As you noticed, it is sufficient to consider the case
$$F=\bigcup_{i=1}^n F_i$$
where $F_1$, $F_2,\dots, F_n$ are disjoint convex figures with nonempty interior.
Let $s$ be mean shadow of $F$.
Denote by $K$ the convex hull of all $F$.
Note that
$$\mathop{\rm length}(\partial K\cap F)\le s.$$
We will prove the following claim: one can bite from $F$ some arbitrary small area $a$ so that mean shadow decrease by amount almost $\ge 2{\cdot}\pi{\cdot}\tfrac{a}s$ (say $\ge 2{\cdot}\pi{\cdot}\tfrac{a}s{\cdot}(1-\tfrac{a}{s^2})$ will do). Once it is proved, we can bite whole $F$ by very small pieces, when nothing remains you will add things up and get the inequality you need.
The claim is easy to prove in case if $\partial F$ has a corner (i.e., the curvature of $\partial F$ has an atom at some point). Note that the total curvature of $\partial K$ is $2{\cdot}\pi$, therefore there is a point $p\in \partial K$ with curvature $\ge 2{\cdot}\pi{\cdot}\tfrac1s$. The point $p$ has to lie on $\partial F$ since $\partial K\backslash \partial F$ is a collection of line segments. Around this point one may bite from $F$ some small area $a$ so that mean shadow decrease by amount almost $\ge 2{\cdot}\pi{\cdot}\tfrac{a}s$.
This way Moreover, if there are no corners, we can bite whole assume that $F$, when nothing remains you will add things up and get the inequality you needp$ is not an end of segment of $\partial K\cap F$.
This proof is a bit technical to formalize, but this is possible. (If I would have to write it down, I would better find an other one.)
