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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.) 4 deleted 15 characters in body 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. Let$s$be mean shadow of$F$. Denote by$K$the convex hull of all$F$. Note thatthe total length of $$\partial \mathop{\rm length}(\partial K\cap F$$ is at least as big as the mean shadow, say$s$of$F$F)\le s.$$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 we can bite whole$F\$, when nothing remains you will add things up and get the inequality you need.

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.)

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