How to show $\lambda_1(\Omega)\ge \lambda_1(B)$ when $|\Omega|=|B|$ $B$ is the unit ball in $\mathbb R^n$ , and $\Omega\subset \mathbb R^n$ is a domain which has same volume with $B$. $\lambda_1$ is the first eigenvalue of Laplace with Dirchlet bound condition. Then, how to show 
$$
\lambda_1(\Omega)\ge \lambda_1(B)
$$
Maybe, this is an open question. If so, any relative reference is welcome.Thanks.
 A: I'll give a slightly more general answer than what you ask for by allowing $\Omega$ to have arbitrary finite measure.
First note that we have the Rayleigh quotient characterization of $\lambda_1$ on any bounded open set $\Omega \subset \mathbb{R}^n$:
$$
\lambda_1(\Omega) = \min \{ \int_\Omega |\nabla u|^2 : u \in H^1_0(\Omega) \text{ and } \int_\Omega |u|^2 =1 \}.
$$
Let $u_\Omega$ denote a minimizer, i.e. the principal eigenfunction.
Since $u_\Omega \in H^1_0(\Omega)$ we can extend it by zero outside $\Omega$ to get $\bar{u}_\Omega \in H^1(\mathbb{R}^n)$ such that 
$$
\lambda_1(\Omega) = \int_\Omega |\nabla u_\Omega|^2 = \int_{\mathbb{R}^n} |\nabla \bar{u}_\Omega|^2.
$$
Now we invoke the Polya-Szego inequality, which tells us that 
$$
\int_{\mathbb{R}^n} |\nabla \bar{u}_\Omega|^2  \ge \int_{\mathbb{R}^n} |\nabla u_\Omega^\ast|^2
$$ 
where $u^\ast_\Omega$ is the symmetric decreasing rearrangement of $\bar{u}_\Omega$.  The definition of $\bar{u}_\Omega$ shows that $u^\ast_\Omega$ is supported in the ball $B$ where the ball's radius is such that $|B| = |\Omega|$.  Moreover, the $u^\ast_\Omega$ and $\bar{u}_\Omega$ are equi-distributed (have the same distribution function), so we have that 
$$
\int_{\mathbb{R}^n} |u^\ast_\Omega|^2 = \int_{\mathbb{R}^n} |\bar{u}_\Omega|^2 = \int_\Omega |u_\Omega|^2 = 1.
$$
Thus 
$$
\lambda_1(\Omega) =  \int_{\mathbb{R}^n} |\nabla \bar{u}_\Omega|^2 \ge \int_{\mathbb{R}^n} |\nabla u_\Omega^\ast|^2 = \int_{B} |\nabla u_\Omega^\ast|^2 \ge \lambda_1(B).
$$
