$x_1,x_2\gt 0,且x_1\ne x_2,则有:\sqrt{x_1x_2} \lt \cfrac{x_1-x_2}{\ln x_1-\ln x_2} \lt\cfrac{x_1+x_2}{2} $
先证明左边的不等式：
$x_1,x_2\gt 0,且x_1\ne x_2,则有:\sqrt{x_1x_2} \lt \cfrac{x_1-x_2}{\ln x_1-\ln x_2} $
$不妨设x_1\gt x_2,\ln x_1-\ln x_2 \lt \cfrac{x_1-x_2}{\sqrt{x_1x_2} } = \cfrac{x_1}{\sqrt{x_1x_2} }-\cfrac{x_2}{\sqrt{x_1x_2} }=\sqrt{\cfrac{x_1}{x_2} } -\sqrt{\cfrac{x_2}{x_1} }$

$令t=\sqrt{\cfrac{x_1}{x_2} } 换元,比t=\cfrac{x_1}{x_2}好$
$\ln \cfrac{x_1}{x_2} \lt \sqrt{\cfrac{x_1}{x_2} } -\sqrt{\cfrac{x_2}{x_1}}\Rightarrow $
$2\ln t \lt t-\cfrac{1}{t} \quad(t\gt1)$
构造函数$f(t)=t-\cfrac{1}{t}-2\ln t \quad (t\gt 1)$
${f}' (x)=1+\cfrac{1}{t^2}-\cfrac{2}{t} =\cfrac{t^2-2t+1}{t^2}$
${f}' (t)\gt 0\quad (t\gt 1) \therefore f(t)\gt f(1)=0$

再证明右边的不等式：
$\cfrac{x_1-x_2}{\ln x_1-\ln x_2} \lt\cfrac{x_1+x_2}{2} $
$不妨设x_1\gt x_2; $

$\cfrac{1}{\ln x_1-\ln x_2} \lt \cfrac{(x_1+x_2)}{2(x_1-x_2)} \Rightarrow $

$\ln x_1-\ln x_2\gt \cfrac{2(x_1-x_2)}{x_1+x_2} \Rightarrow $

$\ln \cfrac{x_1}{x_2}\gt \cfrac{2(\cfrac{x_1}{x_2}-1)}{\cfrac{x_1}{x_2} +1}\quad$ 右式分子分母除以$\cfrac{1}{2} x_2 $
换元令$t=\cfrac{x_1}{x_2}\quad (t\gt 1)$
$\ln t \gt \cfrac{2t-2}{t+1} =2-\cfrac{4}{t+1} \qquad$
$\Rightarrow \ln t -2+\cfrac{4}{t+1} \gt 0\quad $不去分母构造函数，法一：
构造$f(t)=\ln t+\cfrac{4}{t+1} -2\quad (t\gt 1)$
${f}' (t)=\cfrac{1}{t} -\cfrac{4}{(t+1)^2} =\cfrac{(t+1)^2-4t}{t(t+1)^2} =\cfrac{(t-1)^2}{t(t+1)^2}$
${f}' (t)\gt 0 \quad\therefore f(t)\nearrow f(t)\gt f(1)=0$

去分母再构造函数，法二：
$\ln t \gt \cfrac{2t-2}{t+1} \Rightarrow (t+1)\ln t\gt 2t-2$
构造$g(t)=(t+1)\ln t-2t +2$
${g}' (t)=\ln t+\cfrac{t+1}{t}-2=\ln t +\cfrac{1}{t} -1 $
${g}'' (t)=\cfrac{1}{t}-\cfrac{1}{t^2} =\cfrac{t-1}{t^2} \quad t\gt 1$
${g}'' (t)\gt0\Rightarrow {g}' (t)\nearrow \Rightarrow {g}' (t)\gt {g}' (1)=0$
$\Rightarrow g (t)\nearrow \Rightarrow g(t)\gt g(1)=0$