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The conjecture as you state it is false. The variational argument (that can be interpreted in terms of fluid pressure if you like) shows that the surface has constant mean curvature and is orthogonal …

I like a recent proof by Akopyan and Tarasov: A. V. Akopyan, A. S. Tarasov, "A constructive proof of Kirszbraun's theorem"(Russian), Mat. Zametki 84 (2008), no. 5, 781--784; translation in Math. Note …

For any sensible definition of perimeter, an adaptation of following argument proves the claim. Consider the nearest-point projection map $p_A:\mathbb R^n\to A$ (that is, for every $x\in\mathbb R^n$, …

Yes. Here is an elementary proof. Construct a distance non-increasing retraction $p:S^{n-1}\setminus B_r(a)\setminus B_r(b)\to \partial B_r(a)$. (For example: divide the sphere into two-hemispheres b …

The conjecture is false. Let $ABCD$ be a square inscribed in the unit circle. Consider the following piecewise smooth loop in the plane. First it starts from $A$ and moves between $A$ and $B$ along t …

Your inequality is not true in general. You can have $\int_{\partial\Omega} p^2 dS >\int_{\partial\Omega^*} (p^*)^2 dS^* + C(A^*-A)$ for any $C$ while keeping $L$ and $A$ a priori bounded. Indeed, $$ …