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# Sharp power-type Heronian mean bounds for the Sándor and Yang means

Journal of Inequalities and Applications20152015:159

https://doi.org/10.1186/s13660-015-0683-7

• Received: 14 December 2014
• Accepted: 1 May 2015
• Published:

## Abstract

We prove that the double inequalities $$H_{\alpha}(a, b) < X(a, b)<H_{\beta}(a, b)$$ and $$H_{\lambda}(a, b)< U(a, b)<H_{\mu}(a, b)$$ hold for all $$a, b>0$$ with $$a\neq b$$ if and only if $$\alpha\leq1/2$$, $$\beta\geq\log3/(1+\log2)=0.6488\cdots$$, $$\lambda\leq2\log3/(2\log\pi-\log2) =1.3764\cdots$$, and $$\mu\geq2$$, where $$H_{p}(a, b)$$, $$X(a, b)$$, and $$U(a, b)$$ are, respectively, the pth power-type Heronian mean, Sándor mean, and Yang mean of a and b.

## Keywords

• power-type Heronian mean
• Sándor mean
• Yang mean

• 26E60

## 1 Introduction

For $$p\in\mathbb{R}$$, the pth power-type Heronian mean $$H_{p}(a, b)$$ of two positive real numbers a and b is defined by
$$H_{p}(a, b)= \biggl[\frac{a^{p}+(ab)^{p/2}+b^{p}}{3} \biggr]^{1/p} \quad(p\neq 0),\qquad H_{0}(a, b)=\sqrt{ab}.$$
(1.1)

It is well known that $$H_{p}(a, b)$$ is continuous and strictly increasing with respect to $$p\in\mathbb{R}$$ for fixed $$a, b>0$$ with $$a\neq b$$.

Let $$G(a, b)=\sqrt{ab}$$, $$L(a, b)=(a-b)/(\log a-\log b)$$, $$P(a, b)=(a-b)/[2\arcsin((a-b)/(a+b))]$$, $$I(a, b)=(a^{a}/b^{b})^{1/(a-b)}/e$$, $$A(a, b)=(a+b)/2$$, $$T(a, b)=(a-b)/[2\arctan((a-b)/(a+b))]$$, $$Q(a, b)=\sqrt{(a^{2}+b^{2})/2}$$ and $$M_{r}(a, b)=[(a^{r}+b^{r})/2]^{1/r}$$ ($$r\neq0$$), and $$M_{0}(a, b)=\sqrt{ab}$$ be, respectively, the geometric, logarithmic, first Seiffert, identric, arithmetic, second Seiffert, quadratic, and rth power means of two distinct positive real numbers a and b. Then it is well known that the inequalities
\begin{aligned} G(a,b)&=M_{0}(a,b)< L(a,b)<P(a,b)<I(a,b) \\ &<A(a,b)=M_{1}(a,b)<T(a,b)<Q(a,b)=M_{2}(a,b) \end{aligned}
hold for all $$a, b>0$$ with $$a\neq b$$.
Let $$a, b>0$$. Then the Sándor mean $$X(a,b)$$  and Yang mean $$U(a,b)$$  are given by
$$X(a,b)=A(a,b)e^{\frac{G(a,b)}{P(a,b)}-1}$$
(1.2)
and
$$U(a,b)=\frac{a-b}{\sqrt{2}\arctan (\frac{a-b}{\sqrt{2ab}} )} \quad(a\neq b),\qquad U(a, a)=a,$$
(1.3)
respectively.

The Yang mean $$U(a,b)$$ is the special case of the Seiffert type mean $$T_{M, p}(a, b)=(a-b)/[p\arctan((a-b)/(pM(a,b)))]$$ defined by Toader in , where $$M(a, b)$$ is a bivariate mean and p is a positive real number. Indeed, $$U(a,b)=T_{G,\sqrt{2}}(a,b)$$. Recently, the power-type Heronian, Sándor, and Yang means have been the subject of intensive research.

For all $$a, b>0$$ with $$a\neq b$$, Yang  and Sándor  proved that the double inequality
$$M_{1/2}(a,b)< H_{1}(a,b)<I(a,b)$$
holds, and the inequality $$H_{1}(a,b)< M_{2/3}(a,b)$$ can be found in the literature .
Jia and Cao  proved that the inequalities
\begin{aligned} &L(a,b)< H_{p}(a,b)<M_{q}(a,b), \\ &A(a,b)=M_{1}(a,b)<H_{\log3/\log2}(a,b) \end{aligned}
(1.4)
hold for all $$a, b>0$$ with $$a\neq b$$ if $$p\geq1/2$$ and $$q\geq2p/3$$. Inequality (1.4) can also be found in the literature , p.64 and .
In , the authors proved that the double inequality
$$H_{p}(a,b)< T(a,b)<H_{q}(a,b)$$
holds for all $$a, b>0$$ with $$a\neq b$$ if and only if $$p\leq\log3/(\log \pi-\log2)$$ and $$q\geq5/2$$.
Sándor  presented 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)<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
$$M_{p}(a,b)< X(a,b)<M_{q}(a,b)$$
holds for all $$a, b>0$$ with $$a\neq b$$ if and only if $$p\leq1/3$$ and $$q\geq\log2/(1+\log2)$$.
In , Yang established the inequalities
\begin{aligned}& P(a,b)< U(a,b)<T(a,b), \qquad \frac{G(a,b)T(a,b)}{A(a,b)}<U(a,b)<\frac{P(a,b)Q(a,b)}{A(a,b)}, \\& Q^{1/2}(a,b) \biggl[\frac{2G(a,b)+Q(a,b)}{3} \biggr]^{1/2}<U(a,b)<Q^{2/3}(a,b) \biggl[\frac{G(a,b)+Q(a,b)}{2} \biggr]^{1/3}, \\& \frac{G(a,b)+Q(a,b)}{2}<U(a,b)< \biggl[\frac{2}{3} \biggl(\frac {G(a,b)+Q(a,b)}{2} \biggr)^{1/2}+\frac{1}{3}Q^{1/2}(a,b) \biggr]^{2} \end{aligned}
for all $$a, b>0$$ with $$a\neq b$$.
In [13, 14], the authors proved that the double inequalities
\begin{aligned} & \biggl[\frac{2}{3} \biggl(\frac{G(a,b)+Q(a,b)}{2} \biggr)^{p}+ \frac {1}{3}Q^{p}(a,b) \biggr]^{1/p} \\ &\quad< U(a,b)< \biggl[\frac{2}{3} \biggl(\frac{G(a,b)+Q(a,b)}{2} \biggr)^{q}+\frac {1}{3}Q^{q}(a,b) \biggr]^{1/q}, \\ &\frac{2^{1-\lambda}(G(a,b)+Q(a,b))^{\lambda}Q(a,b)+G(a,b)Q^{\lambda }(a,b)}{2^{1-\lambda}(G(a,b)+Q(a,b))^{\lambda}+Q^{\lambda}(a,b)} \\ &\quad<U(a,b)<\frac{2^{1-\mu}(G(a,b)+Q(a,b))^{\mu}Q(a,b)+G(a,b)Q^{\mu }(a,b)}{2^{1-\mu}(G(a,b)+Q(a,b))^{\mu}+Q^{\mu}(a,b)}, \\ &M_{\alpha}(a,b)<U(a,b)<M_{\beta}(a,b) \end{aligned}
hold for all $$a, b>0$$ with $$a\neq b$$ if and only if $$p\leq p_{0}$$, $$q\geq1/5$$, $$\lambda\geq1/5$$, $$\mu\leq p_{1}$$, $$\alpha\leq2\log 2/(2\log\pi-\log2)$$, and $$\beta\geq4/3$$, where $$p_{0}=0.1941\cdots$$ is the unique solution of the equation $$p\log(2/\pi)-\log (1+2^{1-p} )+\log3=0$$ on the interval $$(1/10, \infty)$$, and $$p_{1}=\log(\pi-2)/\log2=0.1910\cdots$$.

The main purpose of this paper is to present the best possible parameter α, β, λ, and μ such that the double inequalities $$H_{\alpha}(a, b)< X(a, b)<H_{\beta}(a, b)$$ and $$H_{\lambda}(a, b)< U(a, b)<H_{\mu}(a, b)$$ hold for all $$a, b>0$$ with $$a\neq b$$.

## 2 Lemmas

In order to prove our main results we need two lemmas, which we present in this section.

### Lemma 2.1

Let $$p\in(0, 1)$$ and
\begin{aligned} f(x)={}&(p-1)x^{3p+2}+(4p-3)x^{2p+2}+(p-3)x^{p+2}+6x^{4p}+6x^{3p}-6x^{p} \\ &{}+2x^{4p-2}+(3-p)x^{3p-2}+(3-4p)x^{2p-2}+(1-p)x^{p-2}-2x^{2}-6. \end{aligned}
(2.1)
Then the following statements are true:
1. (1)

if $$p=1/2$$, then $$f(x)<0$$ for all $$x>1$$;

2. (2)

if $$p=\log3/(1+\log2)=0.6488\cdots$$, then there exists $$\lambda \in(1, \infty)$$ such that $$f(x)>0$$ for $$x\in(1, \lambda)$$ and $$f(x)<0$$ for $$x\in(\lambda, \infty)$$.

### Proof

For part (1), if $$p=1/2$$, then (2.1) becomes
$$f(x)=-\frac{(x-1)(\sqrt {x}-1)^{2}}{2x^{3/2}} \bigl(x^{3}+4x^{5/2}+13x^{2}+16x^{3/2}+13x+4 \sqrt{x}+1 \bigr).$$
(2.2)

Therefore, part (1) follows from (2.2).

For part (2), let $$p=\log3/(1+\log2)$$, $$f_{1}(x)=f^{\prime}(x)/x$$, $$f_{2}(x)=x^{5-p}f_{1}^{\prime}(x)$$, $$f_{3}(x)=f_{2}^{\prime }(x)/(2px)$$, and $$f_{4}(x)=f_{3}^{\prime}(x)/(2x)$$. Then elaborated computations lead to
\begin{aligned}& f(1)=0, \qquad\lim_{x\rightarrow+\infty}f(x)=-\infty, \end{aligned}
(2.3)
\begin{aligned}& \begin{aligned}[b] f_{1}(x)={}&(p-1) (3p+2)x^{3p}+2(p+1) (4p-3)x^{2p}+(p-3) (p+2)x^{p} \\ &{}+24px^{4p-2}+18px^{3p-2}-6px^{p-2}+4(2p-1)x^{4p-4}+(3-p) (3p-2)x^{3p-4} \\ &{}+2(p-1) (3-4p)x^{2p-4}+(1-p) (p-2)x^{p-4}-4, \end{aligned} \\& f_{1}(1)=36(2p-1)>0,\qquad \lim_{x\rightarrow+\infty}f_{1}(x)=- \infty, \end{aligned}
(2.4)
\begin{aligned}& \begin{aligned}[b] f_{2}(x)={}&3p(p-1) (3p+2)x^{2p+4}+4p(p+1) (4p-3)x^{p+4}+48p(2p-1)x^{3p+2} \\ &{}+18p(3p-2)x^{2p+2}+16(2p-1) (p-1)x^{3p}+(3-p) (3p-2) (3p-4)x^{2p} \\ &{}+4(3-4p) (p-1) (p-2)x^{p}+p(p-3) (p+2)x^{4} \\ &{}-6p(p-2)x^{2}+(1-p) (p-2) (p-4), \end{aligned} \\& f_{2}(1)=72(2p-1)^{2}>0,\qquad \lim_{x\rightarrow+\infty}f_{2}(x)=- \infty, \end{aligned}
(2.5)
\begin{aligned}& \begin{aligned}[b] f_{3}(x)={}&3(p-1) (p+2) (3p+2)x^{2p+2}+2(p+1) (p+4) (4p-3)x^{p+2} \\ &{}+24(2p-1) (3p+2)x^{3p}+18(p+1) (3p-2)x^{2p}+24(2p-1) (p-1)x^{3p-2} \\ &{}+(3-p) (3p-2) (3p-4)x^{2p-2}+2(p-1) (p-2) (3-4p)x^{p-2} \\ &{}+2(p-3) (p+2)x^{2}-6(p-2), \end{aligned} \\& f_{3}(1)=12 \bigl(31p^{2}-12p-5 \bigr)>0, \qquad\lim _{x\rightarrow+\infty }f_{3}(x)=-\infty, \end{aligned}
(2.6)
\begin{aligned}& \begin{aligned}[b] f_{4}(x)={}&3(p-1) (p+1) (p+2) (3p+2)x^{2p}+(p+1) (p+2) (p+4) (4p-3)x^{p} \\ &{}+36p(2p-1) (3p+2)x^{3p-2}+18p(p+1) (3p-2)x^{2p-2} \\ &{}+12(p-1) (2p-1) (3p-2)x^{3p-4} \\ &{}+(p-1) (3-p) (3p-2) (3p-4)x^{2p-4}+(p-1) (p-2)^{2}(3-4p)x^{p-4} \\ &{}+2(p-3) (p+2). \end{aligned} \end{aligned}
(2.7)
It follows from (2.7) and $$p=\log3/(1+\log2)=0.6488\cdots$$ together with $$13p^{4}+337p^{3}-80p^{2}-72=-11.3153\cdots<0$$ that
\begin{aligned} f_{4}(x)< {}&3(p-1) (p+1) (p+2) (3p+2)+(p+1) (p+2) (p+4) (4p-3) \\ &{}+36p(2p-1) (3p+2)+18p(p+1) (3p-2)x^{2p-2}+12(p-1) (2p-1) (3p-2) \\ &{}+(p-1) (3-p) (3p-2) (3p-4)x^{2p-4}+(p-1) (p-2)^{2}(3-4p)x^{p-4} \\ &{}+2(p-3) (p+2) \\ ={}&18p(p+1) (3p-2)x^{2p-2}+(p-1) (3-p) (3p-2) (3p-4)x^{2p-4} \\ &{}+(p-1) (p-2)^{2}(3-4p)x^{p-4}+ \bigl(13p^{4}+337p^{3}-80p^{2}-72 \bigr)<0 \end{aligned}
(2.8)
for $$x\in(1, \infty)$$.

Inequality (2.8) implies that $$f_{3}(x)$$ is strictly decreasing on $$[1, \infty)$$. Then (2.6) leads to the conclusion that there exists $$\lambda _{1}\in(1, \infty)$$ such that $$f_{2}(x)$$ is strictly increasing on $$[1, \lambda_{1}]$$ and strictly decreasing on $$[\lambda_{1}, \infty)$$.

It follows from the piecewise monotonicity of $$f_{2}$$ and (2.5) that there exists $$\lambda_{2}\in(\lambda_{1}, \infty)$$ such that $$f_{1}(x)$$ is strictly increasing on $$[1, \lambda_{2}]$$ and strictly decreasing on $$[\lambda_{2}, \infty)$$.

From (2.4) and the piecewise monotonicity of $$f_{1}$$ we clearly see that there exists $$\lambda_{3}\in(\lambda_{2}, \infty)$$ such that $$f(x)$$ is strictly increasing on $$[1, \lambda_{3}]$$ and strictly decreasing on $$[\lambda_{3}, \infty)$$.

Therefore, part (2) follows from (2.3) and the piecewise monotonicity of f. □

### Lemma 2.2

Let $$p\in(1, 2]$$ and
\begin{aligned}& \begin{aligned}[b] g(x)={}&2x^{4p+4}-2x^{4p+2}+10x^{4p}+6x^{4p-2}+(p-1)x^{3p+6}-2x^{3p+4}-2x^{3p+2} \\ &{}+14x^{3p}+(7-p)x^{3p-2}+4(p-1)x^{2p+6}-12x^{2p+4}+12x^{2p}+4(1-p)x^{2p-2} \\ &{}+(p-7)x^{p+6}-14x^{p+4}+2x^{p+2}+2x^{p}+(1-p)x^{p-2}-2 \bigl(3x^{6}+5x^{4}-x^{2}+1 \bigr). \end{aligned} \end{aligned}
(2.9)
Then the following statements are true:
1. (1)

if $$p=2$$, then $$g(x)>0$$ for all $$x>1$$;

2. (2)

if $$p=2\log3/(2\log\pi-\log2)=1.3764\cdots$$, then there exists $$\mu\in(1, \infty)$$ such that $$g(x)<0$$ for $$x\in(1, \mu)$$ and $$g(x)>0$$ for $$x\in(\mu, \infty)$$.

### Proof

For part (1), if $$p=2$$, then (2.9) becomes
$$g(x)=3 \bigl(x^{4}-1 \bigr)^{3}.$$
(2.10)

Therefore, part (1) follows from (2.10).

For part (2), let $$p=2\log3/(2\log\pi-\log2)$$, $$g_{1}(x)=g^{\prime }(x)/x$$, $$g_{2}(x)=g_{1}^{\prime}(x)/x$$, $$g_{3}(x)=g_{2}^{\prime }(x)/x$$, and $$g_{4}(x)=x^{9-p}g_{3}^{\prime}(x)$$. Then elaborated computations lead to
\begin{aligned} &g(1)=0,\qquad \lim_{x\rightarrow\infty}g(x)=+\infty, \end{aligned}
(2.11)
\begin{aligned} &g_{1}(x)=8(p+1)x^{4p+2}-4(2p+1)x^{4p}+40px^{4p-2}+12(2p-1)x^{4p-4}+3(p-1) (p+2)x^{3p+4} \\ &\hphantom{g_{1}(x)=}{}-2(3p+4)x^{3p+2}-2(3p+2)x^{3p}+42px^{3p-2}+(7-p) (3p-2)x^{3p-4} \\ &\hphantom{g_{1}(x)=}{}+8(p-1) (p+3)x^{2p+4} \\ &\hphantom{g_{1}(x)=}{}-24(p+2)x^{2p+2}+24px^{2p-2}-8(1-p)^{2}x^{2p-4}+(p-7) (p+6)x^{p+4} \\ &\hphantom{g_{1}(x)=}{}-14(p+4)x^{p+2}+2(p+2)x^{p}+2px^{p-2}+(1-p) (p-2)x^{p-4}-36x^{4}-40x^{2}+4, \\ &g_{1}(1)=-144(2-p)< 0, \qquad\lim_{x\rightarrow\infty}g_{1}(x)=+ \infty, \end{aligned}
(2.12)
\begin{aligned} &g_{2}(x)=16(p+1) (2p+1)x^{4p}-16p(2p+1)x^{4p-2} +80p(2p-1)x^{4p-4} \\ &\hphantom{g_{2}(x)=}{}+48(p-1)(2p-1)x^{4p-6} \\ &\hphantom{g_{2}(x)=}{}+3(p-1) (p+2) (3p+4)x^{3p+2}-2(3p+2) (3p+4)x^{3p}-6p(3p+2)x^{3p-2} \\ &\hphantom{g_{2}(x)=}{}+42p(3p-2)x^{3p-4} +(7-p) (3p-2) (3p-4)x^{3p-6} \\ &\hphantom{g_{2}(x)=}{}+16(p-1) (p+2) (p+3)x^{2p+2}-48(p+1) (p+2)x^{2p}+48p(p-1)x^{2p-4} \\ &\hphantom{g_{2}(x)=}{}-16(1-p)^{2}(p-2)x^{2p-6}+(p-7) (p+4) (p+6)x^{p+2}-14(p+2) (p+4)x^{p} \\ &\hphantom{g_{2}(x)=}{}+2p(p+2)x^{p-2} +2p(p-2)x^{p-4}+(1-p) (p-2) (p-4)x^{p-6}-144x^{2}-80, \\ &g_{2}(1)=-288 \bigl(2+3p-2p^{2} \bigr)< 0,\qquad \lim _{x\rightarrow\infty }g_{2}(x)=+\infty, \end{aligned}
(2.13)
\begin{aligned} & g_{3}(x)=64p(p+1) (2p+1)x^{4p-2}-32p \bigl(4p^{2}-1 \bigr)x^{4p-4}+320p(p-1) (2p-1)x^{4p-6} \\ &\hphantom{g_{3}(x)=}{}+96(p-1) (2p-1) (2p-3)x^{4p-8}+3(p-1) (p+2) (3p+2) (3p+4)x^{3p} \\ &\hphantom{g_{3}(x)=}{}-6p(3p+2) (3p+4)x^{3p-2} -6p \bigl(9p^{2}-4 \bigr)x^{3p-4}+42p(3p-2) (3p-4)x^{3p-6} \\ &\hphantom{g_{3}(x)=}{}+3(7-p) (p-2) (3p-2) (3p-4)x^{3p-8} \\ &\hphantom{g_{3}(x)=}{}+32 \bigl(p^{2}-1 \bigr) (p+2) (p+3)x^{2p}-96p(p+1) (p+2)x^{2p-2}+96p(p-1) (p-2)x^{2p-6} \\ &\hphantom{g_{3}(x)=}{}-32(1-p)^{2}(p-2) (p-3)x^{2p-8}+(p-7) (p+2) (p+4) (p+6)x^{p} \\ &\hphantom{g_{3}(x)=}{}-14p(p+2) (p+4)x^{p-2} +2p \bigl(p^{2}-4 \bigr)x^{p-4}+2p(p-2) (p-4)x^{p-6} \\ &\hphantom{g_{3}(x)=}{}+(1-p) (p-2) (p-4) (p-6)x^{p-8}-288, \\ &g_{3}(1)=48 \bigl(43p^{3}-100p^{2}+58p-36 \bigr)< 0, \qquad\lim_{x\rightarrow\infty }g_{3}(x)=+ \infty, \end{aligned}
(2.14)
\begin{aligned} &g_{4}(x)= 128p(p+1) \bigl(4p^{2}-1 \bigr)x^{3p+6}-128p \bigl(4p^{2}-1 \bigr) (p-1)x^{3p+4} \\ &\hphantom{g_{4}(x)=}{}+640p(p-1) (2p-1) (2p-3)x^{3p+2}+384(p-1) (p-2) (2p-1) (2p-3)x^{3p} \\ &\hphantom{g_{4}(x)=}{}+9p(p-1) (p+2) (3p+2) (3p+4)x^{2p+8}-6p \bigl(9p^{2}-4 \bigr) (3p+4)x^{2p+6} \\ &\hphantom{g_{4}(x)=}{}-6p \bigl(9p^{2}-4 \bigr) (3p-4)x^{2p+4}+126p(p-2) (3p-2) (3p-4)x^{2p+2} \\ &\hphantom{g_{4}(x)=}{}+3(7-p) (p-2) (3p-2) (3p-4) (3p-8)x^{2p}+64p \bigl(p^{2}-1 \bigr) (p+2) (p+3)x^{p+8} \\ &\hphantom{g_{4}(x)=}{}-192p \bigl(p^{2}-1 \bigr) (p+2)x^{p+6}+192p(p-1) (p-2) (p-3)x^{p+2} \\ &\hphantom{g_{4}(x)=}{}-64(1-p)^{2}(p-2) (p-3) (p-4)x^{p} \\ &\hphantom{g_{4}(x)=}{}+p(p-7) (p+2) (p+4) (p+6)x^{8}-14p \bigl(p^{2}-4 \bigr) (p+4)x^{6}+2p \bigl(p^{2}-4 \bigr) (p-4)x^{4} \\ &\hphantom{g_{4}(x)=}{}+2p(p-2) (p-4) (p-6)x^{2}+(1-p) (p-2) (p-4) (p-6) (p-8) \\ &\hphantom{g_{4}(x)}=:a_{1}x^{3p+6}+a_{4}x^{3p+4} +a_{8}x^{3p+2}+a_{11}x^{3p}+a_{0}x^{2p+8}+a_{3} x^{2p+6}+a_{7}x^{2p+4}+a_{10}x^{2p+2} \\ &\hphantom{g_{4}(x)=}{}+a_{14}x^{2p}+a_{2}x^{p+8}+a_{6}x^{p+6}+a_{13}x^{p+2} +a_{16}x^{p}+a_{5}x^{8}+a_{9}x^{6} \\ &\hphantom{g_{4}(x)=}{}+a_{12}x^{4}+a_{15}x^{2}+a_{17}, \end{aligned}
(2.15)
\begin{aligned} &\sum_{n=0}^{8}a_{n}=2p \bigl(73p^{4}+1306p^{3}-3344p^{2}+3272p-1328 \bigr)>0, \end{aligned}
(2.16)
\begin{aligned} &a_{9}+a_{10}=16p \bigl(70p^{3}-287p^{2}+350p-112 \bigr)>0, \end{aligned}
(2.17)
\begin{aligned} &\sum_{n=11}^{15}a_{n}=-81p^{5}+2839p^{4} -13904p^{3}+25652p^{2}-19600p+4992>0, \end{aligned}
(2.18)
\begin{aligned} &a_{16}+a_{17}=5 \bigl(-13p^{5}+145p^{4}-608p^{3}+1196p^{2}-1104p+384 \bigr)>0. \end{aligned}
(2.19)
Note that
\begin{aligned}[b] 2p+8&>3p+6>p+8>2p+6>3p+4>8>p+6>2p+4\\ &>3p+2>6>2p+2>3p>4>p+2>2p>2>p>1. \end{aligned}
(2.20)
It follows from (2.15)-(2.20) that
$$g_{4}(x)> \Biggl(\sum_{n=0}^{8}a_{n} \Biggr)x^{p+8}+(a_{9}+a_{10})x^{2p+2}+ \Biggl(\sum_{n=11}^{15}a_{n} \Biggr)x^{2p}+(a_{16}+a_{17})x^{p}>0$$
(2.21)
for $$x\in(1, \infty)$$.

Therefore, part (2) follows easily from (2.11)-(2.14) and (2.21). □

## 3 Main results

### Theorem 3.1

The double inequality
$$H_{\alpha}(a, b)< X(a, b)<H_{\beta}(a, b)$$
holds for all $$a, b>0$$ with $$a\neq b$$ if and only if $$\alpha\leq1/2$$ and $$\beta\geq\log3/(1+\log2)=0.6488\cdots$$.

### Proof

Since $$X(a,b)$$ and $$H_{p}(a,b)$$ are symmetric and homogeneous of degree one, we assume that $$a>b$$. Let $$x=\sqrt{a/b}\in (1, \infty)$$ and $$p\in\mathbb{R}$$. Then (1.1) and (1.2) lead to
\begin{aligned} &\log \bigl[X(a,b) \bigr]-\log \bigl[H_{p}(a,b) \bigr] \\ &\quad=\log \biggl(\frac{x^{2}+1}{2} \biggr)+\frac{2x}{x^{2}-1}\arcsin \biggl( \frac{x^{2}-1}{x^{2}+1} \biggr)-\frac{1}{p}\log \biggl(\frac {x^{2p}+x^{p}+1}{3} \biggr)-1:=F(x). \end{aligned}
(3.1)
Simple computations lead to
\begin{aligned}& F(1)=0, \end{aligned}
(3.2)
\begin{aligned}& \lim_{x\rightarrow+\infty}F(x)=\frac{1}{p}\log3-\log2-1, \end{aligned}
(3.3)
\begin{aligned}& F^{\prime}(x)=\frac{2(1+x^{2})}{(x^{2}-1)^{2}}F_{1}(x), \end{aligned}
(3.4)
where
\begin{aligned}& F_{1}(x)=\frac {(x^{2}-1)(2x^{2p}+x^{p+2}+x^{p}+2x^{2})}{2x(1+x^{2}) (x^{2p}+x^{p}+1)}-\arcsin \biggl(\frac{1-x^{2}}{1+x^{2}} \biggr), \\& F_{1}(1)=0,\qquad \lim_{x\rightarrow+\infty}F_{1}(x)=+ \infty, \end{aligned}
(3.5)
\begin{aligned}& F^{\prime}_{1}(x) =-\frac{x^{2}-1}{2(x^{2}+1)^{2}(x^{2p}+x^{p}+1)^{2}}f(x), \end{aligned}
(3.6)
where $$f(x)$$ is defined as in Lemma 2.1.
If $$p=1/2$$, then from Lemma 2.1(1), (3.1), (3.2), and (3.4)-(3.6) we clearly see that
$$X(a,b)>H_{1/2}(a,b)$$
for all $$a, b>0$$ with $$a\neq b$$.
If $$p=\log3/(1+\log2)$$, then (3.3) becomes
$$\lim_{x\rightarrow+\infty}F(x)=0.$$
(3.7)

It follows from Lemma 2.2(2) and (3.6) that there exists $$\lambda\in (1, \infty)$$ such that $$F_{1}(x)$$ is strictly decreasing on $$[1, \lambda ]$$ and strictly increasing on $$[\lambda, \infty)$$.

Equations (3.4) and (3.5) together with the piecewise monotonicity of $$F_{1}$$ lead to the conclusion that there exists $$\lambda^{\ast}\in(1, \infty)$$ such that $$F(x)$$ is strictly decreasing on $$[1, \lambda^{\ast }]$$ and strictly increasing on $$[\lambda^{\ast}, \infty)$$.

Therefore,
$$X(a,b)< H_{\log3/(1+\log2)}(a,b)$$
for all $$a, b>0$$ with $$a\neq b$$ follows from (3.1), (3.2), (3.7), and the piecewise monotonicity of F.

Next, we prove that $$\alpha=1/2$$ and $$\beta=\log3/(1+\log2)$$ are the best possible parameters such that the double inequality $$H_{\alpha}(a, b)< X(a, b)<H_{\beta}(a, b)$$ holds for all $$a, b>0$$ with $$a\neq b$$.

If $$p<\log3/(1+\log2)$$, then (3.3) leads to
$$\lim_{x\rightarrow+\infty}F(x)>0.$$
(3.8)

Equation (3.1) and inequality (3.8) imply that there exists large enough $$T_{0}=T_{0}(p)>1$$ such that $$X(a,b)>H_{p}(a,b)$$ for all $$a, b>0$$ with $$a/b\in(T_{0}, \infty)$$ if $$p<\log3/(1+\log2)$$.

Let $$p>1/2$$, $$x>0$$, and $$x\rightarrow0$$. Then elaborated computations lead to
\begin{aligned} &H{_{p}}(1, 1+x)-X(1, 1+x) \\ &\quad= \biggl[\frac{1+(1+x)^{p/2}+(1+x)^{p}}{3} \biggr]^{1/p}- \biggl(1+ \frac{x}{2} \biggr)e^{\frac{2\sqrt{1+x}\arcsin (\frac{x}{2+x} )}{x}-1} \\ &\quad=\frac{2p-1}{24}x^{2}+o \bigl(x^{2} \bigr). \end{aligned}
(3.9)

Equation (3.9) implies that there exists small enough $$\delta_{0}=\delta _{0}(p)>0$$ such that $$X(1, 1+x)< H{_{p}}(1, 1+x)$$ for $$x\in(0, \delta _{0})$$ if $$p>1/2$$. □

### Theorem 3.2

The double inequality
$$H_{\lambda}(a, b)< U(a, b)<H_{\mu}(a, b)$$
holds for all $$a, b>0$$ with $$a\neq b$$ if and only if $$\lambda\leq2\log 3/(2\log\pi-\log2)=1.3764\cdots$$ and $$\mu\geq2$$.

### Proof

Since $$U(a,b)$$ and $$H_{p}(a,b)$$ are symmetric and homogeneous of degree one, we assume that $$a>b$$. Let $$x=\sqrt{a/b}\in (1, \infty)$$ and $$p\in\mathbb{R}$$. Then (1.1) and (1.3) lead to
\begin{aligned} &\log \bigl[U(a,b) \bigr]-\log \bigl[H_{p}(a,b) \bigr] \\ &\quad=\log \biggl[\frac{x^{2}-1}{\sqrt{2}\arctan (\frac{x^{2}-1}{\sqrt {2}x} )} \biggr]-\frac{1}{p}\log \biggl( \frac{x^{2p}+x^{p}+1}{3} \biggr):=G(x). \end{aligned}
(3.10)
Simple computations lead to
\begin{aligned}& G(1)=0, \end{aligned}
(3.11)
\begin{aligned}& \lim_{x\rightarrow+\infty}G(x)=\frac{1}{p}\log3+ \frac{1}{2}\log2-\log \pi, \end{aligned}
(3.12)
\begin{aligned}& G^{\prime}(x)=\frac {2x^{2p}+x^{p+2}+x^{p}+2x^{2}}{x(x^{2}-1)(x^{2p}+x^{p}+1)\arctan (\frac{x^{2}-1}{\sqrt{2}x} )}G_{1}(x), \end{aligned}
(3.13)
where
\begin{aligned}& G_{1}(x)=\arctan \biggl(\frac{x^{2}-1}{\sqrt{2}x} \biggr)- \frac{\sqrt {2}x(x^{4}-1)(x^{2p}+x^{p}+1)}{(x^{4}+1)(2x^{2p}+x^{p+2}+x^{p}+2x^{2})}, \\& G_{1}(1)=0,\qquad \lim_{x\rightarrow+\infty}G_{1}(x)=- \infty, \end{aligned}
(3.14)
\begin{aligned}& G^{\prime}_{1}(x)=-\frac{\sqrt {2}x^{2}(x^{2}-1)}{(x^{4}+1)^{2}(2x^{2p}+x^{p+2}+x^{p}+2x^{2})^{2}}g(x), \end{aligned}
(3.15)
where $$g(x)$$ is defined as in Lemma 2.2.

If $$p=2\log3/(2\log\pi-\log2)$$, then (3.15) and Lemma 2.2(2) lead to the conclusion that there exists $$\mu\in(1, \infty)$$ such that $$G_{1}(x)$$ is strictly increasing on $$[1, \mu]$$ and strictly decreasing on $$[\mu, \infty)$$.

It follows from (3.13) and (3.14) together with the piecewise monotonicity of $$G_{1}$$ that there exists $$\mu^{\ast}\in(1, \infty)$$ such that $$G(x)$$ is strictly increasing on $$[1, \mu^{\ast}]$$ and strictly decreasing on $$[\mu^{\ast}, \infty)$$.

Note that (3.12) becomes
$$\lim_{x\rightarrow+\infty}G(x)=0.$$
(3.16)
Therefore,
$$U(a,b)>H_{2\log3/(2\log\pi-\log2)}(a,b)$$
for all $$a, b>0$$ with $$a\neq b$$ follows from (3.10), (3.11), and (3.16) together with the piecewise monotonicity of G.
If $$p=2$$, then
$$U(a,b)< H_{2}(a,b)$$
for all $$a, b>0$$ with $$a\neq b$$ follows easily from (3.10), (3.11), and (3.13)-(3.15) together with Lemma 2.2(1).
Next, we prove that $$\lambda=2\log3/(2\log\pi-\log2)$$ and $$\mu=2$$ are the best possible parameters such that the double inequality
$$H_{\lambda}(a, b)< U(a, b)<H_{\mu}(a, b)$$
holds for all $$a, b>0$$ with $$a\neq b$$.
If $$p>2\log3/(2\log\pi-\log2)$$, then (3.12) leads to
$$\lim_{x\rightarrow+\infty}G(x)< 0.$$
(3.17)

Equation (3.10) and inequality (3.17) imply that there exists large enough $$T_{1}=T_{1}(p)>1$$ such that $$U(a, b)< H_{p}(a,b)$$ for all $$a, b>0$$ with $$a/b\in(T_{1}, \infty)$$.

Let $$p<2$$, $$x>0$$, and $$x\rightarrow0$$. Then elaborated computations lead to
\begin{aligned} &U(1, 1+x)-H_{p}(1, 1+x) \\ &\quad=\frac{x}{\sqrt{2}\arctan (\frac{x}{\sqrt{2(1+x)}} )} - \biggl[\frac{1+(1+x)^{p/2}+(1+x)^{p}}{3} \biggr]^{1/p} \\ &\quad=\frac{2-p}{12}x^{2}+o \bigl(x^{2} \bigr). \end{aligned}
(3.18)

Inequality (3.18) implies that there exists small enough $$\delta _{1}=\delta_{1}(p)>0$$ such that $$U(1, 1+x)>H_{p}(1, 1+x)$$ for $$x\in(0, \delta_{1})$$. □

## Declarations

### Acknowledgements

The authors wish to thank the anonymous referees for their careful reading of the manuscript and their fruitful comments and suggestions. The research was supported by the Natural Science Foundation of China under Grants 61374086, 11171307 and 11401191, the Natural Science Foundation of Zhejiang Province under Grant LY13A010004, the Natural Science Foundation of the Open University of China under Grant Q1601E-Y and the Natural Science Foundation of Zhejiang Broadcast and TV University under Grant XKT-13Z04. 