Let us consider the simplest isoperimetric inequality. Consider a smooth simple closed curve given by $r=\rho(\theta)$ in polar coordinates, where $\rho(\theta)>0$ can be regarded as a smooth periodic function with period $2\pi$.

As $dA=rdrd\theta$, the area of the region $\Omega$ enclosed by the closed curve is \begin{align*} A=\iint_{\Omega}rdrd\theta =\frac{1}{2}\int_0^{2\pi}\rho(\theta)^2 d\theta. \end{align*} On the other hand, it is an easy exercise in calculus that the length of the curve is given by \begin{align*} L=&\int_{0}^{2\pi}\sqrt{\rho(\theta)^2+\rho'(\theta)^2}d\theta. \end{align*}

By the isoperimetric inequality, we have $L^2\ge 4\pi A$, so we must have \begin{equation} 2\pi \int_{0}^{2\pi} \rho(\theta)^2 d\theta \le \left(\int_{0}^{2\pi}\sqrt{\rho(\theta)^2+\rho'(\theta)^2}d\theta\right)^2. \end{equation}

This looks similar to the Wirtinger's inequality, which states that \begin{equation} \begin{split} \int_{0}^{2\pi}\rho'(\theta)^2d\theta \ge \int_{0}^{2\pi} \left(\rho(\theta)-\overline \rho\right)^2 d\theta = \int_{0}^{2\pi} \rho(\theta)^2 d\theta-\frac{1}{2\pi} \left(\int_{0}^{2\pi}\rho(\theta)d\theta\right)^2. \end{split} \end{equation}

So \begin{align*} 2\pi \int_{0}^{2\pi} \rho(\theta)^2 d\theta \le& \left(\int_{0}^{2\pi}\rho(\theta)d\theta\right)^2+2\pi\int_{0}^{2\pi}\rho'(\theta)^2d\theta, \end{align*} which isn't quite what I want.

It doesn't seem right to me that we can apply Wirtinger's inequality directly to prove it because the equality case in Wirtinger's inequality holds when $\rho(\theta)=C+a\cos \theta+ b\sin \theta$, but ~~the equality in our inequality can only hold when $\rho$ is constant~~ the equality in our inequality doesn't hold in this case. (However, by geometric consideration, the equality does hold for, say, $\rho(\theta)=\sin \theta$, but only if we restrict $\theta$ to $[0, \pi]$.)

This is where I am stuck. So my question is: Can we show this inequality without using the isoperimetric inequality (say by Fourier analysis or using Wirtinger's inequality more carefully)? Can it be used to prove (at least a simple case of) the isoperimetric inequality? If not, why? (If nothing else, at least I obtain an inequality on circle :-) )

To elaborate further, it is well-known that we can apply Wirtinger's inequality (or Fourier type argument) to prove the isoperimetric inequality on the plane. Indeed, the Wirtinger's inequality and the isoperimetric inequality are equivalent (e.g. Osserman's paper on isoperimetric inequality). Usually, these kinds of proofs involve shifting the center of mass to $0$, applying the Green's theorem and the Wirtinger's (or Cauchy-Schwarz) inequality on some combination of two functions (say $x(s), y(s)$). So as a subquestion, why is there no such argument involving only a single function (say $\rho(\theta)$)?

but the equality in our inequality can only hold when $\rho$ is constantNot really: it holds for circles containing the origin but not all such circles arecenteredat the origin. You have actually noticed it yourself in the next sentence. $\endgroup$ – fedja Jul 22 '17 at 3:17