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Mar 7, 2019 at 13:41 comment added Ben McKay 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.
Mar 6, 2019 at 19:03 comment added Fedor Petrov If R is convex and origin is inside (or even if it is star-shaped with respect to the origin), the triangles cover it without overlaps. Otherwise overlaps are possible. But definitely any point p in R is covered by the segment 0s, where s is the point in which the continuation of the ray 0p meets the boundary of R.
Mar 6, 2019 at 17:25 comment added T. Amdeberhan @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?
Mar 6, 2019 at 13:16 comment added Fedor Petrov 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.
Mar 6, 2019 at 6:06 comment added T. Amdeberhan Good idea there.
Mar 6, 2019 at 1:18 vote accept T. Amdeberhan
Mar 6, 2019 at 1:06 history edited Piotr Hajlasz CC BY-SA 4.0
I added: centered at the origin.
Mar 6, 2019 at 0:21 answer added Piotr Hajlasz timeline score: 7
Mar 5, 2019 at 23:48 comment added RBega2 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?
Mar 5, 2019 at 23:29 history edited YCor CC BY-SA 4.0
added more context in title
Mar 5, 2019 at 23:24 history asked T. Amdeberhan CC BY-SA 4.0