10 Given positive real numbers $x, y$ satisfy $x+2y=4$, then the minimum value of $\frac{1}{x}+\frac{1}{y}$ is
(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 $\frac{3+2 \sqrt{2}}{4}$.