# An isoperimetric inequality for curve in the plane?

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.

• If $C$ is regular enough (e.g. $C^1$) doesn't this this follow from the diveregence theorem applied to $(x,y)$ and the Cauchy-Schwarz inequality? – RBega2 Mar 5 at 23:48
• Good idea there. – T. Amdeberhan Mar 6 at 6:06
• Looking infinitesimally, $\frac12r\Delta s$ is not less than the area of a triangle with vertex at origin and side $\Delta s$. Such triangles cover $R$, thus the inequality. Equality takes place only if the radius-vector is always orthogonal to the tangent line, that means that the derivative of $r$ is zero. – Fedor Petrov Mar 6 at 13:16
• @FedorPetrov: I almost agree, but does this not matter whether the curve is convex or concave? In other words, is the inequality local or an average? – T. Amdeberhan Mar 6 at 17:25
• It generalizes to $\int_{\partial U} r \ge n \operatorname{Vol} U$, for $U \subset \mathbb{R}^n$ a bounded domain with $C^1$ boundary, with the origin in its interior, by the same proof as Piotr Hajlasz's. Maybe even with the origin on the boundary. – Ben McKay Mar 7 at 13:41

Let $$(x(t),y(t))$$, $$t\in [0,1]$$ be a parametrization of $$C$$. From Green's Theorem, $$\int_C-y\,dx+x\,dy=\iint_R2\,dxdy=2\cdot Area(R).$$ From Cauchy-Schwartz 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.