ALG不等式

设$a,b>0$则
$$\sqrt{ab}<\frac{a-b}{\ln a-\ln b}<\frac{a+b}{2}$$

证明：
不妨假设$a>b>0$

右边
$$\begin{array}{rl}\frac{a-b}{\ln a-\ln b}& <\frac{a+b}{2}\\ \ln a-\ln b& >\frac{2\left(a-b\right)}{a+b}\\ \ln \frac{a}{b}& >\frac{2\left(\frac{a}{b}-1\right)}{\frac{a}{b}+1}\\ \left(\frac{a}{b}+1\right)\ln \frac{a}{b}-2\left(\frac{a}{b}-1\right)& >0\end{array}$$
令$t=\frac{a}{b}>1$
$f\left(t\right)=\left(t+1\right)\ln t-2\left(t-1\right)$
$f^{\prime }\left(t\right)=\ln t+\frac{1}{t}-1$
$f^{\prime \prime }\left(t\right)=\frac{t-1}{t^2}>0$
$f^{\prime }\left(t\right)>f^{\prime }\left(1\right)=0$
$f\left(t\right)>f\left(1\right)=0$
因此成立

左边
$$\begin{array}{rl}\sqrt{ab}& <\frac{a-b}{\ln a-\ln b}\\ \ln a-\ln b& <\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\\ 2\ln \sqrt{\frac{a}{b}}-\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}& <0\end{array}$$
令$t=\frac{a}{b}>1$
$g\left(t\right)=2\ln t-t+\frac{1}{t}$
$g^{\prime }\left(t\right)=\frac{2}{t}-1-\frac{1}{t^2}=-\frac{{\left(t-1\right)}^2}{t^2}<0$
$g\left(t\right)<g\left(1\right)=0$
因此成立