## Return to Answer

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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.)

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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. 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.)

Post Undeleted by Anton Petrunin
3 added 385 characters in body
of disjoint union of finite number of convex figures $$F=\bigcup_{i=1}^n F_i$$ where $F_1$, $F_2,\dots, F_n$ . Choose a point $p_i\in F_i$ in each figureare disjoint convex figures. Now set Denote by $F_i(t)=F_i-t{\cdot}\vec p_i$.So K$the convex hull of all$F_i=F_i(0)$.F$.Note that the total length of $$\partial K\cap F$$is at least as big as the mean shadow, say $s$ of $\bigcup_i F_i(t)$ decrease in F\$

The total curvature of $t$and its area stays the same untill \partial K$is$2{\cdot}\pi$,therefore there is a couple of point$F_i(t)$touch eachotherp\in \partial K$ with curvature $\ge 2{\cdot}\pi{\cdot}\tfrac1s$.Once it happen, 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 can exachange the touching figures may bite from $F$ some small area $a$ so that mean shadow decrease by their convex hullamount almost $\ge 2{\cdot}\pi{\cdot}\tfrac{a}s$.

As a result

This way we get can bite whole $n-1$ convex figures with bigger total area F\$, when nothing remains you will add things up and smaller mean shadowget the inequality you need.

This gives proof is a reduction in $n$.bit technical to formalize, but this is possible.(If I would have to write it down, I would better find an other one.)

Post Deleted by Anton Petrunin
2 deleted 4 characters in body

As you noticed, it is sufficient to consider the case of disjoint union of finite number of convex figures $F_1$, $F_2,\dots, F_n$. Choose a point $p_i\in F_i$ in each figure. Now set $F_i(t)=F_i+(1-t){\cdot}\vec F_i(t)=F_i-t{\cdot}\vec p_i$. So $F_i=F_i(0)$.

Note that the mean shadow of $\bigcup_i F_i(t)$ decrease in $t$ and its area stays the same untill a couple of $F_i(t)$ touch eachother. Once it happen, one can exachange the touching figures by their convex hull.

As a result we get $n-1$ convex figures with bigger total area and smaller mean shadow. This gives a reduction in $n$.

1

As you noticed, it is sufficient to consider the case of disjoint union of finite number of convex figures $F_1$, $F_2,\dots, F_n$. Choose a point $p_i\in F_i$ in each figure. Now set $F_i(t)=F_i+(1-t){\cdot}\vec p_i$. So $F_i=F_i(0)$.

Note that the mean shadow of $\bigcup_i F_i(t)$ decrease in $t$ and its area stays the same untill a couple of $F_i(t)$ touch eachother. Once it happen, one can exachange the touching figures by their convex hull.

As a result we get $n-1$ convex figures with bigger total area and smaller mean shadow. This gives a reduction in $n$.
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