899

已知函数 ${\displaystyle f(x)=x^a-{\log }_{b}x}$ (${\displaystyle a>0,b>0}$ 且 ${\displaystyle b\ne 1}$), 若 ${\displaystyle f(x)\ge 1}$ 恒成立，则 ${\displaystyle ab}$ 的最小值为_____.

注意到 ${\displaystyle f(1)=1}$.要使 ${\displaystyle f(x)\ge 1}$ 恒成立，则 $$f_{-}^{\prime }(1)\le 0\le f_{+}^{\prime }(1)\Rightarrow {\displaystyle f^{\prime }(1)=a-\frac{1}{\ln b}=0.}$$这样 $$\begin{array}{r}ab=\frac{b}{\ln b}\ge \frac{b}{b/\mathrm{e}}=\mathrm{e},\end{array}$$
取等当且仅当${\displaystyle b=\mathrm{e},a=1}$.