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 and where \(\cos^{-1}(x)\) and \(\cosh^{-1}(x)=\log(x+\sqrt{x^{2}-1})\) are the inverse cosine and inverse hyperbolic cosine functions, respectively.
$$ M_{p}(a,b)= \textstyle\begin{cases} (\frac{a^{p}+b^{p}}{2} )^{1/p},& p\neq0, \\ \sqrt{ab}, & p=0 \end{cases} $$
(1.1)
$$ \operatorname{SB}(a,b)= \textstyle\begin{cases} \frac{\sqrt{b^{2}-a^{2}}}{\cos^{-1}{(a/b)}}, & a< b, \\ \frac{\sqrt{a^{2}-b^{2}}}{\cosh^{-1}{(a/b)}}, & a>b, \end{cases} $$
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.
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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 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.
$$ M_{p}(a,b)< L(a,b)< M_{q}(a,b),\qquad M_{\lambda}(a,b)< I(a,b)< M_{\mu}(a,b) $$
In [26‐29], the authors proved that the double inequality 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.
$$ M_{p}(a,b)< T^{\ast}(a,b)< M_{q}(a,b) $$
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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 hold for all \(a, b>0\) with \(a\neq b\).
$$ M_{p_{1}}(a,b)< P(a,b)< M_{q_{1}}(a,b),\qquad M_{p_{2}}(a,b)< T(a,b)< M_{q_{2}}(a,b) $$
In [35‐38], the authors proved that the double inequalities 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.
$$\begin{aligned}& M_{\lambda_{1}}(a,b)< M(a,b)< M_{\mu_{1}}(a,b), \\& M_{\lambda_{2}}(a,b)< U(a,b)< M_{\mu_{2}}(a,b), \\& M_{\lambda_{3}}(a,b)< X(a,b)< M_{\mu_{3}}(a,b) \end{aligned}$$
The main purpose of this paper is to present the best possible parameters α, β, λ, and μ such that the double inequalities hold for all \(a, b>0\) with \(a\neq b\).
$$ M_{\alpha}(a,b)< S_{QA}(a,b)< M_{\beta}(a,b), \qquad M_{\lambda }(a,b)< S_{AQ}(a,b)< M_{\mu}(a,b) $$
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
Note that
for \(x>1\), and
$$\begin{aligned}& f(1)=0, \qquad \lim_{x\rightarrow+\infty}f(x)=+\infty, \end{aligned}$$
(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)
$$\begin{aligned}& 2p(p-2) (2p-3)x^{p-1}>2p(p-2) (2p-3)x^{p}, \qquad -9p(p-1)x^{2}>-9p(p-1)x^{3}, \end{aligned}$$
(2.7)
$$\begin{aligned}& p(p-1) (4p-5)x^{3}>p(p-1) (4p-5) \end{aligned}$$
(2.8)
$$\begin{aligned}& 16p^{3}-32p^{2}+18>16\times1.55^{3}-32 \times1.552^{2}+18=0.503472>0, \end{aligned}$$
(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 for \(x>1\).
$$\begin{aligned} f_{3}(x) >&6 \bigl(p^{2}-1 \bigr) (2p-3)x^{p}+2p(p-2) (2p-3)x^{p} \\ &{}+4p \bigl(p^{2}-1 \bigr)x^{3}-9p(p-1)x^{3}+3(p-2) (p-3) \\ =& \bigl(16p^{3}-32p^{2}+18 \bigr)x^{p}+p(p-1) (4p-5)x^{3}+3(p-2) (p-3) \\ >& \bigl(16p^{3}-32p^{2}+18 \bigr)x^{p}+p(p-1) (4p-5)+3(p-2) (p-3) \\ =& \bigl(16p^{3}-32p^{2}+18 \bigr)x^{p}+ \bigl(4p^{3}-6p^{2}-10p+18 \bigr)>0 \end{aligned}$$
(2.11)
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 for \(x>1\).
$$\begin{aligned}& 2p(2p-3)x^{p-1}>2p(2p-3)x^{p}, \\& 2p(p+1)x< 2p(p+1)x^{2}, \\& (p+1) (p-2)>(p+1) (p-2)x^{2} \end{aligned}$$
(2.18)
It follows from (2.17) and (2.18) that for \(x>1\).
$$\begin{aligned} g_{3}(x) >&2(p+1) (2p-1)x^{p}+2p(2p-3)x^{p}+3p(p+1)x^{2} \\ &{}-2p(p+1)x^{2}+(p+1) (p-2)x^{2} \\ =&2 \bigl(4p^{2}-2p-1 \bigr)x^{p}+2\bigl(p^{2}-1 \bigr)x^{2}>0 \end{aligned}$$
(2.19)
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
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
\(\beta\geq5/3\).
$$ M_{\alpha}(a,b)< S_{QA}(a,b)< M_{\beta}(a,b) $$
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
where
where \(f(x)\) is defined by (2.1).
$$\begin{aligned}& F\bigl(1^{+}\bigr)=0, \end{aligned}$$
(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)
$$\begin{aligned}& F_{1}(x)=\frac{\sqrt{2(x^{2}+1)}(x-1)(x^{p-1}+1)}{2(x+1)(x^{p}+1)}-\sinh ^{-1} \biggl( \frac{x-1}{x+1} \biggr), \\& F_{1}(1)=0, \qquad \lim_{x\rightarrow\infty}F_{1}(x)= \frac{\sqrt{2}}{2}-\log (1+\sqrt{2})=-0.1742\ldots< 0, \end{aligned}$$
(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)
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, for all \(a, b>0\) with \(a\neq b\) follows from (3.1)-(3.3) and (3.8) together with the piecewise monotonicity of \(F(x)\).
$$ S_{QA}(a,b)>M_{\log2/[1+\log2-\log(1+\sqrt{2})]}(a,b) $$
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, for all \(a, b>0\) with \(a\neq b\) follows from (3.1)-(3.3), (3.5), (3.6), and the monotonicity of \(F(x)\).
$$ S_{QA}(a,b)< M_{5/3}(a,b) $$
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 for all \(a, b>0\) with \(a/b\in(C_{0}, \infty)\).
$$ S_{QA}(a,b)< M_{p}(a,b) $$
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 for \(x\in(0, \delta_{0})\).
$$ S_{QA}(1, 1+x)>M_{p}(1, 1+x) $$
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
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
\(\beta\geq4/3\).
$$ M_{\lambda}(a,b)< S_{AQ}(a,b)< M_{\mu}(a,b) $$
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
where
where \(g(x)\) is defined by (2.12).
$$\begin{aligned}& G\bigl(1^{+}\bigr)=0, \end{aligned}$$
(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)
$$\begin{aligned}& G_{1}(x)=\frac{(x-1)(x^{p-1}+1)}{2(x^{p}+1)}-\arctan \biggl(\frac {x-1}{x+1} \biggr), \\& G_{1}(1)=0, \qquad \lim_{x\rightarrow+\infty}G_{1}(x)= \frac{1}{2}-\frac{\pi}{4}< 0, \end{aligned}$$
(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)
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, follows from (3.11)-(3.13) and (3.18) together with the piecewise monotonicity of \(G(x)\).
$$ S_{AQ}(a,b)>M_{4\log2/[4+2\log2-\pi]}(a,b) $$
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, follows easily from (3.11)-(3.13), (3.15), (3.16), and the monotonicity of \(G_{1}(x)\).
$$ S_{AQ}(a,b)< M_{4/3}(a,b) $$
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 for all \(a, b>0\) with \(a/b\in(C_{1}, \infty)\).
$$ S_{AQ}(a,b)< M_{p}(a,b) $$
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 for \(x\in(0, \delta_{1})\).
$$ S_{AQ}(1, 1+x)>M_{p}(1,1+x) $$
Therefore, Theorem 3.2 follows easily from Cases 2.1-2.4 and the monotonicity of the function \(p\rightarrow M_{p}(a,b)\). □
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 11301127, 11371125 and 61374086, and the Natural Science Foundation of Hunan Province under Grant 12C0577.
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Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.