The classical isoperimetric inequality can be stated as follows: if $A$ and $B$ are sets in the plane with the same area, and if $B$ is a disk, then the perimeter of $A$ is larger than the perimeter of $B$.


There are several ways to define the perimeter. Here is a unusual one: if $A \subset \mathbb{R}^2$ is a **convex** set, the 
[Cauchy-Crofton formula](http://en.wikipedia.org/wiki/Crofton_formula) says that the perimeter of $A$ equals the measure of the set of lines that hit $A$, or


$$ p(A) = \frac{1}{2} \int\_{S^1} \lambda(P\_{\theta} A) d\theta, $$


where $P\_\theta$ is the orthogonal projection in the direction $\theta \in S^1$, and $\lambda$ the Lebesgue measure on any line.


Now, this definition of $p(A)$ makes sense for non-necessarily convex sets, excepts that it is not the usual notion of perimeter, so let's call it rather "mean shadow". My question if whether the isoperimetric inequality holds for the mean shadow instead of perimeter: if $A,B$ are (open, say) subsets of the plane with equal area, and if $B$ is a disk, is the mean shadow of $A$ larger that the mean shadow of $B$ ?


The inequality is true if $A$ is convex, and we can assume that $A$ is a disjoint union of convex sets (since taking the convex hull of a connected set does not change the mean shadow).



Isoperimetric-like inequality for non-connected sets
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. Moreover, if there are no corners, we can assume that $p$ 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.)


