Example 9 In a regular $\triangle ABC$ with side length 1, points $M, N, P$ lie on sides $BC, CA, AB$ respectively, and $MB + CN + AP = 1$. Find the maximum area of $\triangle MNP$.

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Solution: As shown in Figure 10, let
$$
\begin{array}{l}
B M=x, \\
C N=y, \\
A P=z .
\end{array}
$$

Then $0 \leqslant x, y, z \leqslant 1$,
$$
\begin{array}{l} 
x+y+z=1 . \\
\text { Hence } S_{\triangle M P P}+S_{\triangle B P M}+S_{\triangle C W V} \\
= \frac{1}{2}[x(1-z)+y(1-x)+z(1-y)] \sin 60^{\circ} \\
= \frac{\sqrt{3}}{4}[x(1-z)+y(1-x)+z(1-y)] .
\end{array}
$$

By $(x+y+z)^{2} \geqslant 3(x y+y z+z x)$, it is easy to get
$$
x y+y z+z x \leqslant \frac{1}{3} \text {. }
$$

Thus, $x(1-z)+y(1-x)+z(1-y)$
$$
\begin{array}{l}
=x+y+z-(x y+y z+z x) \\
\geqslant 1-\frac{1}{3}=\frac{2}{3} .
\end{array}
$$

Therefore, $S_{\triangle N P P}+S_{\triangle B P M}+S_{\triangle C M N}$
$$
\geqslant \frac{\sqrt{3}}{4} \times \frac{2}{3}=\frac{\sqrt{3}}{6} \text {. }
$$

Hence $S_{\triangle M N P}=S_{\triangle \triangle B C}-\left(S_{\triangle M P}+S_{\triangle B M}+S_{\triangle O N}\right)$
$$
\leqslant \frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{6}=\frac{\sqrt{3}}{12} \text {. }
$$

When $x=y=z=\frac{1}{3}$, the equality holds.
Therefore, the maximum area of $\triangle M N P$ is $\frac{\sqrt{3}}{12}$.