Circle has smallest shadow to area ratio I believe and would like to show that for an arbitrary two-dimensional geometrical figure, the average (with respect to rotation) one-dimensional shadow that it casts is larger than the shadow of a circle of the same area.
Just to clarify:
Let us define the figure as a set of points in a (x,y) coordinate system. 
We confine the problem to closed and bounded sets, and it is easily seen that any non-convex set has a convex equivalent with the same shadow and larger area, so we only need to consider convex sets.
The shadow with respect to a parallel vertical light source is the largest x-value in the set minus the smallest x-value in the set.
We now calculate the shadow for all rotations of the figure, and average the shadow as a function of the angle with uniform weight to all angles.
For example, the shadow of a circle is its diameter, and the shadow of a rod with length $l$ is $l \sin \phi$, where $\phi$ is the angle between the rod and the y-axis.
 A: 
"for an arbitrary two-dimensional geometrical figure, the average (with respect to rotation) one-dimensional shadow that it casts is larger than the shadow of a circle of the same area"

This is a consequence of Urysohn's inequality, which states that 
among all convex bodies in $\mathbb{R}^n$ with a fixed volume, a Euclidean n-ball has minimal mean width. More precisely:

If $A$ is a convex body in $\mathbb{R}^n$, then
  $$\frac{\text{vol}(A)}{\bar{w}(A)^n} \le \frac{\text{vol}(B)}{\bar{w}(B)^n}$$
  where $B$ denotes the Euclidean unit $n$-ball and $\bar{w}(\cdot)$ denotes the mean width. 

Given a convex body $A\subset\mathbb R^n$ and a direction $v\in\mathcal{S}^{n-1}$, the width $w(A, v)$ is defined as the Euclidean distance between two supporting hyperplanes of $A$ in the directions $v$ and $-v$:
 
The mean width of $A$ is defined as
$$\bar{w}(A) = \int_{\mathcal{S}^{n-1}}w(A, v)\,\sigma(dv)
$$ 
where $\sigma$ is the rotationally invariant probability measure on the unit sphere $\mathcal S^{n-1}$.
Since $\bar{w}(B) = 2$ and $\text{vol}(B) = \frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}$, the inequality can be written
$$\bar{w}(A)^n \ge \left( \frac{2^n\ \Gamma(\frac{n}{2} + 1)}{\pi^{n/2}}   \right)\ \text{vol}(A)
$$
with equality holding iff $A$ is a ball.
In the present case, $n=2$:
$$\bar{w}(A)^2 \ge \left(\frac{4}{\pi}\right)\text{area}(A)
$$
with equality iff $A$ is circular, in which case we have $\bar{w}(A)^2 = \left(\frac{4}{\pi}\right)\pi R^2 = (2R)^2 $ for a circular ball of radius $R$. In this case, the width is what you call the "shadow", and the integral defining the mean width reduces to the average you describe (with the direction $v$ rotating, instead of rotating the figure).   
Hence, among all planar convex figures with a given area, a circular disk has the minimal mean width.
NB: Proofs of Urysohn's inequality can be found on p. 12 ff. of Geometry of Isotropic Convex Bodies by Silouanos Brazitikos, Apostolos Giannopoulos, Petros Valettas, Beatrice-Helen Vritsiou (which can be previewed online), and also as sketched in this MSE answer.
