Example 6 As shown in Figure 1, in the isosceles right triangle $AOB$ with one leg being 1, a point $P$ is randomly taken within this triangle. Parallel lines to the three sides are drawn through $P$, forming three triangles (shaded areas in the figure) with $P$ as a vertex. Find the minimum value of the sum of the areas of these three triangles and the position of point $P$ when the minimum value is achieved.
Parse: Taking $O A, O B$ as the $x$-axis and $y$-axis respectively to establish a rectangular coordinate system as shown in Figure 2, the equation of the line $A B$ is $x+y=1$. Let the coordinates of point $P$ be $P\left(x_{p}, y_{P}\right)$, then the sum of the areas of the three triangles sharing $P$ as a common vertex is $S=\frac{1}{2} x_{P}^{2}+\frac{1}{2} y_{P}^{2}+\frac{1}{2}\left(1-x_{P}-y_{P}\right)^{2}$,

i.e., $2 S=x_{P}^{2}+y_{P}^{2}+\left(1-x_{P}-y_{P}\right)^{2}$. 
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{\left[x_{P}^{2}+y_{P}^{2}+\left(1-x_{P}-y_{P}\right)^{2}\right] \cdot\left(1^{2}+1^{2}+1^{2}\right)} \\
\geqslant\left[x_{P}+y_{P}+\left(1-x_{P}-y_{P}\right)\right]^{2},
\end{array}$$

i.e., $2 S \times 3=6 S \geqslant 1$.
Equality holds if and only if $\frac{x_{P}}{1}=\frac{y_{P}}{1}=\frac{1-x_{P}-y_{P}}{1}$,
i.e., when $x_{P}=y_{P}=\frac{1}{3}$, the area $S$ is minimized, and the minimum value is $S_{\text {min }}=\frac{1}{6}$.