7. If $x, y>0$, and $x+2 y=1$, then the minimum value of $u=\left(x+\frac{1}{x}\right)\left(y+\frac{1}{4 y}\right)$ is

保留源文本的换行和格式，翻译结果如下：

7. If $x, y>0$, and $x+2 y=1$, then the minimum value of $u=\left(x+\frac{1}{x}\right)\left(y+\frac{1}{4 y}\right)$ is
7. Comparison: When $a, b>0, a+b=1$, then $\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right) \geqslant \frac{25}{4}$, with equality if and only if $a=b=\frac{1}{2}$.

It can be seen that, $\left(x+\frac{1}{x}\right)\left(y+\frac{1}{4 y}\right)=\frac{1}{2}\left(x+\frac{1}{x}\right)\left(2 y+\frac{1}{2 y}\right) \geqslant \frac{1}{2} \cdot \frac{25}{4}=\frac{25}{8}$, with equality if and only if $x=\frac{1}{2}, y=\frac{1}{4}$.

Therefore, $u_{\text {min }}=\frac{25}{8}$.