Here is the question. Given any $\alpha>0$ and $u\in H^1(\Omega)$, $\Omega=B(0,1)$ in $n$ dimensions. Then we have $$\int_\Omega |u|^2 dx\leq C(\alpha)\int_\Omega |\nabla u|^2dx $$ provided that the set $M:=\{x\in \Omega, \,\, u(x)=0\}$ has measure more then $\alpha$, i.e., $|M|>\alpha$.
I have a solution but my solution requires that $\alpha$ large enough, i.e., I need $\alpha\geq c>0$ for some constant $c$ and I don't know how to remove this additional assumption...
Anyhow, here is my approach: Let $$u_\Omega:= \frac{1}{|\Omega|} \int_\Omega u dx$$ i.e., the mean value of $u$ in $\Omega$.
Then I have $$|u_\Omega|^2 = \frac{1}{|\Omega|^2} \left(\int_{\Omega\setminus M} u dx\right)^2\leq \frac{|\Omega\setminus M|}{|\Omega|^2} \left(\int_{\Omega\setminus M} u^2 dx\right)\leq \frac{|\Omega\setminus M|}{|\Omega|^2} \left(\int_{\Omega} u^2 dx\right)$$ The Last inequality is by Holder.
Thus I have $$\int_\Omega |u|^2 dx\leq 2\int_\Omega |u-u_\Omega|^2dx + 2\int_\Omega |u_\Omega|^2dx \leq 2\int_\Omega |\nabla u|^2dx + 2\frac{|\Omega\setminus M|}{|\Omega|} \left(\int_{\Omega} u^2 dx\right)$$ and hence I need to make $$2\frac{|\Omega\setminus M|}{|\Omega|} <1$$ which require $\alpha$ large enough... But question state that this is true for any $\alpha>0$. Where am I missing?
Edit: The below answer is really nice. But here let me point out a more generally Poincare inequality which I learned recently. Actually the Poincare inequality hold for any $E\subset \Omega$ such that $|E|>0$, then $$ \int_\Omega |u-u_E|^2 dx\leq C\int_\Omega |\nabla u|^2dx $$ Hence we could take $E$ to be the $0$ set of $u$.
If $u$ vanishes on a set of measure $\alpha$, then $$ \int_\Omega u^2\le C(\alpha) \int_\Omega (u-u_\Omega)^2 $$ because $u_\Omega$ cannot be much larger than $\int u^2$. Indeed, $$ \int_\Omega u^2\ge \int_\Omega (u-u_\Omega)^2 \ge \int_{\{u=0\}} (u-u_\Omega)^2 =\alpha (u_\Omega)^2 $$ hence $$ \int_\Omega u^2 \le 2 (u_\Omega)^2+ 2 \int_\Omega (u-u_\Omega)^2 \le ( 2\alpha^{-1}+2)\int_\Omega (u-u_\Omega)^2 $$