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已知函数 ${\displaystyle f(x)=a{x}^2,g(x)=\ln x}$，两条不同的直线 ${\displaystyle l_1,l_2}$ 均同时与函数 ${\displaystyle f(x),g(x)}$ 的图像相切.

1. 求实数 ${\displaystyle a}$ 的取值范围;
2. 记 ${\displaystyle l_1,l_2}$ 在 ${\displaystyle y}$ 轴上的截距分别为 ${\displaystyle d_1,d_2}$, 证明: ${\displaystyle d_1+d_2<-1}$.

1. 解 设 ${\displaystyle l_1}$ 在 ${\displaystyle f,g}$ 的切点为 ${\displaystyle x_1,x_2}$，${\displaystyle l_2}$ 的为 ${\displaystyle x_3,x_4}$. 一方面，${\displaystyle l_1:y-a{x}_1^2=2a{x}_1(x-x_1)\Rightarrow y=2a{x}_{1}x-a{x}_1^2}$, 另一方面同理 ${\displaystyle l_1:y={\displaystyle \frac{1}{x_2}x+\ln x_2-1}}$.这样$$\begin{array}{r}\{\begin{array}{rl}& 2a{x}_1=\frac{1}{x_2},\\ & -a{x}_1^2=\ln x_2-1.\end{array}\Rightarrow a{x}_1^2=\ln 2a{x}_1+1.\end{array}$$
   记${\displaystyle h(x)=a{x}^2-\ln 2ax-1}$,则${\displaystyle h(x_1)=h(x_3)=0}$.

   • ${\displaystyle a>0}$,则 ${\displaystyle h^{\prime }(x)=2ax-{\displaystyle \frac{1}{x},x>0}}$, ${\displaystyle h(x)}$ 有唯一极小值点 ${\displaystyle {\displaystyle x=\frac{1}{\sqrt{2a}}}}$,则 ${\displaystyle h(x{)}_{min}=h{\displaystyle \left(\frac{1}{\sqrt{2a}}\right)=-\frac{1}{2}(1+\ln 2a)<0\Rightarrow a>\frac{1}{2\mathrm{e}}}}$.此时 ${\displaystyle h{\displaystyle \left(\frac{1}{2a\mathrm{e}}\right)=\frac{1}{4a{\mathrm{e}}^2}>0,h(2)=4a-\ln 4a-1\ge 4a-(4a-1)-1=0}}$,故 ${\displaystyle h(x)}$ 在 ${\displaystyle {\displaystyle (\frac{1}{2a\mathrm{e}},\frac{1}{\sqrt{2a}}),(\frac{1}{\sqrt{2a}},2)}}$ 上各有一零点，也即存在两条公切线.
   • ${\displaystyle a<0}$，则 ${\displaystyle h^{\prime }(x)={\displaystyle 2ax-\frac{1}{x}>0,x<0}}$，故 ${\displaystyle h(x)}$ 在 ${\displaystyle (-\mathrm{\infty },0)}$ 上单调，不存在两个零点.

   综上，${\displaystyle a>{\displaystyle \frac{1}{2\mathrm{e}}}}$.

2. 证明 由 1 得 ${\displaystyle a{x}_i^2=\ln 2a{x}_i+1,i=1,3}$.由 ${\displaystyle a>0}$ 得 ${\displaystyle x_1,x_3>0}$.不妨设 ${\displaystyle x_1>x_3}$.作差得 $$\begin{array}{r}a(x_1^2-x_3^2)=\ln \frac{x_1}{x_3}.\end{array}$$
   利用不等式${\displaystyle {\displaystyle \frac{2(x-1)}{x+1}<\ln x<\frac{1}{2}(x-\frac{1}{x}),x>1}}$，一方面有$$\begin{array}{r}a(x_1^2-x_3^2)>2\frac{x_1-x_3}{x_1+x_3}\Rightarrow a(x_1+x_3{)}^2>2,\end{array}$$
   另一方面有$$\begin{array}{r}a(x_1^2-x_3^2)<\frac{1}{2}\cdot \frac{x_1^2-x_3^2}{x_1{x}_3}\Rightarrow 2a{x}_1{x}_3<1.\end{array}$$
   这两个不等式相减即得$$\begin{array}{r}{d}_1+d_3=-a(x_1^2+x_3^2)<-1.\end{array}$$