Isoperimetric inequality

Let $\gamma$ be a simple closed curved of length $\ell$ enclosing an area $A$. Prove that:

$$4\pi A \leq \ell^2$$

Solution:



We parametrize the curve $\gamma$ with a constant speed $\ell / 2\pi$. We may assume that $\displaystyle \int_{0}^{2\pi} x(t)\, {\rm d} t=0$. Hence:

$$\frac{\ell^2}{2\pi}= \int_{0}^{2\pi}\left \| \gamma(t) \right \|^2 \, {\rm d}t= \int_{0}^{2\pi}\left [ x'(t)^2 + y'(t)^2  \right ]\, {\rm d}t$$

To evaluate the area $A$ we are using Green's theorem. Therefore:

$$\iint \limits_{{\rm int}\gamma}{\rm d}x \, {\rm d}y =\int_{0}^{2\pi} x (t) y'(t)\, {\rm d}t$$

This however means:

$$\begin{aligned}
\ell^2 - 4\pi A &=2\pi \int_{0}^{2\pi}\left [ x'(t)^2 + y'(t)^2 -2x(t)y'(t) \right ]\, {\rm d}t \\
 &=2\pi \int_{0}^{2\pi}\left [ x'(t)^2 -x(t)^2 \right ]\, {\rm d}t +2\pi \int_{0}^{2\pi}\left [ y'(t)- x(t) \right ]^2 \, {\rm d}t\geq 0 
\end{aligned}$$

which completes the proof.