7. Given vectors $\boldsymbol{a}=(x, 1), \boldsymbol{b}=(2, y), \boldsymbol{c}=$ $(1,1)$, and $\boldsymbol{a}-\boldsymbol{b}$ is collinear with $\boldsymbol{c}$. Then the minimum value of $|\boldsymbol{a}|+2|\boldsymbol{b}|$ is $\qquad$ .
7. $3 \sqrt{5}$.
From $a-b=(x-2,1-y)$ being collinear with $c$, we get $x+y=3$.
From the absolute value inequality, we have
$$
\begin{array}{l}
|a|+2|\boldsymbol{b}| \geqslant|\boldsymbol{a}+2 \boldsymbol{b}| \\
=\sqrt{(x+4)^{2}+(1+2 y)^{2}} \\
=\sqrt{(7-y)^{2}+(1+2 y)^{2}} \\
=\sqrt{5 y^{2}-10 y+50} \\
\geqslant \sqrt{50-\frac{10^{2}}{20}}=3 \sqrt{5},
\end{array}
$$

The equality holds if and only if $y=1, x=2$.
Therefore, the minimum value sought is $3 \sqrt{5}$.