907

已知函数 ${\displaystyle f(x)=x{\mathrm{e}}^x-\ln x-1,g(x)=(x-1)\ln x}$. 若 ${\displaystyle f(a)+g(b)={\mathrm{e}}^a\ln b}$, 求证: ${\displaystyle f(x)-{\displaystyle f\left(\frac{1}{b}\right)>x^2-a}}$.

注意到条件式等价于$$\begin{array}{r}({\mathrm{e}}^a-b)(a-\ln b)+ab-\ln ab-1=0,\end{array}$$
其中 ${\displaystyle {\mathrm{e}}^a\ge b⟺a\ge \ln b}$, 因此 ${\displaystyle ({\mathrm{e}}^a-b)(a-\ln b)\ge 0}$. 又 ${\displaystyle ab-\ln ab-1\ge 0}$, 故左边 ${\displaystyle \ge 0}$，故只有$$\begin{array}{r}a=\ln b,ab=1\Rightarrow b\ln b=1.\end{array}$$
因此$$\begin{array}{r}f\left(\frac{1}{b}\right)-a=\frac{1}{b}{\mathrm{e}}^{\frac{1}{b}}-1={\mathrm{e}}^{\ln \frac{1}{b}}{\mathrm{e}}^{\frac{1}{b}}-1=e^{\frac{1}{b}-\ln b}-1=0,\end{array}$$
而$$\begin{array}{r}f(x)-x^2\ge x(x+1)-(x-1)-1-x^2=0,\end{array}$$
从而$$\begin{array}{r}f(x)-x^2\ge 0>f\left(\frac{1}{b}\right)-a.\end{array}$$