0 引言

对所有$a, b>0$且$a\ne b$, Schwab-Borchardt平均$SB({a, b})$^[1-4]定义为

$ SB(a, b) = \left\{ \begin{array}{l} \frac{{\sqrt {{b^2}-{a^2}} }}{{{\rm{arccos}}(a/b)}}, a < b, \\ \frac{{\sqrt {{a^2}-{b^2}} }}{{{\rm{arccosh}}(a/b)}}, \quad a > b, \end{array} \right. $

其中$\mathrm {arccos}(x)$和$\mathrm {arccosh} (x)=\log ({x+\sqrt{x^{2}-1}} )$分别是反余弦函数和反双曲余弦函数.

设$G({a, b})=\sqrt{ab}$, $A({a, b})=( {a+b})/2$, $Q({a, b})=\sqrt{( {a^{2}+b^{2}} )/2}$, $C({a, b})=( {a^{2}+b^{2}} )/( a$ $+b)$和$M_{p} ({a, b})=[{( {a^{p}+b^{p}} )/2}]^{1/p}$分别是两个正数$a$和$b$的几何平均, 算术平均, 二次平均, 反调和平均和$p$阶幂平均.

我们熟知Schwab-Borchardt平均$SB({a, b})$关于正数$a$和$b$都是严格单调递增的, 并且关于$a$和$b$是非对称和一阶齐次的.许多对称二元平均都是Schwab-Borchardt平均的特殊情形.例如: $P({a, b})=(a-b)/[{2\arcsin ( {(a-b)/(a+b)} )}]=SB[{G({a, b}), A({a, b})}]$是第一类Seiffert平均, $T({a, b})=(a-b)/[{2\arctan ( {(a-b)/(a+b)} )}]=SB[{A({a, b}), Q({a, b})}]$是第二类Seiffert平均, $M({a, b})=(a-b)/[{2\mathrm {arcsinh} ( {(a-b)/(a+b)} )}]=SB[{Q({a, b}), A({a, b})}]$是Neuman-Sándor平均, $L({a, b})=(a-b)/[{2\mathrm {arctanh} ( {(a-b)/(a+b)} )}]=SB[{A({a, b}), G({a, b})}]$是对数平均.

我们熟知, 对所有$a, b>0$且$a\ne b$, 不等式

$ G\left( {a, b} \right) < L\left( {a, b} \right) < P\left( {a, b} \right) < A\left( {a, b} \right) < M\left( {a, b} \right) < T\left( {a, b} \right) < Q\left( {a, b} \right) < C\left( {a, b} \right) $

成立.

在文献[5]中, 杨镇杭证明了$S({a, b})=b\mathrm e^{a/SB({a, b})-1}$是一个关于正数$a$和$b$的平均, 并且介绍了两个Sándor-Yang平均如下.

$ \begin{array}{l} {S_{QA}}(a, b) = S[Q(a, b), A(a, b)]\\ \;\;\;\;\;\;\;\;\;\;\;\;\; = A(a, b){{\rm{e}}^{Q(a, b)/SB[Q(a, b), A(a, b)] -1}} = A(a, b){{\rm{e}}^{Q(a, b)/M(a, b) -1}}, \end{array} $

(0.1)

$ \begin{array}{l} {S_{AQ}}(a, b) = S[A(a, b), Q(a, b)]\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; = Q(a, b){{\rm{e}}^{A(a, b)/SB[A(a, b), Q(a, b)] -1}} = Q(a, b){{\rm{e}}^{A(a, b)/T(a, b) -1}}. \end{array} $

(0.2)

在文献[6-7]中, 杨镇杭等证明了对所有$a, b>0$且$a\ne b$,

$ {M_1}\left( {a, b} \right) < {S_{AQ}}\left( {a, b} \right) < {M_2}\left( {a, b} \right) $

和

$ {S_{AQ}}(a, b) > {\lambda _p}{M_p}(a, b) $

成立, 且当$p\geqslant 3/4$时的最佳参数是$\lambda_{p} =\mathrm e^{\pi /4}2^{1/p-1/2}$.

最近, 赵铁洪、钱伟茂和宋迎清^[8]证明了对所有$a, b>0$且$a\ne b$, 双向不等式

$ \begin{array}{l} {M_\alpha }(a, b) < {S_{QA}}(a, b) < {M_\alpha }(a, b), \\ {M_\lambda }(a, b) < {S_{AQ}}(a, b) < {M_\mu }(a, b) \end{array} $

成立当且仅当$\alpha \leqslant \ln 2/[{1+\ln 2-\sqrt{2}\ln ( {1+\sqrt{2}} )}]\approx 1.551\, 7$, $\beta \geqslant 5/3$, $\lambda \leqslant 4\ln 2/[{4+2\ln 2-\pi }]\approx 1.235\, 1 $和$\mu \geqslant 4/3$.

本文的主要目的是给出最佳参数$\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha _{4}, \beta_{1}, \beta_{2}, \beta_{3}, \beta_{4} \in ({0, 1})$，使得对所有$a, b>0$且$a\ne b$, 双向不等式

$ \begin{array}{l} {\alpha _1}Q(a, b) + (1-{\alpha _1})A(a, b) < {S_{QA}}(a, b) < {\beta _1}Q(a, b) + (1-{\beta _1})A(a, b), \\ {\alpha _2}Q(a, b) + (1-{\alpha _2})A(a, b) < {S_{AQ}}(a, b) < {\beta _2}Q(a, b) + (1 - {\beta _2})A(a, b), \\ {\alpha _3}C(a, b) + (1 - {\alpha _3})A(a, b) < {S_{QA}}(a, b) < {\beta _3}C(a, b) + (1 - {\beta _3})A(a, b), \\ {\alpha _4}C(a, b) + (1 - {\alpha _4})A(a, b) < {S_{AQ}}(a, b) < {\beta _4}C(a, b) + (1 - {\beta _4})A(a, b) \end{array} $

成立.

1 引理

为了证明我们的主要结果, 本节给出我们需要的四个引理.

引理1.1 设$p\in ( {0, 1} )$,

$ f(x) = \frac{{\sqrt {{x^2}- 1} [(1-p)x + p]}}{{px + (1 -p)}} -{\rm{arcsinh}}(\sqrt {{x^2} -1} ), $

(1.1)

则以下结论成立.

(1) 若$p=2/3$, 则当$x\in ( {1, \sqrt{2}} )$时有$f(x)>0$;

(2) 若$p=( {1+\sqrt{2}} )[{( {1+\sqrt{2}} )^{\sqrt{2}}-\mathrm e}]/\mathrm e\approx 0.674\, 7$, 则存在$\lambda_{1} \in ( {1, \sqrt{2}} )$, 使得当$x\in ( {1, \lambda_{1} } )$时, $f(x) < 0$; 当$x\in ( {\lambda_{1}, \sqrt{2}} )$时, $f(x)>0$.

证 明 简单计算可得

$ \begin{array}{l} f(1) = 0, \\ f(\sqrt 2 ) = \frac{{\sqrt 2- (\sqrt 2- 1)p}}{{1 + (\sqrt 2- 1)p}} - \log (1 + \sqrt 2 ), \\ f'(x) = \frac{{\sqrt {{x^2} - 1} }}{{{{[px + (1-p)]}^2}}}{f_1}(x), \end{array} $

其中

$ \begin{array}{l} {f_1}(x) = p(1-p)x + {p^2}-4p + 2, \\ {{f'}_1}(x) = p(1-p) > 0. \end{array} $

(1) 若$p=2/3$, 则对于所有$x\in ( {1, \sqrt{2}} )$有

$ \begin{array}{l} {f_1}(x) = \frac{2}{9}(x-1) > 0, \\ f'(x) > 0, \\ f(x) > f(1) = 0. \end{array} $

(2) 若$p=( {1+\sqrt{2}} )[{( {1+\sqrt{2}} )^{\sqrt{2}}-\mathrm e}]/\mathrm e\approx0.674\, 7$, 数值计算可得

$ \begin{array}{l} f(\sqrt 2 ) \approx 0.005{\mkern 1mu} 5 > 0, \\ {f_1}(1) = 2- 3p \approx- 0.024{\mkern 1mu} 1 < 0, \\ {f_1}(\sqrt 2 ) =- (\sqrt 2 - 1)[{p^2} + \sqrt 2 (3 + \sqrt 2 )p-2(1 + \sqrt 2 )] \approx 0.066{\mkern 1mu} 7 > 0. \end{array} $

我们由上述系列等式和不等式, 并结合$f(x)$在分段区间上的单调性容易得到引理1.1(2).

引理1.2 设$p\in ( {0, 1} )$,

$ g(x) = \frac{{\sqrt {{x^2}- 1} [(1-p)x + p]}}{{x[px + (1-p)]}} -\arctan (\sqrt {{x^2} -1} ), $

(1.2)

则以下结论成立.

(1) 若$p=1/3$, 则当$x\in ( {1, \sqrt{2}} )$时有$g(x)>0$;

(2) 若$p=( {\sqrt{2}\mathrm e^{\frac{\pi }{4}-1}-1} )/( {\sqrt{2}-1} )\approx0.340\, 5$, 则存在$\lambda_{2} \in ( {1, \sqrt{2}} )$, 使得当$x\in ( {1, \lambda_{2} } )$时, $g(x) < 0$; 当$x\in ( {\lambda_{2}, \sqrt{2}} )$时, $g(x)>0$.

证 明 简单计算可得

$ \begin{array}{l} g(1) = 0, \\ g(\sqrt 2 ) = \frac{{\sqrt 2- (\sqrt 2- 1)p}}{{\sqrt 2 + (2- \sqrt 2 )p}} - \frac{\pi }{4}, \\ g'(x) = - \frac{{\sqrt {{x^2} - 1} }}{{{x^2}{{[px + (1-p)]}^2}}}{g_1}(x), \end{array} $

其中

$ \begin{array}{l} {g_1}(x) = ({p^2} + 2p-1)x + p(1-p), \\ {{g'}_1}(x) = ({p^2} + 2p-1). \end{array} $

(1) 若$p=1/3$, 则对于所有$x\in ( {1, \sqrt{2}} )$有

$ \begin{array}{l} {g_1}(x) =-\frac{2}{9}(x-1) < 0, \\ g'(x) > 0, \\ g(x) > g(1) = 0. \end{array} $

(2) 若$p=( {\sqrt{2}\mathrm e^{\frac{\pi }{4}-1}-1} )/( {\sqrt{2}-1} )\approx0.340\, 5 $, 对于所有$x\in ( {1, \sqrt{2}} )$, 数值计算可得

$ \begin{array}{l} g(\sqrt 2 ) \approx 0.003{\mkern 1mu} 5 > 0, \\ {{g'}_1}(x) = ({p^2} + 2p-1) \approx-0.202{\mkern 1mu} 8 < 0, \\ {g_1}(1) = 3p-1 \approx 0.021{\mkern 1mu} 7 > 0, \\ {g_1}(\sqrt 2 ) = (\sqrt 2 - 1){p^2} + (2\sqrt 2 + 1)p - \sqrt 2 \approx - 0.062{\mkern 1mu} 2 < 0. \end{array} $

我们由上述系列等式和不等式, 并结合$g(x)$在分段区间上的单调性, 容易得到引理1.2(2).

引理1.3 设$p\in ( {0, 1} )$,

$ h(x) = \frac{{x\sqrt {{x^2}- 1} [-p{x^2} + (1 + p)]}}{{p{x^2} + (1 -p)}} -{\rm{arcsinh}}(\sqrt {{x^2} -1} ), $

(1.3)

则以下结论成立.

(1) 若$p=1/3$, 则当$x\in ( {1, \sqrt{2}} )$时有$h(x) < 0$;

(2) 若$p=( {1+\sqrt{2}} )^{\sqrt{2}}/\mathrm e-1\approx0.279\, 4 $, 则存在$\lambda_{3} \in ( {1, \sqrt{2}} )$使得当$x\in ( {1, \lambda _{3} } )$时，$h(x)>0$；当$x\in ( {\lambda_{3}, \sqrt{2}} )$时，$h(x) < 0$.

证 明 简单计算可得

$ \begin{array}{l} h(1) = 0, \\ h(\sqrt 2 ) = \frac{{\sqrt 2 (1- p)}}{{1 + p}}- \log (1 + \sqrt 2 ), \\ h'(x) =- \frac{{\sqrt {{x^2} - 1} }}{{{{[p{x^2} + (1-p)]}^2}}}{h_1}(x), \end{array} $

其中

$ {h_1}(x) = {p^2}{x^4} + p(2-p){x^2} + p-1. $

(1) 若$p=1/3$, 则对于所有$x\in ( {1, \sqrt{2}} )$整理可得

$ \begin{array}{l} {h_1}(x) = \frac{1}{9}({x^2}-1)({x^2} + 6) > 0, \\ h'(x) < 0, \\ h(x) < h(1) = 0. \end{array} $

(2) 若$p=( {1+\sqrt{2}} )^{\sqrt{2}}/\mathrm e-1\approx0.279\, 4 $, 对于所有$x\in ( {1, \sqrt{2}} )$, 我们有

$ \begin{array}{l} h(\sqrt 2 ) \approx-0.084{\mkern 1mu} 9 < 0, \\ {h_1}(1) = 3p-1 \approx-0.161{\mkern 1mu} 5 < 0, \\ {h_1}(\sqrt 2 ) = 2{p^2} + 5p - 1 \approx {\mkern 1mu} 0.553{\mkern 1mu} 5 > 0, \\ {{h'}_1}(x) = 4{p^2}{x^3} + 2p(2 - p)x > 0. \end{array} $

我们由上述系列等式和不等式, 并结合$h(x)$在分段区间上的单调性, 容易得到引理1.3(2).

引理1.4 设$p\in ( {0, 1} )$,

$ k(x) = \frac{{\sqrt {{x^2}- 1} [-p{x^2} + (1 + p)]}}{{p{x^2} + (1 -p)}} -\arctan (\sqrt {{x^2} -1} ), $

(1.4)

则以下结论成立.

(1) 若$p=1/6$, 则当$x\in ( {1, \sqrt{2}} )$时有$k(x)< 0$;

(2) 若$p=\sqrt{2}\mathrm e^{\frac{\pi }{4}-1}-1\approx 0.141\, 0 $, 则存在$\lambda_{4} \in ( {1, \sqrt{2}} )$, 使得当$x\in ( {1, \lambda_{4} } )$时, $k(x)>0$; 当$x\in ( {\lambda_{4}, \sqrt{2}} )$时, $k(x)< 0$.

证 明 简单计算可得

$ \begin{array}{l} k(1) = 0, \\ k(\sqrt 2 ) = \frac{{1- p}}{{1 + p}}- \frac{\pi }{4}, \\ k'(x) =- \frac{{\sqrt {{x^2} - 1} }}{{{{[p{x^2} + (1-p)]}^2}}}{k_1}(x), \end{array} $

其中

$ {k_1}(x) = {p^2}{x^4} + 4p{x^2}-{(1-p)^2}. $

(1) 若$p=1/6$, 则对于所有$x\in ( {1, \sqrt{2}} )$, 有

$ \begin{array}{l} {k_1}(x) = \frac{1}{{36}}({x^2}-1)({x^2} + 25) > 0, \\ k'(x) < 0, \\ k(x) < k(1) = 0. \end{array} $

(2) 若$p=\sqrt{2}\mathrm e^{\frac{\pi }{4}-1}-1\approx0.141\, 0 $, 对于所有$x\in ( {1, \sqrt{2}} )$, 我们有

$ \begin{array}{l} k(\sqrt 2 ) \approx-0.032{\mkern 1mu} 6 < 0, \\ {k_1}(1) = 6p-1 \approx-0.153{\mkern 1mu} 5 < 0\\ {k_1}(\sqrt 2 ) = 3{p^2} + 10p - 1 \approx 0.470{\mkern 1mu} 4 > 0, \\ {{k'}_1}(x) = 4{p^2}{x^3} + 8px > 0. \end{array} $

我们由上述系列等式和不等式, 并结合$k(x)$在分段区间上的单调性, 容易得到引理1.4(2).

2 主要结果

定理2.1 双向不等式

$ {\alpha _1}Q(a, b) + (1-{\alpha _1})A(a, b) < {S_{QA}}(a, b) < {\beta _1}Q(a, b) + (1-{\beta _1})A(a, b) $

(2.1)

对所有$a, b>0$且$a\ne b$成立的充分必要条件是$\alpha_{1} \leqslant 2/3$, $\beta _{1} \geqslant ( {1+\sqrt{2}} )[{( {1+\sqrt{2}} )^{\sqrt{2}}-\mathrm e}]/\mathrm e\approx0.674\, 7$.

证 明 设$p_{1} =( {1+\sqrt{2}} )[{( {1+\sqrt{2}} )^{\sqrt{2}}-\mathrm e}]/\mathrm e\approx0.674\, 7 $.因$S_{QA} ({a, b})$, $Q({a, b})$和$A( {a, b} )$是对称且一阶齐次的, 不失一般性, 我们不妨设$a>b>0$.令$v=(a-b)/(a+b)\in ( {0, 1} )$, $x=\sqrt{1+v^{2}}\in ( {1, \sqrt{2}} )$以及$p\in [{0, 1}]$.则从等式(0.1) 和$Q({a, b})=A({a, b})\sqrt{1+v^{2}}$可得

$ \frac{{{S_{QA}}(a, b)-A(a, b)}}{{Q(a, b)-A(a, b)}} = \frac{{{{\rm{e}}^{\frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}-1}} - 1}}{{\sqrt {1 + {v^2}} - 1}}, $

(2.2)

$ \mathop {\lim }\limits_{v \to {0^ + }} = \frac{{{{\rm{e}}^{\frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}-1}}-1}}{{\sqrt {1 + {v^2}}-1}} = \frac{2}{3}, $

(2.3)

$ \begin{array}{l} \mathop {\lim }\limits_{v \to {1^-}} = \frac{{{{\rm{e}}^{\frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}-1}}-1}}{{\sqrt {1 + {v^2}} - 1}} = \frac{1}{{\rm{e}}}{(1 + \sqrt 2 )^{1 + \sqrt 2 }} - (1 + \sqrt 2 )\\ \;\;\;\;\;\; \approx 0.6747, \end{array} $

(2.4)

$ \begin{array}{l} \log \frac{{{S_{QA}}(a, b)}}{{pQ(a, b) + (1- p)A(a, b)}} = \frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}- 1- \log [p\sqrt {1 + {v^2}} + (1-p)]\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{{x{\rm{arcsinh}}(\sqrt {{x^2} - 1} )}}{{\sqrt {{x^2} - 1} }} - 1 - \log [px + (1-p)]. \end{array} $

(2.5)

设

$ F(x) = \frac{{x{\rm{arcsinh}}(\sqrt {{x^2}- 1} )}}{{\sqrt {{x^2}- 1} }}- 1 - \log [px + (1-p)]. $

(2.6)

简单计算可得

$ F(1) = 0, $

(2.7)

$ F(\sqrt 2 ) = (1 + \sqrt 2 )\log (1 + \sqrt 2 )-\log (1 + \sqrt 2 + p)-1, $

(2.8)

$ F'(x) = \frac{1}{{{{({x^2}-1)}^{3/2}}}}f(x), $

(2.9)

其中$f(x)$的定义由引理1.1给出.

我们分两种情形证明.

情形1 若$p=2/3$.则从等式(2.5)-(2.7) 和(2.9), 结合引理1.1(1) 可得结论

$ {S_{QA}}(a, b) > \frac{2}{3}Q(a, b) + \frac{1}{3}A(a, b). $

(2.10)

情形2 若$p=p_{1} $.则从等式(2.9) 和引理1.1(2) 可得结论：存在$\lambda_{1} \in ( {1, \sqrt{2}} )$, 使得当$x\in ( {1, \lambda_{1} }]$时, $F(x)$严格单调递减; 当$x\in [{\lambda_{1}, \sqrt{2}} )$时, $F(x)$严格单调递增.注意从等式(2.8) 可推得

$ F(\sqrt 2 ) = 0. $

(2.11)

我们从等式(2.5)-(2.7) 和(2.11), 结合$F(x)$的分段单调性, 可得

$ {S_{QA}}\left( {a, b} \right) < {p_1}Q\left( {a, b} \right) + \left( {1-{p_1}} \right)A\left( {a, b} \right). $

(2.12)

所以, 我们从等式(2.3)-(2.4) 和不等式(2.10)、(2.12), 并结合不等式(2.1) 等价(2.13) 的事实，容易得到定理2.1,

$ {\alpha _1} < \frac{{{{\rm{e}}^{\frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}-1}}-1}}{{\sqrt {1 + {v^2}}-1}} < {\beta _1}. $

(2.13)

定理2.2 双向不等式

$ {\alpha _2}Q(a, b) + (1-{\alpha _2})A(a, b) < {Y_{AQ}}(a, b) < {\beta _2}Q(a, b) + (1-{\beta _2})A(a, b) $

(2.14)

对所有$a, b>0$且$a\ne b$成立的充分必要条件是$\alpha_{2} \leqslant 1/3$, $\beta _{2} \geqslant ( {\sqrt{2}\mathrm e^{\frac{\pi }{4}-1}-1} )/( {\sqrt{2}-1} )\approx0.340\, 5$.

证 明 设$p_{2} =( {\sqrt{2}\mathrm e^{\frac{\pi }{4}-1}-1} )/( {\sqrt{2}-1} )\approx0.340\, 5 $.因为$S_{AQ} ({a, b})$, $Q({a, b})$和$A({a, b})$是对称且一阶齐次的, 不失一般性, 我们不妨设$a>b$.令$v=(a-b)/(a+b)\in ( {0, 1} )$, $x=\sqrt{1+v^{2}}\in ( {1, \sqrt{2}} )$和$p\in [{0, 1}]$.从等式(0.2) 可得

$ \frac{{{S_{AQ}}(a, b)-A(a, b)}}{{Q(a, b)-A(a, b)}} = \frac{{\sqrt {1 + {v^2}} {{\rm{e}}^{\frac{{\arctan (v)}}{v}-1}} - 1}}{{\sqrt {1 + {v^2}} - 1}}, $

(2.15)

$ \mathop {\lim }\limits_{v \to {0^ + }} \frac{{\sqrt {1 + {v^2}} {{\rm{e}}^{\frac{{\arctan (v)}}{v}-1}}-1}}{{\sqrt {1 + {v^2}}-1}} = \frac{1}{3}, $

(2.16)

$ \mathop {\lim }\limits_{v \to {1^-}} \frac{{\sqrt {1 + {v^2}} {{\rm{e}}^{\frac{{\arctan (v)}}{v}-1}}-1}}{{\sqrt {1 + {v^2}} - 1}} = \frac{{\sqrt 2 {{\rm{e}}^{\frac{\pi }{4} - 1}} - 1}}{{\sqrt 2 - 1}} \approx 0.340{\mkern 1mu} 5, $

(2.17)

$ \begin{array}{l} \log \frac{{{S_{AQ}}(a, b)}}{{pQ(a, b) + (1- p)A(a, b)}} = \log \sqrt {1 + {v^2}} + \frac{{\arctan (v)}}{v}- 1- \log [p\sqrt {1 + {v^2}} + (1-p)]\\ = \log x + \frac{{\arctan (\sqrt {{x^2} - 1} )}}{{\sqrt {{x^2} - 1} }} - 1 - \log [px + (1-p)]. \end{array} $

(2.18)

设

$ G(x) = \log x + \frac{{\arctan (\sqrt {{x^2}- 1} )}}{{\sqrt {{x^2}- 1} }}- 1 - \log [px + (1-p)]. $

(2.19)

简单计算可得

$ G{\rm{(1) = 0, }} $

(2.20)

$ G(\sqrt 2 ) = \log \sqrt 2 + \frac{\pi }{4}- 1- \log [(\sqrt 2-1)p + 1], $

(2.21)

$ G'(x) = \frac{x}{{{{({x^2}-1)}^{3/2}}}}g(x), $

(2.22)

其中$g(x)$的定义由引理1.2给出.

我们分两种情形证明.

情形1 若$p=1/3$.则从等式(2.18)-(2.20) 和(2.22) 以及引理1.2(1) 可得结论

$ {S_{AQ}}(a, b) > \frac{1}{3}Q(a, b) + \frac{2}{3}A(a, b). $

(2.23)

情形2 若$p=p_{2} $.则从引理1.2(2) 和等式(2.22) 可知存在$\lambda _{2} \in ( {1, \sqrt{2}} )$, 使得当$x\in ( {1, \lambda_{2} }]$时$G(x)$是严格单调递减的; 当$x\in [{\lambda_{2}, \sqrt{2}} )$时$G(x)$严格单调递增的.注意从等式(2.21) 可推得

$ G(\sqrt 2 ) = 0. $

(2.24)

从等式(2.18)-(2.20) 和(2.24) 结合$G(x)$的分段单调性, 可得

$ {S_{QA}}\left( {a, b} \right) < {p_2}Q\left( {a, b} \right) + \left( {1-{p_2}} \right)A\left( {a, b} \right). $

(2.25)

所以, 我们从等式(2.16)-(2.17) 和不等式(2.23)、(2.25), 并结合不等式(2.14) 等价于(2.26) 的事实, 容易得到定理2.2,

$ {\alpha _2} < \frac{{\sqrt {1 + {v^2}} {{\rm{e}}^{\frac{{\arctan (v)}}{v}-1}}-1}}{{\sqrt {1 + {v^2}}-1}} < {\beta _2}. $

(2.26)

定理2.3 双向不等式

$ {\alpha _3}C(a, b) + (1-{\alpha _3})A(a, b) < {S_{QA}}(a, b) < {\beta _3}C(a, b) + (1-{\beta _3})A(a, b) $

(2.27)

对所有$a, b>0$且$a\ne b$成立的充分必要条件是$\alpha_{3} \leqslant ( {1+\sqrt{2}} )^{\sqrt{2}}/\mathrm e-1\approx0.279\, 4$, $\beta_{3} \geqslant 1/3$.

证 明 不失一般性, 我们不妨设$a>b>0$.令$v=(a-b)/(a+b)\in ( {0, 1} )$, $x=\sqrt{1+v^{2}}\in ( {1, \sqrt{2}} )$和$p\in [{0, 1}]$.从等式(0.1) 和$C({a, b})=A({a, b})( {1+v^{2}} )$可得

$ \frac{{{S_{QA}}(a, b)-A(a, b)}}{{C(a, b)-A(a, b)}} = \frac{{{{\rm{e}}^{\frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}-1}} - 1}}{{{v^2}}}, $

(2.28)

$ \mathop {\lim }\limits_{v \to {0^ + }} \frac{{{{\rm{e}}^{\frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}-1}}-1}}{{{v^2}}} = \frac{1}{3}, $

(2.29)

$ \mathop {\lim }\limits_{x \to {1^-}} \frac{{{{\rm{e}}^{\frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}-1}}-1}}{{{v^2}}} = \frac{{{{(1 + \sqrt 2 )}^{\sqrt 2 }}}}{{\rm{e}}} - 1 \approx 0.279{\mkern 1mu} 4, $

(2.30)

$ \begin{array}{l} \log \frac{{{S_{QA}}(a, b)}}{{pC(a, b) + (1- p)A(a, b)}} = \frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}- 1- \log [p(1 + {v^2}) + (1-p)]\\ = \frac{{x{\rm{arcsinh}}(\sqrt {{x^2} - 1} )}}{{\sqrt {{x^2} - 1} }} - 1 - \log [p{x^2} + (1-p)]. \end{array} $

(2.31)

设

$ H(x) = \frac{{x{\rm{arcsinh}}(\sqrt {{x^2}- 1} )}}{{\sqrt {{x^2}- 1} }}- 1 - \log [p{x^2} + (1-p)]. $

(2.32)

简单计算可得

$ H{\rm{(1) = 0, }} $

(2.33)

$ H(\sqrt 2 ) = \sqrt 2 \log (1 + \sqrt 2 )-1-\log (1 + p), $

(2.34)

$ H'(x) = \frac{1}{{{{({x^2}-1)}^{3/2}}}}h(x), $

(2.35)

其中$h(x)$的定义由引理1.3给出.

我们分两种情形证明.

情形1 若$p=( {1+\sqrt{2}} )^{\sqrt{2}}/\mathrm e-1\approx0.279\, 4 $.则从引理1.3(2) 和等式(2.35) 可知存在$\lambda_{3} \in ( {1, \sqrt{2}} )$, 使得当$x\in ( {1, \lambda_{3} }]$时$H(x)$是严格单调递增的; 当$[{\lambda_{3}, \sqrt{2}} )$时$H(x)$是严格单调递减的.注意从等式(2.34) 可推得

$ H(\sqrt 2 ) = 0. $

(2.36)

从等式(2.31)-(2.33) 和(2.36) 结合$H(x)$的分段单调性, 可得

$ {S_{AQ}}(a, b) > [{(1 + \sqrt 2 )^{\sqrt 2 }}/{\rm{e}}-1]C(a, b) + [2-{(1 + \sqrt 2 )^{\sqrt 2 }}/{\rm{e}}]A(a, b). $

(2.37)

情形2 若$p=1/3$.则从等式(2.31)-(2.33) 和(2.35), 并结合引理1.3(1), 可得结论

$ {S_{AQ}}(a, b) < \frac{1}{3}C(a, b) + \frac{2}{3}A(a, b). $

(2.38)

所以, 我们从等式(2.29)-(2.30) 和不等式(2.37)、(2.38), 并结合不等式(2.27) 等价于(2.39) 的事实, 容易得到定理2.3,

$ {\alpha _3} < \frac{{{{\rm{e}}^{\frac{{\sqrt {1 + {v^2}} {\rm{arcsinh}}(v)}}{v}-1}}-1}}{{{v^2}}} < {\beta _3}. $

(2.39)

定理2.4 双向不等式

$ {\alpha _4}C(a, b) + (1-{\alpha _4})A(a, b) < {S_{AQ}}(a, b) < {\beta _4}C(a, b) + (1-{\beta _4})A(a, b) $

(2.40)

对所有$a, b>0$且$a\ne b$成立的充分必要条件是$\alpha_{4} \leqslant \sqrt{2}\mathrm e^{\frac{\pi }{4}-1}-1\approx0.141\, 0$, $\beta_{4} \geqslant 1/6$.

证 明 不失一般性，我们不妨设$a>b>0$.设$v=(a-b)/(a+b)\in ( {0, 1} )$, $x=\sqrt{1+v^{2}}\in ( {1, \sqrt{2}} )$和$p\in [{0, 1}]$.从等式(0.2) 可得

$ \frac{{{S_{AQ}}(a, b)-A(a, b)}}{{C(a, b)-A(a, b)}} = \frac{{\sqrt {1 + {v^2}} {{\rm{e}}^{\frac{{\arctan (v)}}{v}-1}} - 1}}{{{v^2}}}, $

(2.41)

$ \mathop {\lim }\limits_{v \to {0^ + }} \frac{{\sqrt {1 + {v^2}} {{\rm{e}}^{\frac{{\arctan (v)}}{v}-1}}-1}}{{{v^2}}} = \frac{1}{6}, $

(2.42)

$ \mathop {\lim }\limits_{v \to {1^-}} \frac{{\sqrt {1 + {v^2}} {{\rm{e}}^{\frac{{\arctan (v)}}{v}-1}}-1}}{{{v^2}}} = \sqrt 2 {{\rm{e}}^{\frac{\pi }{4} - 1}} - 1 \approx 0.141{\mkern 1mu} 0, $

(2.43)

$ \begin{array}{l} \log \frac{{{S_{AQ}}(a, b)}}{{pC(a, b) + (1- p)A(a, b)}} = \log \sqrt {1 + {v^2}} + \frac{{\arctan (v)}}{v}- 1- \log [p(1 + {v^2}) + (1-p)]\\ = \log x + \frac{{\arctan (\sqrt {{x^2} - 1} )}}{{\sqrt {{x^2} - 1} }} - 1 - \log [p{x^2} + (1-p)]. \end{array} $

(2.44)

设

$ J(x) = \log x + \frac{{\arctan (\sqrt {{x^2}- 1} )}}{{\sqrt {{x^2}- 1} }}- 1 - \log [p{x^2} + (1-p)]. $

(2.45)

简单计算可得

$ J{\rm{(1) = 0, }} $

(2.46)

$ J(\sqrt 2 ) = \log \sqrt 2 + \frac{\pi }{4}-1-\log (1 + p), $

(2.47)

$ J'(x) = \frac{x}{{{{({x^2}-1)}^{3/2}}}}k(x), $

(2.48)

其中$k(x)$的定义由引理1.4给出.

我们分两种情形证明.

情形1 $p=\sqrt{2}\mathrm e^{\frac{\pi }{4}-1}-1\approx0.141\, 0$.则从引理1.4(2) 和等式(2.48) 可知存$\lambda_{4} \in ( {1, \sqrt{2}} )$, 使得当$x\in ( {1, \lambda_{4} }]$时$J(x)$是严格单调递增的; 当$x\in [{\lambda_{4}, \sqrt{2}} )$时$J(x)$是严格单调递减的.注意从等式(2.47) 可推得

$ J(\sqrt 2 ) = 0. $

(2.49)

从等式(2.44)-(2.46) 和(2.49) 结合$J(x)$的分段单调性, 可得

$ {S_{AQ}}(a, b) > (\sqrt 2 {{\rm{e}}^{\frac{\pi }{4}-1}}-1)Q(a, b) + (2-\sqrt 2 {{\rm{e}}^{\frac{\pi }{4} - 1}})A(a, b). $

(2.50)

情形2 若$p=1/6$.从等式(2.44)--(2.46) 和(2.48) 并结合引理1.4(1), 可得结论

$ {S_{AQ}}(a, b) < \frac{1}{6}Q(a, b) + \frac{5}{6}A(a, b). $

(2.51)

所以, 我们从等式(2.42)-(2.43) 和不等式(2.50)、(2.51), 并结合不等式(2.40) 等价于(2.52) 的事实, 容易得到定理2.4,

$ {\alpha _4} < \frac{{\sqrt {1 + {v^2}} {{\rm{e}}^{\frac{{{\rm{arctan}}(v)}}{v}-1}}-1}}{{{v^2}}} < {\beta _4}. $

(2.52)