Example 4 Given a direction $\vec{l}$ in plane $\alpha$, and $F$ is a convex set in plane $\alpha$ with area $S(F)$. Among all triangles inscribed in $F$ and having one side parallel to $\vec{l}$, the one with the largest area is denoted as $\triangle$, and its area is denoted as $S(\Delta)$. Find the maximum positive real number $c$ such that for any convex figure $F$ in plane $\alpha$, we have $S(\Delta) \geqslant c \cdot S(F)$.
Solve as shown in Figure 6-29, draw two supporting lines $l_{1}, l_{2}$ of $F$ parallel to $l$ ($l_{1}, l_{2}$ intersect the boundary of $F$ at least at one point $E, G$ respectively, and $F$ is between $l_{1}$ and $l_{2}$), then draw a line $l_{3}$ parallel to $l_{1}, l_{2}$ and equidistant from them. $A B, C D$ are two chords of $F$ and $A B$ is parallel and equidistant to $l_{1}, l_{3}$, $C D$ is parallel and equidistant to $l_{2}, l_{3}$. Draw supporting lines of $F$ through $A, B, C, D$ intersecting $l_{1}, l_{2}, l_{3}$ to form two trapezoids. Let the distance between adjacent parallel lines be $h$, and assume $C D \geqslant A B$. Thus, $S(F)$ is not greater than the sum of the areas of the two trapezoids, i.e.,
$$
S(F) \leqslant A B \cdot 2 h + C D \cdot 2 h \leqslant C D \cdot 4 h = \frac{8}{3}\left(\frac{1}{2} C D \cdot \frac{3}{2} h\right) = \frac{8}{3} S_{\triangle C D E} \leqslant \frac{8}{3} S(\triangle).
$$

Therefore, $S(\triangle) \geqslant \frac{3}{8} S(F)$.
On the other hand, take $F$ as a regular hexagon $A B C D E$ with side length $a$ in the plane $\alpha$, such that $A B \parallel \vec{l}$, and $\triangle P Q R$ is the largest inscribed triangle in $F$ with one side $P Q \parallel \vec{l}$. Clearly, $P, Q, R$ must be on the boundary of the regular hexagon (Figure 6-30), and let $\frac{B Q}{Q C}=\frac{\lambda}{1-\lambda}(0 \leqslant \lambda<1)$. It is easy to calculate that the height from side $P Q$ of $\triangle P Q R$ is
$$
h=\sqrt{3} a - \frac{\sqrt{3}}{2} \lambda a = \sqrt{3} a \left(1 - \frac{1}{2} \lambda\right) \text{ and } P Q = a + \lambda a = (1 + \lambda) a,
$$

so
$$
\begin{aligned}
S(\triangle P Q R) & = \frac{1}{2} h \cdot P Q = \frac{1}{2} \sqrt{3} a \left(1 - \frac{1}{2} \lambda\right) \cdot (1 + \lambda) a \\
& = \frac{\sqrt{3}}{4} a^{2} \left(2 + \lambda - \lambda^{2}\right) = \frac{\sqrt{3}}{4} a^{2} \left[\frac{9}{4} - \left(\lambda - \frac{1}{2}\right)^{2}\right] \leqslant \frac{9 \sqrt{3}}{16} a^{2}.
\end{aligned}
$$

And $S(F) = \frac{3 \sqrt{3}}{2} a^{2}$, so $S(\triangle P Q R) \leqslant \frac{3}{8} S(F)$ and the equality holds when $\lambda = \frac{1}{2}$.
In summary, the maximum value of $c$ is $\frac{3}{8}$.