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