Largest area given a fixed perimeter

The question is simple: given a fixed perimeter of say P, find the shape that gives the largest possible area. The answer is somewhat equally simple: a circle gives the largest area for a fixed perimeter. The proof, is not so clear. The definition is a trickier problem, which I shall not care to resolve (I define the center of the circle to be the outside of the perimeter...).

Browsing the internet yields more than just one proof and references to proofs that exist. One of which I trust to be rather unambiguously true is referenced to as the proof by calculus of variations, which I suppose works, since calculus is such an overpowered sledgehammer. However, yet another interesting proof presents itself, of which the outline is presented below:

Define X to be a shape that maximizes the area for a given P.

1. X is convex.

Proof: If X is not convex, then let C be a point outside of X, and A and B be points on the perimeter of X such that C lies on AB. Moreover, let there be no points on the outline of X, say Q, such that Q lies on line segment AB. In that case, if we reflect the outline AB about the line AB, the area increases. This contradicts the maximality of area X. Q.E.D..

Attack: What if we cannot pick A and B such that there are no points of the outline of X, Q, such that Q lies between A and B? (a.k.a. fractals)

2. Any polygon with a fixed number of sides N has maximum area when it is regular. (more or less indisputable).

Corollary: A regular polygon with i sides has greater area than one with j sides given a fixed perimeter iff i>j. (Think of degenerate polygons)

3. Therefore, as the number of sides N goes to infinity, the area increases, and we get a circle, and therefore a circle maximizes area for a fixed perimeter P.

Attack: Where did that come from? Can we even apply rule number 2 to "polygons with infinite sides" (a.k.a. shapes with curves)?

Is there a reasonable patch to the leap from part 2 to part 3? I would agree that it is intuitively very probably true that the argument *sort of* works, and just need formalisation, but with infinities I would not be very sure, ever since the last time I saw 1=1+0+0+...=0+1+0+...=0+0+1+...= ... =0+0+0+...=0. Therefore 1=0. Instead I would probably work towards a more geometry-based idea, described handwaving-ly all the way through, since I have not worked out a full "proof" yet:

Lemma 1 remains, despite the attack. Fractals can be dealt with separately, ... some other time... some other person.

2. The outline of X is smooth. (i.e. no line segments anywhere) random rantings.

Imagine the outline as a wire. Now divide the wire into 2 parts of equal length, and call the parts, with direction, AB and BA respectively (i.e. starting with A and going say clockwise to B and starting with B and going clockwise to A respectively). Define the area enclosed by arc AB and line segment BA be called ABA and BAB be defined similarly. if ABA > BAB, then copy ABA over with symmetry on like AB. X clearly does not have this property since otherwise the operation would produce a shape with the same perimeter and larger area. Hence for X, ABA=BAB for every choice of A and B. Now notice that the reflection does not change the area for X, even though it might change the shape. This does not bring us to an immediate conclusion, since X might not be a unique shape. Also, reflecting about AB must result in another convex shape. This fixes X as a circle... somehow? Good night.