31.已知函数f(x)=lnx+ax(a∈R)f(x)=lnx+\frac{a}{x}(a\in R)f(x)=lnx+xa(a∈R).(1)讨论f(x)f(x)f(x)的单调性;(2)函数g(x)=xf(x)−ax2−xg(x)=xf(x)-ax^{2}-xg(x)=xf(x)−ax2−x有两个不同的极值点x1x_{1}x1,x2(x1<x2)x_{2}(x_{1}<x_{2})x2(x1<x2),证明:lnx1+2lnx2>3lnx_{1}+2lnx_{2}>3lnx1+2lnx2>3.