1607

已知 ${\displaystyle a,b}$ 为正实数. 若 ${\displaystyle a+4b=a^2{b}^3}$, 求 ${\displaystyle \frac{1}{a}+\frac{1}{8b}}$ 的最小值.

将条件式转化为 ${\displaystyle \frac{1}{4b}+\frac{1}{a}=\frac{a{b}^2}{4}}$. 则 $$\begin{array}{r}{(\frac{1}{a}+\frac{1}{8b})}^2=\frac{1}{64b^2}+\frac{1}{a}(\frac{1}{a}+\frac{1}{4b})=\frac{1}{64b^2}+\frac{b^2}{4}\ge \frac{1}{8},\end{array}$$ 因此 ${\displaystyle \frac{1}{a}+\frac{1}{8b}\ge \frac{\sqrt{2}}{4}}$. 取等条件为 ${\displaystyle (a,b)=(\frac{\sqrt{2}-1}{4},\frac{1}{2})}$.