Clearly the projection is a (possibly degenerate) hexagon consisting of three parallelograms touching each other at one side and having one common point, the projection of one of the vertices of $C$. Now let $P,Q,R$ be the three vertices adjacent to this one vertex. Then $PQR$ is equilateral with area $\frac{\sqrt{3}}{2}$ and the image of $PQR$ fills exactly half of each parallelogram and hence exactly half of the hexagon. But of course the image of $PQR$ has at most the area of $PQR$ hence the area of the hexagon is at most $\sqrt{3}$ with the maximum being achieved when projecting on $PQR$.

Let $F_1,F_2,F_3$ be the areas of the three parallelograms of which the hexagon (as in the solution above) consists. Using coordinates/analytic geometry it is easily shown that $F_1^2+F_2^2+F_3^2=1$. Then $F_1+F_2+F_3 \le \sqrt{3}$ follows directly from the inequality between arithmetic and quadratic mean.