I am wondering whether there is a ‘simple’ proof of the Isoperimetric Theorem in the plane, i.e. that any simple closed curve in $\mathbb{R}^2$ with length $L$ and enclosed area $A$ fulfils $$4\pi A \leq L^2\ .$$
I should clarify what I mean by ‘simple’: I am looking for something that does not use very advanced mathematics (i.e. something like undergraduate-level mathematics, perhaps), but is also relatively short and understandable.
In particular, I was wondering whether it can be obtained easily from some other standard results, e.g. from Complex Analysis (Cauchy Integral Theorem, Riemann Mapping Theorem), but, if possible, avoiding things like Green's Theorem and Fourier series.
If you're wondering about these oddly specific requirements, I'm thinking about how to formalise this in a theorem prover.
In Dido's problem Lagrange multiplier $\lambda$ that connects maximum Area $A$ for given boundary length $L$ is recognized as a curve of constant radius as the parameter $\lambda =r$ for the Circle. The proof is given in all books of Variational calculus.
Consequently when parameter $r$ is eliminated between area and circumference $$ A = \pi r^2,\, L= 2 \pi r ,\,$$ we obtain $$ 4 \pi A = L^2 $$ as a relationship for the maximum condition.. the inequality $$ 4 \pi A < L^2 $$ for all possible variations falling out from the full circle optimal situation.
Hope it serves as a simple enough proof.