Question 1 Find the minimum value of the function $y=2 \sqrt{(x-1)^{2}+4}+$ $\sqrt{(x-8)^{2}+9}$.
Let $a, b$ be undetermined constants, and $a, b$ are not both zero.
Then by the Cauchy-Schwarz inequality for two variables $\left(p^{2}+q^{2}\right)\left(x^{2}+y^{2}\right) \geqslant (p x+q y)^{2}$ (equality holds if and only if $q x=p y$), we have
$$\begin{aligned}
\sqrt{a^{2}+b^{2}} y= & 2 \sqrt{\left(a^{2}+b^{2}\right)\left[(x-1)^{2}+2^{2}\right]} \\
& +\sqrt{\left(b^{2}+a^{2}\right)\left[(8-x)^{2}+3^{2}\right]} \\
\geqslant 2[a(x-1)+ & 2 b]+[b(8-x)+3 a] \\
= & (2 a-b) x+a+12 b
\end{aligned}$$
Equality holds if and only if
( I) $\left\{\begin{array}{l}b(x-1)=2 a, \\ a(8-x)=3 b, \\ a(x-1)+2 b \geqslant 0, \\ b(8-x)+3 a \geqslant 0 .\end{array}\right.$
In (1), let $2 a-b=0$,
i.e., $b=2 a$ (where $a, b$ are not both zero),
substituting into (I) yields $\left\{\begin{array}{l}x=2, \\ a+2 b \geqslant 0, \\ 3 a+6 b \geqslant 0, \\ (b=2 a) .\end{array}\right.$
In particular, let $a=1, b=2$,
then (1) becomes $\sqrt{5} y \geqslant 25, y \geqslant 5 \sqrt{5}$.
Therefore, when $x=2$, $y_{\text {min }}=5 \sqrt{5}$.