 $$(\tau_{0}, 1)$$. □

## 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)

\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)

\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)

\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)

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

## References

1. Sándor, J: Two sharp inequalities for trigonometric and hyperbolic functions. Math. Inequal. Appl. 15(2), 409-413 (2012)

2. Sándor, J: On two new means of two variables. Notes Number Theory Discrete Math. 20(1), 1-9 (2014)

3. Yang, Z-H, Wu, L-M, Chu, Y-M: Sharp power mean bounds for Sándor mean. Abstr. Appl. Anal. 2014, Article ID 172867 (2014)

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## 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.

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Correspondence to Yu-Ming Chu.

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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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