Maybe it's time to post your solution, since you say it's easy...

Hint : u should solve it with circle packing

Maybe the best idea is to break it in small squares, or any regular figure..., but I think the problem is the function is not continuous....... Also, here arises a question, is A a contour of the figure, or is A the entire figure, because some continuity problems appear.... What do you think, sam-n ????

Take a point P inside A. Take all the distances from P to all points of the perimeter of A. we take $k=max[d(P,f(B))/d(P,B)]$ where B is a point of the perimeter. We know that k<1 and all points of f(A) are in A' where A' is A dilatate about P of the coefficient k. So S[f(A)]<=S(A') but S(A')=kS(A)<S(A) and so S[f(A)]<S(A) c.v.d.

what is circle packing?

Well for any figure $B$ you must first put some cirlces in $B$ that they are distinct and sum of their areas is $S(B)- \epsilon$.

Well consider circles with centers $f(A_1),f(A_2),...,f(A_n)(A_i \in A)$ with radius $r_1,...,r_n$ that are distinct and sum of ther areas is $S(f(A))- \epsilon$ Now draw circles with centers $A_1,...,A_n$ and radius $r_1,...,r_n$ . Prove they are distinct and are in $A$ so for every $\epsilon$: $$S(A) \geq S(f(A))-\epsilon$$ And this proves the problem

For circle packing you must use this that for every $\epsilon$ there are finite squares in $A$ sum of their area is at least $S(A) - \epsilon$

Simo_the_Wolf wrote: Take a point P inside A. Take all the distances from P to all points of the perimeter of A. we take $k=max[d(P,f(B))/d(P,B)]$ where B is a point of the perimeter. We know that k<1 and all points of f(A) are in A' where A' is A dilatate about P of the coefficient k. So S[f(A)]<=S(A') but S(A')=kS(A)<S(A) and so S[f(A)]<S(A) c.v.d. But the figure may not be convex so why f(A) is in A'?