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已知函数 ${\displaystyle f(x)={\displaystyle \frac{\mathrm{e}(\ln x+1)}{x}+(1-a)\ln x,h(x)=\frac{\mathrm{e}x}{{\mathrm{e}}^x}}}$.

1. 当 ${\displaystyle x>1}$ 时，求证: ${\displaystyle h(x)>{\displaystyle -\frac{1}{2}x+\frac{3}{2}}}$;
2. 函数 ${\displaystyle f(x)}$ 有两个极值点 ${\displaystyle x_1,x_2}$, 其中 ${\displaystyle x_1<x_2}$, 求证: ${\displaystyle {\displaystyle \frac{x_2}{x_1}>{\mathrm{e}}^{3a}}}$.

1. 当 ${\displaystyle x\ge 3,h(x)>0\le {\displaystyle -\frac{1}{2}x+\frac{3}{2}}}$.当 ${\displaystyle 1<x<3}$,即证 ${\displaystyle {\displaystyle \frac{x}{{\mathrm{e}}^{x-1}(3-x)}-\frac{1}{2}>0}}$.记左侧函数为 ${\displaystyle g(x)}$,则$$\begin{array}{r}{g}^{\prime }(x)=\frac{x^2-3x+3}{{\mathrm{e}}^{x-1}(3-x{)}^2}>0,\end{array}$$
   从而 ${\displaystyle g(x)>g(1)=0}$，不等式得证.
2. 求导得 ${\displaystyle f^{\prime }(x)={\displaystyle -\frac{\mathrm{e}\ln x}{x^2}+\frac{1-a}{x}}}$,从而$$\begin{array}{r}\frac{1-a}{\mathrm{e}}=\frac{\ln x_1}{x_1}=\frac{\ln x_2}{x_2}.\end{array}$$
   简单分析${\displaystyle u(x)={\displaystyle \frac{\ln x}{x}}}$,可知${\displaystyle u(x)>0}$当且仅当${\displaystyle x>1}$,且在${\displaystyle (1,\mathrm{e})}$上增，${\displaystyle (\mathrm{e},+\mathrm{\infty })}$上减，因此存在相异极值点的必要条件是${\displaystyle 1<x_1<\mathrm{e}<x_2}$.记${\displaystyle t=\ln x_2-\ln x_1>0}$，解得${\displaystyle \ln x_1={\displaystyle \frac{t}{{\mathrm{e}}^t-1},\ln x_2=\frac{t{\mathrm{e}}^t}{{\mathrm{e}}^t-1}}}$.利用第一问的结论：$$\begin{array}{r}1-a=\frac{\mathrm{e}\ln x_2}{{\mathrm{e}}^{\ln x_2}}>-\frac{1}{2}\ln x_2+\frac{3}{2}\Rightarrow 2a<\ln x_2-1.\end{array}$$
   因此只需证$$\begin{array}{rl}& \ln x_2-\ln x_1=t>3a=2a+a\\ ⟸& t>\ln x_2-1+a\\ ⟺& 1-a>\frac{t}{{\mathrm{e}}^t-1}=\ln x_1=\frac{1-a}{\mathrm{e}}{x}_1\\ ⟺& x_1<\mathrm{e},\end{array}$$
   而这已经被说明.因此结论得证.