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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.

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    $\begingroup$ 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? $\endgroup$
    – RBega2
    Commented Mar 5, 2019 at 23:48
  • $\begingroup$ Good idea there. $\endgroup$
    – T. Amdeberhan
    Commented Mar 6, 2019 at 6:06
  • $\begingroup$ 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. $\endgroup$
    – Fedor Petrov
    Commented Mar 6, 2019 at 13:16
  • $\begingroup$ @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? $\endgroup$
    – T. Amdeberhan
    Commented Mar 6, 2019 at 17:25
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    $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Commented Mar 7, 2019 at 13:41

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Expanding on the comment of RBega2:

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.

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