Let $f(x,y)=0$ be a (smooth) simple closed curve $C$ on the plane and $R$ the region bounded by $C$ (appropriately oriented). Assume the origin lies in the interior of $R$.



> 
> **QUESTION.** Let $r=\sqrt{x^2+y^2}$. Is this true?
>  $$\int\_Cr\,ds\geq 2\cdot Area(R).$$
>  Equality iff $C$ is a circle centered at the origin.
> 
> 
> 



An isoperimetric inequality for curve in the plane?
Expanding on the comment of RBega2:


Let $(x(t),y(t))$, $t\in [0,1]$ be a parametrization of $C$. From [Green's Theorem](http://mathworld.wolfram.com/GreensTheorem.html), 
$$\int\_C-y\,dx+x\,dy=\iint\_R2\,dxdy=2\cdot Area(R).$$
From [Cauchy-Schwartz](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality) inequality,
$$\vert\langle x,y\rangle\cdot\langle -\dot y,\dot x\rangle\vert
\leq \Vert\langle x,y\rangle\Vert\,\,\Vert\langle -\dot y,\dot x\rangle\Vert
=(x^2+y^2)^{1/2}(\dot x^2+\dot y^2)^{1/2}.$$
Therefore, we have
\begin{align\*}
2\, Area(R)=
\int\_Cx\, dy-y\, dx&=\int\_0^1x(t)\dot y(t)-y(t)\dot x(t)\, dt \\
&\leq
\int\_0^1(x^2+y^2)^{1/2}(\dot x^2+\dot y^2)^{1/2}\, dt \\
&=\int\_Cr\, ds.
\end{align\*}
Equality holds if and only if $\langle x,y\rangle$ is parallel to $\langle\dot y,-\dot x\rangle$, that is if $\langle x,y\rangle$ is orthogonal to the velocity vector $\langle\dot x,\dot y\rangle$ which is exactly when $C$ is a circle centered at the origin.


