0 引言

对于$a$, $b>0$且$a\neq b$, Neuman-Sándor平均^[1]定义为

$ M(a, b) =\frac{a-b}{2{\rm arcsinh}\big[(a-b)/(a+b)\big]}, $

(1)

其中${\rm arcsinh}(x)=\ln(x+\sqrt{x^2+1})$是反双曲正弦函数.

设$\overline{H}(a, b)=\sqrt{2}ab/\sqrt{a^2+b^2}$, $H(a, b)=2ab/(a+b)$, $G(a, b)=\sqrt{ab}$, $L(a, b)=(a-b)/(\ln a-\ln b)$, $P(a, b)=(a-b)/\big[2\arcsin\big((a-b)/(a+b)\big)\big]$, $A(a, b)=(a+b)/2$, $T(a, b)=(a-b)/\big[2\arctan\big((a-b)/(a+b)\big)\big]$, $E(a, b)=2(a^2+ab+b^2)/\big[3(a+b)\big]$, $Q(a, b)=\sqrt{(a^2+b^2)/2}$, $C(a, b)=(a^2+b^2)/(a+b)$和$D(a, b)=(a^3+b^3)/(a^2+b^2)$分别是调和根平方平均、调和平均、几何平均、对数平均、第一类Seiffert平均、算术平均、第二类Seiffert平均、形心平均、二次平均、反调和平均和第二类反调和平均.我们有熟知不等式

$ \overline{H}(a, b) <H(a, b)< G(a, b)<L(a, b)<P(a, b)<A(a, b) \\ <M(a, b)<T(a, b)<E(a, b)<Q(a, b)<C(a, b)<D(a, b), $

(2)

对所有$a$, $b>0$且$a\neq b$成立.

近年来, Neuman-Sándor平均和其他二元平均得到了深入的研究.特别地, 从有关Neuman-Sándor平均与其他二元平均和它们的各类组合比较中发现了许多重要不等式, 参见文献[2-6].

Neuman和Sándor^{[1, 7]}证明了不等式

$ A(a, b) <M(a, b)<\frac{A(a, b)}{\ln(1+\sqrt{2})}, \frac{\pi}{4}T(a, b)<M(a, b)<T(a, b), \\ M(a, b)<\frac{A^{2}(a, b)}{P(a, b)}, M(a, b)<\frac{2A(a, b)+Q(a, b)}{3}, \\ \sqrt{A(a, b)T(a, b)}<M(a, b)<\sqrt{\frac{A^{2}(a, b)+T^{2}(a, b)}{2}}, $

(3)

对所有$a$, $b>0$且$a\neq b$成立.

在文献[8]中, Neuman证明了不等式

$ \alpha Q(a, b)+(1-\alpha)A(a, b) <M(a, b) <\beta Q(a, b)+(1-\beta)A(a, b), \\ \lambda C(a, b)+(1-\lambda)A(a, b)<M(a, b)<\mu C(a, b)+(1-\mu)A(a, b) $

(4)

对所有$a$, $b>0$且$a\neq b$成立当且仅当$\alpha \leqslant \big[1-\ln(1+\sqrt{2})\big]/\big[(\sqrt{2}-1)\ln(1+\sqrt{2})\big]=0.324~9\cdots$, $\beta \geqslant 1/3$, $\lambda \leqslant \big[1-\ln(1+\sqrt{2})\big]/\ln(1+\sqrt{2})=0.134~5\cdots$和$\mu \geqslant 1/6$.

赵铁洪等^[9]发现了最佳参数$\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$, $\beta_{1}$, $\beta_{2}$, $\beta_{3}\in(0, 1)$使得双边不等式

$ \alpha_{1} H(a, b)+(1-\alpha_{1})Q(a, b)<M(a, b)<\beta_{1} H(a, b)+(1-\beta_{1})Q(a, b), \\ \alpha_{2} G(a, b)+(1-\alpha_{2})Q(a, b)<M(a, b)<\beta_{2} G(a, b)+(1-\beta_{2})Q(a, b), \\ \alpha_{3} H(a, b)+(1-\alpha_{3})C(a, b)<M(a, b)<\beta_{3} H(a, b)+(1-\beta_{3})C(a, b) $

(5)

对所有$a$, $b>0$且$a\neq b$成立.

钱伟茂和禇玉明^[10]发现了最大值$\alpha_{1}$, $\lambda_{1}$, $\alpha_{2}$, $\lambda_{2}$和最小值$\beta_{1}$, $\mu_{1}$, $\beta_{2}$, $\mu_{2}$使得双边不等式

$ \alpha_{1} E(a, b)+(1-\alpha_{1})A(a, b)<M(a, b)<\beta_{1} E(a, b)+(1-\beta_{1})A(a, b), \\ E^{\lambda_{1}}(a, b)A^{1-\lambda_{1}}(a, b)<M(a, b)<E^{\mu_{1}}(a, b)A^{1-\mu_{1}}(a, b), \\ \alpha_{2} Q(a, b)+(1-\alpha_{2})\overline{H}(a, b)<M(a, b)<\beta_{2} Q(a, b)+(1-\beta_{2})\overline{H}(a, b), \\ \lambda_{2} C(a, b)+(1-\lambda_{2})\overline{H}(a, b)<M(a, b)<\mu_{2} C(a, b)+(1-\mu_{2})\overline{H}(a, b) $

(6)

对所有$a$, $b>0$且$a\neq b$成立.

在文献[11]中, 作者证明了双边不等式

$ \lambda_{1} D(a, b)+(1-\lambda_{1})A(a, b) <T^*(a, b)<\mu_{1} D(a, b)+(1-\mu_{1})A(a, b), \\ D\big[\lambda_{2}a+(1-\lambda_{2})b, \lambda_{2}b+(1-\lambda_{2})a\big] <T^*(a, b) <D\big[\mu_{2}a+(1-\mu_{2})b, \mu_{2}b+(1-\mu_{2})a\big] $

(7)

对所有$a, b>0$且$a\neq b$成立的充要条件是$\lambda_{1} \leqslant 1/8$, $\mu_{1} \geqslant 4/\pi -1=0.273~2\cdots$, $\lambda_{2} \leqslant 1/2+\sqrt{2}/8=0.676~7\cdots$和$\mu_{2} \geqslant 1/2+\sqrt{(4-\pi)/(3 \pi -4)}/2=0.698~8\cdots$.其中$T^*(a, b)=\frac{2}{\pi} \int_0^{\pi/2}\sqrt{a^2 \cos^2\theta+b^2 \sin^2 \theta}{\rm{d}}\theta$是两个正数$a$和$b$的Toader平均.

在文献[12-13]中作者发现了最佳参数$\alpha_{1}, \alpha_{2}, \beta_{1}, \beta_{2} \in(0, 1)$使得双边不等式

$ \alpha_{1} D(a, b)+(1-\alpha_{1})A(a, b)<T(a, b)<\beta_{1} D(a, b)+(1-\beta_{1})A(a, b), \\ \alpha_{2} D(a, b)+(1-\alpha_{2})H(a, b)<T(a, b)<\beta_{2} D(a, b)+(1-\beta_{2})H(a, b) $

(8)

对所有$a, b>0$且$a\neq b$成立.

杨月英等^[14]找到了最大值$\alpha_{1}, \alpha_{2}$和最小值$\beta_{1}, \beta_{2}$使得双边不等式

$ \alpha_{1} D(a, b)+(1-\alpha_{1})G(a, b)<P(a, b)<\beta_{1} D(a, b)+(1-\beta_{1})G(a, b), \\ \alpha_{2} D(a, b)+(1-\alpha_{2})G(a, b)<T(a, b)<\beta_{2} D(a, b)+(1-\beta_{2})G(a, b) $

(9)

对所有$a, b>0$且$a\neq b$成立.

本文的主要结果是找到了最大值$\lambda_{1}, \lambda_{2}$和最小值$\mu_{1}, \mu_{2}$使得双边不等式

$ \lambda_{1} D(a, b)+(1-\lambda_{1})\overline{H}(a, b)<M(a, b)<\mu_{1} D(a, b)+(1-\mu_{1})\overline{H}(a, b), $

(10)

$ \lambda_{2} D(a, b)+(1-\lambda_{2})H(a, b)<M(a, b)<\mu_{2} D(a, b)+(1-\mu_{2})H(a, b) $

(11)

对所有$a, b>0$且$a\neq b$成立.

1 引理

为了证明主要结论, 我们需要以下引理.

引理1.1  设$p\in(0, 1)$,

$ f(x)= (1-p)^2 x^5 + (7p^2-11p+1)x^4+6(2p^2-1)x^3 \\ -2(2p^2-11p+3)x^2+12p(1-p)x-4p^2, $

(12)

则下列结论成立.

(1) 若$p=10/21$, 则当$x\in(1, \sqrt{2})$时有$f(x) < 0$;

(2) 若$p=1/\big[2\ln(1+\sqrt{2})\big]$, 则存在$x_1 \in(1, \sqrt{2})$使得当$x\in(1, x_1)$时有$f(x)>0$, 而当$x\in(x_1, \sqrt{2})$时有$f(x) < 0$.

证明 (1)当$p=10/21, x\in(1, \sqrt{2})$时, 等式(12)变成

$ f(x)= \frac{1}{441}(x-1)(121x^4-1~048x^3-2~494x^2-920x+400) \\ < \frac{1}{441}(x-1)(121x^4-1~048-2~494-920+400) \\ = -\frac{1}{441}(x-1)(4~062-121x^4)<0, $

(13)

所以, 我们从不等式(13)容易证得(1).

(2) 当$p=1/\big[2\ln(1+\sqrt{2})\big]$时, 则由数值计算知

$ 7p^2-11p+1 =-2.987~4\cdots<0, $

(14)

$ 2p^2-1 =-0.356~3\cdots<0, $

(15)

$ 2p^2-11p+3 =-2.596~6\cdots<0, $

(16)

$ 27p^2-4p-8 =-1.579~9\cdots<0, $

(17)

$ f(1)=21p-10 =1.913~2\cdots>0, $

(18)

$ f\big(\sqrt{2}\big) =4(1+\sqrt{2})\big[4p^2+\sqrt{2}(\sqrt{2}-1)p-2\big]=-3.673~3\cdots<0 $

(19)

及

$ f'(x)= 5(1-p)^2 x^4 + 4(7p^2-11p+1)x^3+18(2p^2-1)x^2 \\ -4(2p^2-11p+3)x+12p(1-p). $

(20)

从不等式(14)-(17)和等式(20), 可知

$ f'(x) < 5(1-p)^2 x^4 + 4(7p^2-11p+1)x^2+18(2p^2-1)x^2\\ -4(2p^2-11p+3)x^2+12p(1-p)x^2\\ = x^2\big[5(1-p)^2 x^2+(44p^2+12p-26)\big]\\ < x^2\big[5(1-p)^2 (\sqrt{2})^2 +(44p^2+12p-26)\big]\\ = 2(27p^2-4p-8)x^2 <0 $

(21)

对所有$x\in(1, \sqrt{2})$成立.故从不等式(21)可知函数$f(x)$在区间$(1, \sqrt{2})$上严格单调递减.所以, 由不等式(18)和(19)以及函数$f(x)$的单调性知, 存在$x_1\in(1, \sqrt{2})$, 使得当$x\in(1, x_1)$时, 有$f(x)>0$; 当$x\in(x_1, \sqrt{2})$时, 有$f(x) < 0$.

引理1.2  设$p\in(0, 1)$和

$ g(x)= (1-p)^2 x^7 - p(1-p)x^6-(3p+4)(1-p)x^5+(3p^2-2p-4)x^4 \\ +6px^3+12px^2+4p(1-p)x-4p^2, $

(22)

则下列结论成立.

(1) 若$p=7/18$, 则当$x\in(1, \sqrt{2})$时有$g(x) < 0$;

(2) 若$p=1/\big[2\ln(1+\sqrt{2})\big]$, 则存在$x_2 \in(1, \sqrt{2})$, 使得当$x\in(1, x_2)$时有$g(x)>0$, 而当$x\in(x_2, \sqrt{2})$时有$g(x) < 0$.

证明  (1)当$p=7/18, x\in(1, \sqrt{2})$时, 等式(22)变成

$ g(x)= \frac{1}{324}(x-1)(121x^6+44x^5-979x^4-2~380x^3-1624x^2-112x+196) \\ <\frac{1}{324}(x-1)(121x^6+44x^6-979x-2~380x-1~624x-112x+196x^6) \\ = -\frac{1}{324}x(x-1)(5~095-361x^5)<0, $

(23)

所以, 由不等式(23)容易证得(1).

(2) 当$p=1/\big[2\ln(1+\sqrt{2})\big]$时, 则由数值计算知

$ 3p^2-2p-4 =-4.169~1\cdots<0, $

(24)

$ 4p^2+\sqrt{2}p-2\sqrt{2} =-0.738~8\cdots<0, $

(25)

$ 57p^2-19p-8 =-0.434~5\cdots <0, $

(26)

$ g(1) =18p-7=3.211~3\cdots>0, $

(27)

$ g(\sqrt{2}) =4(1+\sqrt{2})(4p^2+\sqrt{2}p-2\sqrt{2})=-7.134~9\cdots<0 $

(28)

及

$ g'(x)= 7(1-p)^2 x^6 -6 p(1-p)x^5-5(3p+4)(1-p)x^4 \\ +4(3p^2-2p-4)x^3+18px^2+24px+4p(1-p). $

(29)

从不等式(24)、(26)和等式(29), 可知

$ g'(x)< 7(1-p)^2 x^6 -6 p(1-p)x^2-5(3p+4)(1-p)x^2 \\ +4(3p^2-2p-4)x^2+18px^2+24px^2+4p(1-p)x^2\\ = x^2\big[7(p^2-2p+1)x^4+(29p^2+37p-36)\big]\\ <x^2\big[7(p^2-2p+1)(\sqrt{2})^4+(29p^2+37p-36)\big]\\ = (57p^2-19p-8)x^2<0 $

(30)

对所有$x\in(1, \sqrt{2})$成立.故从不等式(30)可知函数$g(x)$在区间$(1, \sqrt{2})$上严格单调递减.所以, 由不等式(27)和(28)以及函数$g(x)$的单调性知, 存在$x_2\in(1, \sqrt{2})$, 使得当$x\in(1, x_2)$时有$g(x)>0$; 当$x\in(x_2, \sqrt{2})$时, 有$g(x) < 0$.

2 主要结果

定理2.1  双边不等式

$ \lambda_{1} D(a, b)+(1-\lambda_{1})\overline{H}(a, b)<M(a, b)<\mu_{1} D(a, b)+(1-\mu_{1})\overline{H}(a, b) $

(31)

对所有$a, b>0$且$a\neq b$成立的充要条件是$\lambda_1\leqslant 10/21$和$\mu_1\geqslant1/\big[2\ln(1+\sqrt{2})\big]=0.5672\cdots$.

证明  首先, 我们证明不等式

$ M(a, b)>\frac{10}{21}D(a, b)+\frac{11}{21}\overline{H}(a, b), $

(32)

$ M(a, b)<\frac{1}{2\ln(1+\sqrt{2})}D(a, b)+\Big[1-\frac{1}{2\ln(1+\sqrt{2})}\Big]\overline{H}(a, b) $

(33)

对所有$a, b>0$且$a\neq b$成立.

根据$M(a, b)$, $D(a, b)$和$\overline{H}(a, b)$是对称且一阶齐次的, 不失一般性, 我们假设$a>b>0$.设$v=(a-b)/(a+b)$, $x=\sqrt{1+v^2}$, $p\in(0, 1)$.则有$v\in(0, 1)$, $x\in(1, \sqrt{2})$和

$ \frac{M(a, b)}{A(a, b)}=\frac{v}{{\rm arcsinh}(v)}, \frac{D(a, b)}{A(a, b)}=\frac{1+3v^2}{1+v^2}, \frac{\overline{H}(a, b)}{A(a, b)}=\frac{1-v^2}{\sqrt{1+v^2}}. $

(34)

我们从等式(34)得到

$ \frac{M(a, b)-\overline{H}(a, b)}{D(a, b)-\overline{H}(a, b)}=\frac{\big[v\sqrt{1+v^2}-(1-v^2){\rm arcsinh}(v)\big]\sqrt{1+v^2}} {\big[(1+3v^2)-(1-v^2)\sqrt{1+v^2}\big]{\rm arcsinh}(v)} $

(35)

和

$ \lim\limits_{v\rightarrow 0^{+}}\frac{\big[v\sqrt{1+v^2}-(1-v^2){\rm arcsinh}(v)\big]\sqrt{1+v^2}}{\big[(1+3v^2)-(1-v^2)\sqrt{1+v^2}\big]{\rm arcsinh}(v)}=\frac{10}{21}, $

(36)

$ \lim\limits_{v\rightarrow 1^{-}}\frac{\big[v\sqrt{1+v^2}-(1-v^2){\rm arcsinh}(v)\big]\sqrt{1+v^2}} {\big[(1+3v^2)-(1-v^2)\sqrt{1+v^2}\big]{\rm arcsinh}(v)}=\frac{1}{2\ln(1+\sqrt{2})}, $

(37)

$ M(a, b)-\big[pD(a, b)+(1-p)\overline{H}(a, b)\big] \\ =A(a, b)\bigg\{\frac{v}{{\rm arcsinh}(v)}-\Big[p\Big(\frac{1+3v^2} {1+v^2}\Big)+(1-p)\Big(\frac{1-v^2}{\sqrt{1+v^2}}\Big)\Big]\bigg\} \\ =\frac{A(a, b)\big[p(1+3v^2)+(1-p)(1-v^2)\sqrt{1+v^2} \big]}{(1+v^2){\rm arcsinh}(v)}F(x), $

(35)

其中

$ F(x)=\frac{x^2\sqrt{x^2-1}}{(p-1)x^3+3px^2+2(1-p)x-2p}-{\rm arcsinh}\big(\sqrt{x^2-1}\big). $

(39)

经简单计算得

$ F(1) =0, $

(40)

$ F(\sqrt{2}) =\frac{1}{2p}-\ln(1+\sqrt{2}), $

(41)

$ F'(x) =-\frac{(x-1)}{\sqrt{x^2-1}\big[(p-1)x^3+3px^2+2(1-p)x-2p\big]^2}f(x), $

(42)

其中函数$f(x)$由式(12)给出.

现在, 我们分两种情形证明.

情形1  当$p=10/21$时, 则从等式(42)和引理1.1(1)可看到函数$F(x)$在区间$(1, \sqrt{2})$上严格单调递增.所以, 不等式(32)容易从等式(38)-(40)及$F(x)$的单调性得到.

情形2  当$p=1/\big[2\ln(1+\sqrt{2})\big]$时, 则从等式(42)和引理1.1(2)得:存在$x_1 \in(1, \sqrt{2})$, 使得当$x\in(1, x_1)$时有$F'(x) < 0$; 当$x\in(x_1, \sqrt{2})$时有$F'(x)>0$.这样, 函数$F(x)$在区间$(1, x_1)$上严格递减, 在区间$(x_1, \sqrt{2})$上严格递增.注意到此时等式(41)变成

$ F(\sqrt{2})=0, $

(43)

所以, 不等式(33)可从等式(38)-(40)和(43)及函数$F(x)$的分段单调性得到.因此, 定理2.1容易由不等式(32)和(33)以及下面的事实得到.

•若$p>10/21$, 则由等式(35)和(36)得, 存在一个$0 < \delta_1 < 1$, 使得

$ M(a, b) <pD(a, b)+(1-p)\overline{H}(a, b) $

对所有$a$, $b>0$且$(a-b)/(a+b)\in(0, \delta_1)$成立.

•若$p < 1/\big[2\ln(1+\sqrt{2})\big]$, 则由等式(35)和(37)得, 存在一个$0 < \delta_2 < 1$, 使得

$ M(a, b)>pD(a, b)+(1-p)\overline{H}(a, b) $

对所有$a, b>0$且$(a-b)/(a+b)\in(1-\delta_2, 1)$成立.

定理2.2  双边不等式

$ \lambda_{2} D(a, b)+(1-\lambda_{2})H(a, b) <M(a, b) <\mu_{2} D(a, b)+(1-\mu_{2})H(a, b) $

(44)

对所有$a, b>0$且$a\neq b$成立的充要条件是$\lambda_2\leqslant 7/18$和$\mu_2\geqslant1/\big[2\ln(1+\sqrt{2})\big]=0.5672\cdots$.

证明  首先, 我们证明不等式

$ M(a, b)>\frac{7}{18}D(a, b)+\frac{11}{18}H(a, b), $

(45)

$ M(a, b) <\frac{1}{2\ln(1+\sqrt{2})}D(a, b)+ \Big[1-\frac{1}{2\ln(1+\sqrt{2})}\Big]H(a, b) $

(46)

对所有$a, b>0$且$a\neq b$成立.

根据$M(a, b)$, $D(a, b)$和$H(a, b)$是对称且一阶齐次的, 不失一般性, 我们假设$a>b>0$.设$v=(a-b)/(a+b)$, $x=\sqrt{1+v^2}$, $p\in(0, 1)$.则有$v\in(0, 1)$, $x\in(1, \sqrt{2})$和

$ \frac{M(a, b)}{A(a, b)}=\frac{v}{{\rm arcsinh}(v)}, \frac{D(a, b)}{A(a, b)}=\frac{1+3v^2}{1+v^2}, \frac{H(a, b)}{A(a, b)}=1-v^2. $

(47)

从等式(47)得到

$ \frac{M(a, b)-H(a, b)}{D(a, b)-H(a, b)}=\frac{(1+v^2)\big[v-(1-v^2){\rm arcsinh}(v)\big]}{v^2(v^2+3){\rm arcsinh}(v)} $

(48)

和

$ \lim\limits_{v\rightarrow 0^{+}}\frac{(1+v^2)\big[v-(1-v^2){\rm arcsinh}(v)\big]}{v^2(v^2+3){\rm arcsinh}(v)}=\frac{7}{18}, $

(49)

$ \lim\limits_{v\rightarrow 1^{-}}\frac{(1+v^2)\big[v-(1-v^2){\rm arcsinh}(v)\big]}{v^2(v^2+3){\rm arcsinh}(v)}=\frac{1}{2\ln(1+\sqrt{2})}, $

(50)

$ M(a, b)-\big[pD(a, b)+(1-p)H(a, b)\big] \\ =A(a, b)\bigg\{\frac{v}{{\rm arcsinh}(v)}-\Big[p\Big(\frac{1+3v^2}{1+v^2}\Big)+(1-p)(1-v^2)\Big]\bigg\} \\ =\frac{A(a, b)\big[p(1+3v^2)+(1-p)(1-v^4)\big]}{(1+v^2){\rm arcsinh}(v)}G(x), $

(51)

其中

$ G(x)=\frac{x^2\sqrt{x^2-1}}{(p-1)x^4+(p+2)x^2-2p}-{\rm arcsinh}\big(\sqrt{x^2-1}\big). $

(52)

经简单计算得

$ G(1) =0, $

(53)

$ G(\sqrt{2}) =\frac{1}{2p}-\ln(1+\sqrt{2}), $

(54)

$ G'(x) =-\frac{(x-1)}{\sqrt{x^2-1}\big[(p-1)x^4+(p+2)x^2-2p\big]^2}g(x) $

(55)

其中函数$g(x)$由式(22)定义.

现在, 我们分两种情形证明.

情形1  当$p=7/18$时, 则从等式(55)和引理1.2(1)知函数$G(x)$在区间$(1, \sqrt{2})$上是严格单调递增的.所以, 不等式(45)容易从等式(51)-(53)及$G(x)$的单调性得到.

情形2  当$p=1/\big[2\ln(1+\sqrt{2})\big]$时, 则从等式(55)和引理1.2(2)知, 存在$x_2 \in(1, \sqrt{2})$使得当$x\in(1, x_2)$时有$G'(x) < 0$; 当$x\in(x_2, \sqrt{2})$时有$G'(x)>0$.这样, 函数$G(x)$在区间$(1, x_2)$上严格递减, 在区间$(x_2, \sqrt{2})$上严格递增.注意到此时等式(54)变成

$ G(\sqrt{2})=0, $

(56)

所以, 不等式(46)可从等式(51)-(53)和(56)以及函数$G(x)$的分段单调性得到.因此, 定理2.2容易从不等式(45)和(46)协同下面事实得到.

•若$p>7/18$, 则由等式(48)和(49)知, 存在$0 < \delta_3 < 1$, 使得

$ M(a, b) <pD(a, b)+(1-p)H(a, b) $

对所有$a, b>0$且$(a-b)/(a+b)\in(0, \delta_3)$成立.

•若$p < 1/\big[2\ln(1+\sqrt{2})\big]$, 则由等式(48)和(50)知, 存在$0 < \delta_4 < 1$, 使得

$ M(a, b)>pD(a, b)+(1-p)H(a, b) $

对所有$a, b>0$且$(a-b)/(a+b)\in(1-\delta_4, 1)$成立.