Example 5 Find the values of real numbers $x, y$ such that $(y-1)^{2}+(x+$ $y-3)^{2}+(2 x+y-6)^{2}$ reaches the minimum value. (Example 8 in [1])
$$\begin{array}{l}
\text { To solve, introduce parameters } u, v, w, \text { by Cauchy-Schwarz inequality we have } \\
\left(u^{2}+v^{2}+w^{2}\right)\left[(y-1)^{2}+(x+y-3)^{2}+(2 x\right. \\
\left.+y-6)^{2}\right] \geqslant[u(y-1)+v(x+y-3)+w(2 x+y \\
-6)]^{2}=[(v+2 w) x+(u+v+w) y-u-3 v- \\
6 w]^{2},
\end{array}$$

To find the minimum value of the expression, we can set $v+2 w=0, u+$ $v+w=0$. Let's take $u=1, v=-2, w=1$, and substitute these into the inequality to get $(y-1)^{2}+(x+y-3)^{2}+(2 x+y-$ $6)^{2} \geqslant \frac{1}{6}$. Therefore, the minimum value is $\frac{1}{6}$. At this point, $x=\frac{5}{2}, y=\frac{5}{6}$.