Example 4 Given that $\triangle ABC$ is an equilateral triangle with side length 1, $D$ is a point on side $BC$ such that $BD=p$, and $r_{1}, r_{2}$ are the radii of the inscribed circles of $\triangle ABD$ and $\triangle ADC$ respectively. Express $r_{1} r_{2}$ in terms of $p$, and find the maximum value of $r_{1} r_{2}$.
(2000, North Korea Mathematical Olympiad)
Solution: As shown in Figure 4, in $\triangle A B D$, by the cosine rule we have
$$
\begin{array}{l}
A D^{2}=p^{2}-p+1, \\
A D=\sqrt{p^{2}-p+1} .
\end{array}
$$

The area of a triangle is equal to the product of its semiperimeter and the inradius, so we have
$$
\begin{array}{l}
\frac{1+p+\sqrt{p^{2}-p+1}}{2} \cdot r_{1}=S_{\triangle A B D} \\
=\frac{1}{2} \times 1 \times p \sin 60^{\circ} .
\end{array}
$$

Thus, $r_{1}=\frac{\sqrt{3}\left(1+p-\sqrt{p^{2}-p+1}\right)}{6}$.
Similarly, from $\triangle A D C$ we get
$$
r_{2}=\frac{\sqrt{3}\left(2-p-\sqrt{p^{2}-p+1}\right)}{6} .
$$

Therefore, $r_{1} r_{2}=\frac{1-\sqrt{p^{2}-p+1}}{4}$.
When $p=\frac{1}{2}$, the maximum value of $r_{1} r_{2}$ is $\frac{2-\sqrt{3}}{8}$.