most recent 30 from http://mathoverflow.net 2013-06-19T03:28:16Z http://mathoverflow.net/feeds/question/22131 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22131/name-for-an-inequality-of-isoperimetric-type AN 2010-04-22T00:37:10Z 2010-04-27T03:07:17Z

<p>I want to know if the following fact has a standard name and/or reference</p> <blockquote> <p>Let $X$ be a subset of $\mathbb R^2$ and $B$ be a disc of the same area as $X$. Set $X_\epsilon$ to be the $\epsilon$-neigborhood of $X$. Then $$area\,X_\epsilon\ge area\,B_\epsilon.$$</p> </blockquote>

http://mathoverflow.net/questions/22131/name-for-an-inequality-of-isoperimetric-type/22671#22671 Gjergji Zaimi 2010-04-27T03:07:17Z 2010-04-27T03:07:17Z

<p>This inequality is essentially equivalent to the <a href="http://en.wikipedia.org/wiki/Isoperimetric_inequality" rel="nofollow">Classical isoperimetric inequality</a>. If you have a measurable body $X$ in $\mathbb{R}^n$ and a ball $B\subset \mathbb R^n$ of same volume then you have the following: $$Area(X)=\lim_{\epsilon \to 0} \frac{\operatorname{Vol}(X_{\epsilon})-\operatorname{Vol}(X)}{\epsilon}$$ $$Area(B)=\lim_{\epsilon \to 0} \frac{\operatorname{Vol}(B_{\epsilon})-\operatorname{Vol}(B)}{\epsilon}$$ Proving that $Area(X)\geq Area(B)$ follows from $\operatorname{Vol}(X_{\epsilon})\geq \operatorname{Vol}(B_{\epsilon})$, which is your inequality. ($n=2$) Now this follows from the <a href="http://en.wikipedia.org/wiki/Brunn%E2%80%93Minkowski_theorem" rel="nofollow">Brunn Minkowski inequality</a> because $$\operatorname{Vol}(X_{\epsilon})=\left(\operatorname{Vol}(X+\epsilon B)^{1/n}\right)^n \geq \left(\operatorname{Vol}(X)^{1/n}+\epsilon \operatorname{Vol}(B)^{1/n}\right)^n=\operatorname{Vol}(B_{\epsilon})$$</p>