[函数] 还是来自群……

已知函数$f(x)=a\ln x-(x-1)^2-ax$，设$a>0$，如果对于$f(x)$的图象上两点$P_1(x_1,f(x_1)),P_2(x_2,f(x_2))$ $(x_1<x_2)$，存在$x_0\in(x_1,x_2)$，使得$f(x)$的图象在$x=x_0$处的切线$m//P_1P_2$，求证
\[x_0<\frac{x_1+x_2}2.\]

由已知得
\[f'(x_0)=\frac{f(x_1)-f(x_2)}{x_1-x_2},\]
代入数据解得
\[2x_{0}-\frac{a}{x_{0}}=x_{1}+x_{2}-\frac{a(\ln x_{1}-\ln x_{2})}{x_{1}-x_{2}},\]
下面证明
\[\frac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}>\frac{2}{x_{1}+x_{2}},（*）\]
令$\frac{x_{1}}{x_{2}}=t\in(0,1)$，则上式等价于
\[\ln t<2-\frac{4}{t+1},\]
令$g(t)=\ln t-2+\frac{4}{t+1}$，则$g(1)=0$且
\[g'(t)=\frac{(1-t)^{2}}{t(1+t)^{2}}>0,\]
从而当$t\in(0,1)$时必有$g(t)<0$，式（*）得证，于是由$a>0$得到
\[2x_{0}-\frac{a}{x_{0}}=x_{1}+x_{2}-\frac{a(\ln x_{1}-\ln x_{2})}{x_{1}-x_{2}}<2\cdot \frac{x_{1}+x_{2}}{2}-\frac{a}{\frac{x_{1}+x_{2}}{2}},\]
显然当$a>0,x>0$时$h(x)=2x-\frac ax$为增函数，从而由上式有
\[h(x_0)<h\left(\frac{x_1+x_2}2\right) \iff x_0<\frac{x_1+x_2}2.\]