Problem 745

(10) Since $\frac{1}{4} x+\frac{1}{2} y=1$, thus
$$\begin{aligned} \frac{1}{x}+\frac{1}{y} & =\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{4} x+\frac{1}{2} y\right) \\ & =\frac{3}{4}+\frac{x}{4 y}+\frac{y}{2 x} \\ & \geqslant \frac{3}{4}+\frac{\sqrt{2}}{2} \\ & =\frac{3+2 \sqrt{2}}{4}, \end{aligned}$$

equality holds if and only if
$$\left\{\begin{array}{l} x+2 y=4, \\ \frac{x}{4 y}=\frac{y}{2 x} \end{array}\right.$$

which is
$$\left\{\begin{array}{l} x=4 \sqrt{2}-4 \\ y=4-2 \sqrt{2} \end{array}\right.$$

when the equality holds.
Therefore, the minimum value of $\frac{1}{x}+\frac{1}{y}$ is $\frac{3+2 \sqrt{2}}{4}$.