Optimal bounds for two Sándor-type means in terms of power means

Abstract

In the article, we prove that the double inequalities $M_{\alpha }(a,b)< S_{QA}(a,b)< M_{\beta}(a,b)$ and $M_{\lambda }(a,b)< S_{AQ}(a,b)< M_{\mu}(a,b)$ hold for all $a, b>0$ with $a\neq b$ if and only if $\alpha\leq\log 2/[1+\log2-\sqrt{2}\log(1+\sqrt{2})]=1.5517\ldots$ , $\beta\geq5/3$, $\lambda\leq4\log2/[4+2\log2-\pi]=1.2351\ldots$ and $\mu\geq4/3$, where $S_{QA}(a,b)=A(a,b)e^{Q(a,b)/M(a,b)-1}$ and $S_{AQ}(a,b)=Q(a,b)e^{A(a,b)/T(a,b)-1}$ are the Sándor-type means, $A(a,b)=(a+b)/2$, $Q(a,b)=\sqrt{(a^{2}+b^{2})/2}$, $T(a,b)=(a-b)/[2\arctan((a-b)/(a+b))]$, and $M(a,b)=(a-b)/[2\sinh ^{-1}((a-b)/(a+b))]$ are, respectively, the arithmetic, quadratic, second Seiffert, and Neuman-Sándor means.

1 Introduction

For $p\in\mathbb{R}$ and $a,b>0$ with $a\neq b$, the pth power mean $M_{p}(a,b)$ and Schwab-Borchardt mean $\operatorname{SB}(a,b)$ [1, 2] of a and b are, respectively, given by

$$ M_{p}(a,b)= \textstyle\begin{cases} (\frac{a^{p}+b^{p}}{2} )^{1/p},& p\neq0, \\ \sqrt{ab}, & p=0 \end{cases} $$

(1.1)

and

$$ \operatorname{SB}(a,b)= \textstyle\begin{cases} \frac{\sqrt{b^{2}-a^{2}}}{\cos^{-1}{(a/b)}}, & a< b, \\ \frac{\sqrt{a^{2}-b^{2}}}{\cosh^{-1}{(a/b)}}, & a>b, \end{cases} $$

where $\cos^{-1}(x)$ and $\cosh^{-1}(x)=\log(x+\sqrt{x^{2}-1})$ are the inverse cosine and inverse hyperbolic cosine functions, respectively.

It is well known that the power mean $M_{p}(a,b)$ is continuous and strictly increasing with respect to $p\in\mathbb{R}$ for fixed $a, b>0$ with $a\neq b$, the Schwab-Borchardt mean $\operatorname{SB}(a,b)$ is strictly increasing in both a and b, nonsymmetric and homogeneous of degree 1 with respect to a and b. Many symmetric bivariate means are special cases of the Schwab-Borchardt mean. For example, $P(a,b)=(a-b)/[2\arcsin((a-b)/(a+b))]=\operatorname{SB}[G(a,b), A(a,b)]$ is the first Seiffert mean, $T(a,b)=(a-b)/[2\arctan ((a-b)/(a+b))]=\operatorname{SB}[A(a,b), Q(a,b)]$ is the second Seiffert mean, $M(a,b)=(a-b)/[2\sinh^{-1}((a-b)/(a+b))]=\operatorname{SB}[Q(a,b), A(a,b)]$ is the Neuman-Sándor mean, $L(a,b)=(a-b)/[2\tanh ^{-1}((a-b)/(a+b))]=\operatorname{SB}[A(a,b), G(a,b)]$ is the logarithmic mean, where $\sinh^{-1}(x)=\log(x+\sqrt{1+x^{2}})$ is the inverse hyperbolic sine function, $\tanh^{-1}(x)=\log[(1+x)/(1-x)]/2$ is the inverse hyperbolic tangent function, and $G(a,b)=\sqrt{ab}$, $A(a,b)=(a+b)/2$, and $Q(a,b)=\sqrt{(a^{2}+b^{2})/2}$ are the geometric, arithmetic, and quadratic means of a and b, respectively.

The Sándor mean $X(a,b)=A(a,b)e^{G(a,b)/P(a,b)-1}$ [3] can be rewritten as $X(a,b)= A(a,b)e^{G(a,b)/\operatorname{SB}[G(a,b), A(a,b)]-1}$. Yang [4] proved that $S(a,b)=be^{a/\operatorname{SB}(a,b)-1}$ is a mean of a and b, and introduced two Sándor-type means $S_{QA}(a,b)$ and $S_{AQ}(a,b)$ as follows:

$$\begin{aligned}& S_{QA}(a,b)\triangleq S\bigl[Q(a,b), A(a,b)\bigr] \\& \hphantom{S_{QA}(a,b)}=A(a,b)e^{Q(a,b)/\operatorname{SB}[Q(a,b), A(a,b)]-1}=A(a,b)e^{Q(a,b)/M(a,b)-1}, \end{aligned}$$

(1.2)

$$\begin{aligned}& S_{AQ}(a,b)\triangleq S\bigl[A(a,b), Q(a,b)\bigr] \\& \hphantom{S_{AQ}(a,b)}=Q(a,b)e^{A(a,b)/\operatorname{SB}[A(a,b), Q(a,b)]-1}=Q(a,b)e^{A(a,b)/T(a,b)-1}. \end{aligned}$$

(1.3)

Recently, the bounds involving the power mean for certain bivariate means and Gaussian hypergeometric function have attracted the attention of many researchers [5–21].

Radó [22] (see also [23–25]) proved that the double inequalities

$$ M_{p}(a,b)< L(a,b)< M_{q}(a,b),\qquad M_{\lambda}(a,b)< I(a,b)< M_{\mu}(a,b) $$

hold for all $a, b>0$ with $a\neq b$ if and only if $p\leq0$, $q\geq 1/3$, $\lambda\leq2/3$, and $\mu\geq\log2$, where $I(a,b)=(b^{b}/a^{a})^{1/(b-a)}/e$ is the identric mean of a and b.

In [26–29], the authors proved that the double inequality

$$ M_{p}(a,b)< T^{\ast}(a,b)< M_{q}(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $p\leq3/2$ and $q\geq\log2/(\log\pi-\log2)$, where $T^{\ast}(a,b)=\frac{2}{\pi}\int _{0}^{\pi/2}\sqrt{a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}\, d\theta$ is the Toader mean of a and b.

Jagers [30], Hästö [31, 32], Costin and Toader [33], and Li et al. [34] proved that $p_{1}=\log2/\log\pi$, $q_{1}=2/3$, $p_{2}=\log2/(\log\pi-\log2)$, and $q_{2}=5/3$ are the best possible parameters such that the double inequalities

$$ M_{p_{1}}(a,b)< P(a,b)< M_{q_{1}}(a,b),\qquad M_{p_{2}}(a,b)< T(a,b)< M_{q_{2}}(a,b) $$

hold for all $a, b>0$ with $a\neq b$.

In [35–38], the authors proved that the double inequalities

$$\begin{aligned}& M_{\lambda_{1}}(a,b)< M(a,b)< M_{\mu_{1}}(a,b), \\& M_{\lambda_{2}}(a,b)< U(a,b)< M_{\mu_{2}}(a,b), \\& M_{\lambda_{3}}(a,b)< X(a,b)< M_{\mu_{3}}(a,b) \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$ if and only if $\lambda_{1}\leq \log2/\log[2\log(1+\sqrt{2})]$, $\mu_{1}\geq4/3$, $\lambda_{2}\leq 2\log2/(2\log\pi-\log2)$, $\mu_{2}\geq4/3$, $\lambda_{3}\leq1/3$, and $\mu_{3}\geq\log2/(1+\log2)$, where $U(a,b)=(a-b)/ [\sqrt {2}\arctan (\frac{a-b}{\sqrt{2ab}} ) ]$ is the Yang mean of a and b.

The main purpose of this paper is to present the best possible parameters α, β, λ, and μ such that the double inequalities

$$ M_{\alpha}(a,b)< S_{QA}(a,b)< M_{\beta}(a,b), \qquad M_{\lambda }(a,b)< S_{AQ}(a,b)< M_{\mu}(a,b) $$

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 $p\in\mathbb{R}$ and

$$ f(x)=(p-1)x^{p+1}-3x^{p}+3x^{p-2}+(1-p)x^{p-3}+3x^{2p-2}+x^{2p-3}-x-3. $$

(2.1)

Then the following statements are true:

• (1) $f(x)>0$ for all $x\in(1, \infty)$ if $p=5/3$;

• (2) there exists $\sigma\in(1, \infty)$ such that $f(x)<0$ for $x\in (1, \sigma)$ and $f(x)>0$ for $x\in(\sigma, \infty)$ if $p=\log2/[1+\log2-\sqrt{2}\log(1+\sqrt{2})]=1.5517\ldots$ .

Proof

For part (1), if $p=5/3$, then (2.1) leads to

$$ f(x)=\frac{ (x^{\frac{2}{3}}-1 ) (x^{\frac{1}{3}}-1 )^{2}}{3x^{\frac{4}{3}}} \bigl(2x^{\frac{8}{3}}+4x^{\frac{7}{3}}+8x^{2}+3x^{\frac {5}{3}}+9x^{\frac{4}{3}}+3x+8x^{\frac{2}{3}}+4x^{\frac{1}{3}}+2 \bigr). $$

(2.2)

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

For part (2), let $p=\log2/[1+\log2-\sqrt{2}\log(1+\sqrt {2})]=1.5517\ldots$ , $f_{1}(x)=f^{\prime}(x)$, $f_{2}(x)=x^{5-p}f^{\prime}_{1}(x)$ and $f_{3}(x)=f^{\prime}_{2}(x)$. Then simple computations lead to

$$\begin{aligned}& f(1)=0, \qquad \lim_{x\rightarrow+\infty}f(x)=+\infty, \end{aligned}$$

(2.3)

$$\begin{aligned}& f_{1}(1)=12 \biggl(p-\frac{5}{3} \biggr)< 0, \qquad \lim _{x\rightarrow+\infty }f_{1}(x)=+\infty, \end{aligned}$$

(2.4)

$$\begin{aligned}& f_{2}(1)=24 \biggl(p-\frac{3}{2} \biggr) \biggl(p- \frac{5}{3} \biggr)< 0,\qquad \lim_{x\rightarrow+\infty}f_{2}(x)=+ \infty, \end{aligned}$$

(2.5)

$$\begin{aligned}& f_{3}(x)=6 \bigl(p^{2}-1 \bigr) (2p-3)x^{p}+2p(p-2) (2p-3)x^{p-1} \\& \hphantom{f_{3}(x)={}}{}+4p \bigl(p^{2}-1 \bigr)x^{3}-9p(p-1)x^{2}+3(p-2) (p-3). \end{aligned}$$

(2.6)

Note that

$$\begin{aligned}& 2p(p-2) (2p-3)x^{p-1}>2p(p-2) (2p-3)x^{p}, \qquad -9p(p-1)x^{2}>-9p(p-1)x^{3}, \end{aligned}$$

(2.7)

$$\begin{aligned}& p(p-1) (4p-5)x^{3}>p(p-1) (4p-5) \end{aligned}$$

(2.8)

for $x>1$, and

$$\begin{aligned}& 16p^{3}-32p^{2}+18>16\times1.55^{3}-32 \times1.552^{2}+18=0.503472>0, \end{aligned}$$

(2.9)

$$\begin{aligned}& 4p^{3}-6p^{2}-10p+18>4\times1.5^{3}-6 \times1.6^{2}-10\times1.6+18=0.14>0. \end{aligned}$$

(2.10)

It follows from (2.6)-(2.10) that

$$\begin{aligned} f_{3}(x) >&6 \bigl(p^{2}-1 \bigr) (2p-3)x^{p}+2p(p-2) (2p-3)x^{p} \\ &{}+4p \bigl(p^{2}-1 \bigr)x^{3}-9p(p-1)x^{3}+3(p-2) (p-3) \\ =& \bigl(16p^{3}-32p^{2}+18 \bigr)x^{p}+p(p-1) (4p-5)x^{3}+3(p-2) (p-3) \\ >& \bigl(16p^{3}-32p^{2}+18 \bigr)x^{p}+p(p-1) (4p-5)+3(p-2) (p-3) \\ =& \bigl(16p^{3}-32p^{2}+18 \bigr)x^{p}+ \bigl(4p^{3}-6p^{2}-10p+18 \bigr)>0 \end{aligned}$$

(2.11)

for $x>1$.

Inequality (2.11) implies that $f_{2}(x)$ is strictly increasing on $(1, \infty)$. Then from (2.5) we know that there exists $\sigma_{1}>1$ such that $f_{1}(x)$ is strictly decreasing on $(1, \sigma_{1}]$ and strictly increasing on $[\sigma_{1}, \infty)$.

It follows from (2.4) and the piecewise monotonicity of $f_{1}(x)$ that there exists $\sigma_{2}>1$ such that $f(x)$ is strictly decreasing on $(1, \sigma_{2}]$ and strictly increasing on $[\sigma_{2}, \infty)$.

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

Lemma 2.2

Let $p\in\mathbb{R}$, and

$$ g(x)=(p-1)x^{p+1}-(p+1)x^{p}+(p+1)x^{p-1}+(1-p)x^{p-2}+x^{2p-1}+x^{2p-2}-x-1. $$

(2.12)

Then the following statements are true:

• (1) $g(x)>0$ for all $x\in(1, \infty)$ if $p=4/3$;

• (2) there exists $\tau\in(1, \infty)$ such that $g(x)<0$ for $x\in (1, \tau)$ and $g(x)>0$ for $x\in(\tau, \infty)$ if $p=4\log2/[4+2\log2-\pi]=1.2351\ldots$ .

Proof

For part (1), if $p=4/3$, then (2.12) becomes

$$ g(x)=\frac{ (x^{1/3}-1 )^{3}}{3x^{2/3}} \bigl(x^{2}+3x^{5/3}+9x^{4/3}+12x+9x^{2/3}+3x^{1/3}+1 \bigr). $$

(2.13)

Therefore, part (1) follows from (2.13).

For part (2), let $p=4\log2/[4+2\log2-\pi]=1.2351\ldots$ , $g_{1}(x)=g^{\prime}(x)$, $g_{2}(x)=x^{4-p}g^{\prime}_{1}(x)/(p-1)$, and $g_{3}(x)=g^{\prime}_{2}(x)$. Then simple computations lead to

$$\begin{aligned}& g(1)=0, \qquad \lim_{x\rightarrow+\infty}g(x)=+\infty, \end{aligned}$$

(2.14)

$$\begin{aligned}& g_{1}(1)=6 \biggl(p-\frac{4}{3} \biggr)< 0,\qquad \lim _{x\rightarrow+\infty }g_{1}(x)=+\infty, \end{aligned}$$

(2.15)

$$\begin{aligned}& g_{2}(1)=12 \biggl(p-\frac{4}{3} \biggr)< 0, \qquad \lim _{x\rightarrow+\infty }g_{2}(x)=+\infty, \end{aligned}$$

(2.16)

$$\begin{aligned}& g_{3}(x)=2(p+1) (2p-1)x^{p}+2p(2p-3)x^{p-1} \\& \hphantom{g_{3}(x)={}}{}+3p(p+1)x^{2}-2p(p+1)x+(p+1) (p-2). \end{aligned}$$

(2.17)

Note that

$$\begin{aligned}& 2p(2p-3)x^{p-1}>2p(2p-3)x^{p}, \\& 2p(p+1)x< 2p(p+1)x^{2}, \\& (p+1) (p-2)>(p+1) (p-2)x^{2} \end{aligned}$$

(2.18)

for $x>1$.

It follows from (2.17) and (2.18) that

$$\begin{aligned} g_{3}(x) >&2(p+1) (2p-1)x^{p}+2p(2p-3)x^{p}+3p(p+1)x^{2} \\ &{}-2p(p+1)x^{2}+(p+1) (p-2)x^{2} \\ =&2 \bigl(4p^{2}-2p-1 \bigr)x^{p}+2\bigl(p^{2}-1 \bigr)x^{2}>0 \end{aligned}$$

(2.19)

for $x>1$.

Inequality (2.19) implies that $g_{2}(x)$ is strictly increasing on $(1, \infty)$. Then from (2.16) we know that there exists $\tau_{1}\in(1, \infty)$ such that $g_{1}(x)$ is strictly decreasing on $(1, \tau_{1}]$ and strictly increasing on $[\tau_{1}, \infty)$.

It follows from (2.15) and the piecewise monotonicity of $g_{1}(x)$ that there exists $\tau_{2}\in(1, \infty)$ such that $g(x)$ is strictly decreasing on $(1, \tau_{2}]$ and strictly increasing on $[\tau_{2}, \infty)$.

Therefore, part (2) follows from (2.14) and the piecewise monotonicity of $g(x)$. □

3 Main results

Theorem 3.1

The double inequality

$$ M_{\alpha}(a,b)< S_{QA}(a,b)< M_{\beta}(a,b) $$

holds for all $a,b>0$ with $a\neq b$ if and only if $\alpha\leq\log 2/[1+\log2-\log(1+\sqrt{2})]=1.5517\ldots$ and $\beta\geq5/3$.

Proof

Since both $S_{QA}(a,b)$ and $M_{p}(a,b)$ are symmetric and homogeneous of degree one, we assume that $a>b$. Let $x=a/b>1$ and $p>0$. Then (1.1) and (1.2) lead to

$$\begin{aligned}& \log \bigl[S_{QA}(a,b) \bigr]-\log \bigl[M_{p}(a,b) \bigr] \\& \quad =\log \biggl(\frac{x+1}{2} \biggr)+\frac{\sqrt{2(x^{2}+1)}\sinh^{-1} (\frac{x-1}{x+1} )}{x-1}- \frac{1}{p}\log \biggl(\frac {x^{p}+1}{2} \biggr)-1. \end{aligned}$$

(3.1)

Let

$$ F(x)=\log \biggl(\frac{x+1}{2} \biggr)+\frac{\sqrt{2(x^{2}+1)}\sinh ^{-1} (\frac{x-1}{x+1} )}{x-1}- \frac{1}{p}\log \biggl(\frac {x^{p}+1}{2} \biggr)-1. $$

(3.2)

Then elaborated computations lead to

$$\begin{aligned}& F\bigl(1^{+}\bigr)=0, \end{aligned}$$

(3.3)

$$\begin{aligned}& \lim_{x\rightarrow+\infty}F(x)=\sqrt{2}\log(1+\sqrt{2})-(1+\log 2)+ \frac{1}{p}\log2, \end{aligned}$$

(3.4)

$$\begin{aligned}& F^{\prime}(x)=\frac{2(x+1)}{(x-1)^{2}\sqrt{2(x^{2}+1)}}F_{1}(x), \end{aligned}$$

(3.5)

where

$$\begin{aligned}& F_{1}(x)=\frac{\sqrt{2(x^{2}+1)}(x-1)(x^{p-1}+1)}{2(x+1)(x^{p}+1)}-\sinh ^{-1} \biggl( \frac{x-1}{x+1} \biggr), \\& F_{1}(1)=0, \qquad \lim_{x\rightarrow\infty}F_{1}(x)= \frac{\sqrt{2}}{2}-\log (1+\sqrt{2})=-0.1742\ldots< 0, \end{aligned}$$

(3.6)

$$\begin{aligned}& F_{1}^{\prime}(x)=-\frac{x(x-1)}{(x+1)^{2}(x^{p}+1)^{2}\sqrt{2(x^{2}+1)}}f(x), \end{aligned}$$

(3.7)

where $f(x)$ is defined by (2.1).

We divide the proof into four cases.

Case 1.1. $p=\log2/[1+\log2-\log(1+\sqrt{2})]$. Then it follows from Lemma 2.1(2) and (3.7) that there exists $\sigma\in(1, \infty)$ such that $F_{1}(x)$ is strictly increasing on $(1, \sigma]$ and strictly decreasing on $[\sigma, \infty)$.

Equations (3.5) and (3.6) together with the piecewise monotonicity of $F_{1}(x)$ lead to the conclusion that there exists $\sigma_{0}\in(1, \infty)$ such that $F(x)$ is strictly increasing on $(1, \sigma_{0}]$ and strictly decreasing on $[\sigma_{0}, \infty)$.

Note that (3.4) becomes

$$ \lim_{x\rightarrow+\infty}F(x)=0. $$

(3.8)

Therefore,

$$ S_{QA}(a,b)>M_{\log2/[1+\log2-\log(1+\sqrt{2})]}(a,b) $$

for all $a, b>0$ with $a\neq b$ follows from (3.1)-(3.3) and (3.8) together with the piecewise monotonicity of $F(x)$.

Case 1.2. $p=5/3$. Then it follows from Lemma 2.1(1) and (3.7) that $F_{1}(x)$ is strictly decreasing on $(1, \infty)$.

Therefore,

$$ S_{QA}(a,b)< M_{5/3}(a,b) $$

for all $a, b>0$ with $a\neq b$ follows from (3.1)-(3.3), (3.5), (3.6), and the monotonicity of $F(x)$.

Case 1.3. $p>\log2/[1+\log2-\log(1+\sqrt{2})]$. Then (3.4) leads to

$$ \lim_{x\rightarrow+\infty}F(x)< 0. $$

(3.9)

Equations (3.1) and (3.2) together with inequality (3.9) imply that there exists large enough $C_{0}>1$ such that

$$ S_{QA}(a,b)< M_{p}(a,b) $$

for all $a, b>0$ with $a/b\in(C_{0}, \infty)$.

Case 1.4. $1< p<5/3$. Let $x>0$, $x\rightarrow0$, then making use of (1.1) and (1.2) together with the Taylor expansion we get

$$\begin{aligned}& \begin{aligned}[b] &S_{QA}(1, 1+x)-M_{p}(1,1+x) \\ &\quad = \biggl(1+\frac{x}{2} \biggr)e^{\sqrt{2(x^{2}+2x+2)}\sinh ^{-1}[x/(2+x)]/x-1}- \biggl[ \frac{1+(1+x)^{p}}{2} \biggr]^{1/p} \\ &\quad =\frac{5-3p}{24}x^{2}+o\bigl(x^{2}\bigr). \end{aligned} \end{aligned}$$

(3.10)

Equation (3.10) implies that there exists small enough $\delta_{0}>0$ such that

$$ S_{QA}(1, 1+x)>M_{p}(1, 1+x) $$

for $x\in(0, \delta_{0})$.

Therefore, Theorem 3.1 follows easily from Cases 1.1-1.4 and the monotonicity of the function $p\rightarrow M_{p}(a,b)$. □

Theorem 3.2

The double inequality

$$ M_{\lambda}(a,b)< S_{AQ}(a,b)< M_{\mu}(a,b) $$

holds for all $a,b>0$ with $a\neq b$ if and only if $\lambda\leq4\log 2/[4+2\log2-\pi]=1.2351\ldots$ and $\beta\geq4/3$.

Proof

Since both $S_{AQ}(a,b)$ and $M_{p}(a,b)$ are symmetric and homogeneous of degree one, we assume that $a>b$. Let $x=a/b>1$ and $p>0$. Then (1.1) and (1.3) lead to

$$\begin{aligned}& \log \bigl[S_{AQ}(a,b) \bigr]-\log \bigl[M_{p}(a,b) \bigr] \\& \quad =\frac{1}{2}\log \biggl(\frac{x^{2}+1}{2} \biggr)+ \frac{x+1}{x-1}\arctan \biggl(\frac{x-1}{x+1} \biggr)-\frac{1}{p}\log \biggl(\frac {x^{p}+1}{2} \biggr)-1. \end{aligned}$$

(3.11)

Let

$$ G(x)=\frac{1}{2}\log \biggl(\frac{x^{2}+1}{2} \biggr)+ \frac {x+1}{x-1}\arctan \biggl(\frac{x-1}{x+1} \biggr)-\frac{1}{p}\log \biggl(\frac {x^{p}+1}{2} \biggr)-1. $$

(3.12)

Then elaborated computations lead to

$$\begin{aligned}& G\bigl(1^{+}\bigr)=0, \end{aligned}$$

(3.13)

$$\begin{aligned}& \lim_{x\rightarrow+\infty}G(x)=\frac{\pi}{4}-\frac{1}{2}\log2-1+ \frac {1}{p}\log2, \end{aligned}$$

(3.14)

$$\begin{aligned}& G^{\prime}(x)=\frac{2}{(x-1)^{2}}G_{1}(x), \end{aligned}$$

(3.15)

where

$$\begin{aligned}& G_{1}(x)=\frac{(x-1)(x^{p-1}+1)}{2(x^{p}+1)}-\arctan \biggl(\frac {x-1}{x+1} \biggr), \\& G_{1}(1)=0, \qquad \lim_{x\rightarrow+\infty}G_{1}(x)= \frac{1}{2}-\frac{\pi}{4}< 0, \end{aligned}$$

(3.16)

$$\begin{aligned}& G^{\prime}_{1}(x)=-\frac{x-1}{2(x^{2}+1)^{2}(x^{p}+1)^{2}}g(x), \end{aligned}$$

(3.17)

where $g(x)$ is defined by (2.12).

We divide the proof into four cases.

Case 2.1. $p=4\log2/[4+2\log2-\pi]$. Then it follows from Lemma 2.2(2) and (3.17) that there exists $\tau\in(1, \infty)$ such that $G_{1}(x)$ is strictly increasing on $(1, \tau]$ and strictly decreasing on $[\tau, \infty)$.

Equations (3.15) and (3.16) together with the piecewise monotonicity of $G_{1}(x)$ lead to the conclusion that there exists $\tau_{0}\in(1, \infty)$ such that $G(x)$ is strictly increasing on $(1, \tau_{0}]$ and strictly decreasing on $[\tau_{0}, \infty)$.

Note that (3.14) becomes

$$ \lim_{x\rightarrow+\infty}G(x)=0. $$

(3.18)

Therefore,

$$ S_{AQ}(a,b)>M_{4\log2/[4+2\log2-\pi]}(a,b) $$

follows from (3.11)-(3.13) and (3.18) together with the piecewise monotonicity of $G(x)$.

Case 2.2. $p=4/3$. Then Lemma 2.2(2) and (3.17) imply that $G_{1}(x)$ is strictly decreasing on $(1, \infty)$.

Therefore,

$$ S_{AQ}(a,b)< M_{4/3}(a,b) $$

follows easily from (3.11)-(3.13), (3.15), (3.16), and the monotonicity of $G_{1}(x)$.

Case 2.3. $p>4\log2/[4+2\log2-\pi]$. Then (3.14) leads to

$$ \lim_{x\rightarrow+\infty}G(x)< 0. $$

(3.19)

Equations (3.11) and (3.12) and inequality (3.19) imply that there exists large enough $C_{1}>1$ such that

$$ S_{AQ}(a,b)< M_{p}(a,b) $$

for all $a, b>0$ with $a/b\in(C_{1}, \infty)$.

Case 2.4. $0< p<4/3$. Let $x>0$ and $x\rightarrow0$. Then making use of (1.1) and (1.3) together with the Taylor expansion we get

$$\begin{aligned}& S_{AQ}(1, 1+x)-M_{p}(1,1+x) \\& \quad =\sqrt{\frac{1+(1+x)^{2}}{2}}e^{(2+x)\arctan[x/(2+x)]/x-1}- \biggl[\frac {1+(1+x)^{p}}{2} \biggr]^{1/p} \\& \quad =\frac{4-3p}{24}x^{2}+o\bigl(x^{2}\bigr). \end{aligned}$$

(3.20)

Equation (3.20) implies that there exists small enough $\delta_{1}>0$ such that

$$ S_{AQ}(1, 1+x)>M_{p}(1,1+x) $$

for $x\in(0, \delta_{1})$.

Therefore, Theorem 3.2 follows easily from Cases 2.1-2.4 and the monotonicity of the function $p\rightarrow M_{p}(a,b)$. □