Based in this post G-N-S $\Rightarrow$ Isoperimetric Inequality in Euclidean Space. With the Isoperimetric Inequality in $\mathbb{R}^{n}$. How can answer the following question: Between all open bounded subsets of $\mathbb{R}^{n}$ with fixed volume ,which one has the smallest surface area? I know which the answer is the sphere, but i don't know how prove it. Any tips?

With the tips of Giuseppe and Guy , i imagine which the answer of my answer is prove that the inequality in the tagged post with the constant



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    $\begingroup$ I like the notes of Brian Weber on this topic: math.upenn.edu/~brweber/Courses/2015/760/Math760.html (Supplementary Notes). The connection between Sobolev's inequalities and isoperimetric inequalities is the very first topic. $\endgroup$ – Giuseppe Negro Dec 12 '17 at 15:31
  • $\begingroup$ it seems the solution below it not clear to you $\endgroup$ – Guy Fsone Dec 14 '17 at 7:58

From this book by Cedric Villani Old and New page 571 you have that $$ \frac{|\partial B^n|}{|B^n|^{\frac{n-1}{n}}}\le \frac{|\partial A|}{|A|^{\frac{n-1}{n}}} $$

Now what happens if $|A|= |B^n(r)|?$ clearly it follows that $$\color{red}{|\partial B^n(r)|\le |\partial A| ~~~\text{which prove your statement}}$$

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  • $\begingroup$ I imagine that this inequality in the book of Villani is the same inequality in the tagged post, yes? In this the constant $C=\frac{\mathcal{L}^{n-1}(\mathbb{B}^{n}) }{\mathcal{L}^{n}(\mathbb{B}^{n})}$? $\endgroup$ – C. Junior Dec 12 '17 at 23:04
  • $\begingroup$ @C.Junior yes it is the same inequality $\endgroup$ – Guy Fsone Dec 13 '17 at 7:44

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