Let $\Omega = \{ (x,y): 0<x<a, 0<y<b \}$. If there exists a constant $\lambda > 0$ s.t. $$\int_\Omega u^2 dxdy \le \lambda \int_\Omega |\partial u|^2 dxdy$$ for all $u \in H^1_0(\Omega).$ Then we must have $\lambda \ge \frac{a^2 b^2}{(a^2+b^2)\pi}$.
I want to set $\lambda=\sup \lbrace \frac{\int_\Omega u^2 dxdy}{\int_\Omega |\partial u|^2 dxdy}:u \in H^1_0(\Omega) \rbrace$. How can I make use of maximum principle? Is it possible to find a $u$ directly s.t. $\frac{\int_\Omega u^2 dxdy}{\int_\Omega |\partial u|^2 dxdy}=\frac{a^2 b^2}{(a^2+b^2)\pi}$?
To estimate the supremum from below, an example suffices. $$u(x, y) = \sin(\pi x/a) \sin(\pi y/b)$$ is such an example. Here $$ \int_\Omega |\nabla u(x, y)|^2 = \left(\frac{\pi^2}{a^2}+\frac{\pi^2}{b^2}\right) \frac{ab}{4} $$ while $$ \int_\Omega | u(x, y)|^2 = \frac{ab}{4} $$ (both follow from the fact that the average value of sine squared or cosine squared is $1/2$.)