Isoperimetric inequality

Remarks

The inequality follows from the Brunn–Minkowski theorem. Let $B$ be the closed unit ball and $\bar\Omega$ the closure of $\Omega$. Then $\abs{\bar\Omega + \varepsilon B} \ge (\abs{\Omega}^{1/n} + \varepsilon \omega_n^{1/n})^n$. For a $C^1$ boundary, $\abs{\partial \Omega}$ is the limit of $(\abs{\bar\Omega + \varepsilon B} - \abs{\Omega}) / \varepsilon$ as $\varepsilon \to 0$, and this limit is at least $n \omega_n^{1/n} \abs{\Omega}^{(n-1)/n}$. Raising to the power $n$ gives the inequality; the equality case needs a separate argument. For $n = 2$ it reads $L^2 \ge 4 \pi A$ for a region of area $A$ bounded by curves of total length $L$. Example: a square of side $1$ has $L = 4$ and $A = 1$, and indeed $16 \ge 4\pi$.

Going deeper

For a student

The inequality has a version for sets of finite perimeter, which is the natural setting in geometric measure theory. It is equivalent to the sharp Sobolev inequality $\norm{u}_{n/(n-1)} \le C_n \norm{\nabla u}_1$ for smooth compactly supported $u$, through the coarea formula. In the plane it is often proved with Fourier series or by Steiner symmetrisation.