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 \(p\rightarrow M_{p}(a,b)\). □