I want to know if the following fact has a standard name and/or reference



> 
> 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 $$area\\,X\_\epsilon\ge area\\,B\_\epsilon.$$
> 
> 
> 



Name for an inequality of isoperimetric type
This inequality is essentially equivalent to the [Classical isoperimetric inequality](http://en.wikipedia.org/wiki/Isoperimetric_inequality). 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 [Brunn Minkowski inequality](http://en.wikipedia.org/wiki/Brunn%E2%80%93Minkowski_theorem) because
$$\operatorname{Vol}(X\_{\epsilon})=\left(\operatorname{Vol}(X+\epsilon B)^{1/n}\right)^n \geq \left(\operatorname{Vol}(X)^{1/n}+\epsilon \operatorname{Vol}(B)^{1/n}\right)^n=\operatorname{Vol}(B\_{\epsilon})$$


