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已知 ${\displaystyle x_1,x_2,x_3}$ 是关于 ${\displaystyle x}$ 的方程 ${\displaystyle |x\ln x|=a}$ 的三个不同的根，且 ${\displaystyle x_1<x_2<x_3}$.

1. 求 ${\displaystyle a}$ 的取值范围;
2. 证明: ${\displaystyle x_1{x}_2{x}_3\le {\displaystyle {\mathrm{e}}^{-\frac{5}{3}}}}$.

1. 解 令$${\displaystyle f(x)=|x\ln x|-a=\{\begin{array}{rl}& -x\ln x-a,0<x\le 1,\\ & x\ln x-a,x>1,\end{array}}$$
   有 ${\displaystyle x>1}$ 时 ${\displaystyle f^{\prime }(x)=1+\ln x>0}$; ${\displaystyle x<1}$ 时 ${\displaystyle f^{\prime }(x)=-(1+\ln x)}$，故 ${\displaystyle f(x)}$ 在 ${\displaystyle (0,1)}$ 上有唯一最大值 ${\displaystyle {\displaystyle f({\mathrm{e}}^{-1})={\mathrm{e}}^{-1}-a}}$.

   • 当 ${\displaystyle a\le 0}$，仅在 ${\displaystyle a=0}$ 时有解 ${\displaystyle x=1}$.
   • 当 ${\displaystyle a\ge {\mathrm{e}}^{-1}}$. 在 ${\displaystyle (0,1),(1,+\mathrm{\infty })}$ 上各至多有一解；当 ${\displaystyle 0<a<{\mathrm{e}}^{-1}}$, 则 ${\displaystyle x<1}$ 时 ${\displaystyle f(x)={\displaystyle 2x\ln \sqrt{\frac{1}{x}}-a\le 2x\frac{1}{\mathrm{e}\sqrt{x}}-a=\frac{2\sqrt{x}}{\mathrm{e}}-a}}$, 因此取 ${\displaystyle x_1=min{\displaystyle \{{\left(\frac{a\mathrm{e}}{2}\right)}^2,\frac{1}{\mathrm{e}}\}}}$ 得 ${\displaystyle f(x_1)<0}$. 又 ${\displaystyle f({\mathrm{e}}^{-1})={\mathrm{e}}^{-1}-a>0,f(1)=-a<0,f(\mathrm{e})=\mathrm{e}-a>0}$, 故 ${\displaystyle f(x)}$ 在 ${\displaystyle (x_1,{\mathrm{e}}^{-1}),({\mathrm{e}}^{-1},1),(1,\mathrm{e})}$ 上各有一零点.

   综上，${\displaystyle a\in (0,{\mathrm{e}}^{-1})}$.

2. 证明 由1得 ${\displaystyle 0<x_1<{\mathrm{e}}^{-1}<x_2<1<x_3}$.从而 ${\displaystyle x_1\ln x_1=x_2\ln x_2}$，从而$$\begin{array}{r}\frac{\ln x_1}{\frac{1}{x_1}}=\frac{\ln x_2}{\frac{1}{x_2}}=\frac{\ln \frac{x_1}{x_2}}{\frac{1}{x_1}-\frac{1}{x_2}}=\frac{\ln x_1{x}_2}{\frac{1}{x_1}+\frac{1}{x_2}},\end{array}$$
   进而$$\begin{array}{r}\ln x_1{x}_2=\frac{x_1+x_2}{x_2-x_1}\ln \frac{x_1}{x_2}<\frac{x_1+x_2}{x_2-x_1}\frac{2(\frac{x_1}{x_2}-1)}{\frac{x_1}{x_2}+1}=-2\Rightarrow x_1{x}_2\le {\mathrm{e}}^{-2}.\end{array}$$
   又${\displaystyle {\displaystyle \ln 3<\frac{1}{2}(3-\frac{1}{3})=\frac{4}{3}\Rightarrow \frac{1}{3}{\mathrm{e}}^{\frac{1}{3}}-a=f({\mathrm{e}}^{\frac{1}{3}})>f\left(\frac{1}{\mathrm{e}}\right)}}$,从而${\displaystyle x_3<{\mathrm{e}}^{\frac{1}{3}}}$,这样${\displaystyle x_1{x}_2{x}_3<{\mathrm{e}}^{-\frac{5}{3}}}$.