Example 4. On the sides $AB$, $BC$, and $CA$ of an equilateral triangle $ABC$, there are moving points $D$, $E$, and $F$ respectively, such that $|AD| + |BE| + |CF| = |AB|$. If $|AB| = 1$, when does the area of $\triangle DEF$ reach its maximum value? What is this maximum value?

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Translating the text as requested, while preserving the original formatting and line breaks.
Let $|A D|=x,|B E|=y,|C F|=z$, then $|D B|=1-x,|E C|=1-y$, $|F A|=1-z$.
$\therefore S_{\triangle D B P}=S_{\triangle A B C}-\left(S_{\triangle A D P}+\tilde{S}_{\triangle E B D}+S_{\triangle C F I}\right)$.
$S_{\triangle D R}=\sqrt{3}-\frac{\sqrt{3}}{4}[x(1-z)+y(1$
$$
\begin{array}{c}
-x)+z(1-y) \\
=\frac{\sqrt{3}}{4}(x y \\
+y z+z x) . \\
\therefore x y+z y+z^{x} \\
\leqslant x^{2}+y^{2}+z^{2} .
\end{array}
$$
(when and only when $x=y=z$
equality holds)
$$
\begin{aligned}
\therefore 3 & (x y+y z+z x) \leqslant x^{2}+y^{2}+z^{2} \\
& +2(x y+y z+z x) \\
& =(x+y+z)^{2}=1,
\end{aligned}
$$

i.e., $x y+y z+z x \leqslant \frac{1}{3}$.
$$
\therefore S_{\triangle D B F} \leqslant \frac{\sqrt{3}}{4} \cdot \frac{1}{3}=\frac{\sqrt{3}}{12} \text {. }
$$

i.e., when $x=y=z=\frac{1}{3}$, the area of $\triangle D E F$ is maximized, with the maximum area being $\frac{\sqrt{3}}{12}$ square units.