## Proof of the Isoperimetric Inequality 3

I will present here a third proof for the planar Isoperimetric Inequality, using some simple notions of differential curves. For this suppose that the simple closed plane curve has length and encloses area . Then

and equality holds if and only if is a circle.

Choose parallel lines and tangent to and enclosing , as seen in the picture. We draw a circle of radius tangent to and put the origin at its center, with the -axis parallel to . We parametrize by arclength taking and . We consider defined by

Note that the image of is the circle , but does not need to be a parametrization of since it may cover some parts multiple times. Also note that .

We use Green’s formula to obtain that the areas of and , respectively are

We add these equations and apply the Cauchy-Schwarz inequality:

because is always on the circle of radius centered at the origin, and since is parametrized by arclength.

Therefore by applying the AM-GM inequality we have

which is equivalent to

Suppose now that the equality holds. Then we have and , so does not depend on the orientation of the parallel lines . Furthermore, the equality holds in the Cauchy-Schwarz inequality and we have (because one vector has length and the other one has length ). This implies .

Now consider the case when the direction of is changed by an angle equal to . Then and change places in the parametrization and in the new coordinates. We could pick the position of the circle such that is tangent to all four lines and their rotates with . Therefore the change of coordinates is just swapping and . Reasoning the same way we get , and therefore and is a circle.