Interesting question in analysis I am trying to prove this :
Consider $\Omega \subset R^n$ ( $n \geq 2$) a bounded and open set and $u $ a smooth function defined in $\overline{\Omega}$. Suppose that $u(y) = 0$ for $y \in \partial \Omega$ and suppose that exists a $\alpha >0$ such that $|\nabla u (x)| = \sqrt{\displaystyle\sum_{i=1}^{n} (\frac{\partial u}{ \partial x_i }(x)} )^2\geq \alpha >0$ for all $x \in \Omega$, then   
$$ |u(x)| \geq \alpha |x-y|$$
for all $x \in \Omega$ and for all $y \in \partial \Omega$.
drawing a picture is easy to see the affirmation..
 i am trying to prove this. but nothing ... My professor said that this is true....
Someone can give me a hint ?

I think there is something wrong here. 
If $u$ is zero in $\partial \Omega$, then $u$ has a minimum or a maximum in $\Omega$. So there is a point $P$ in $\Omega$ such that $\nabla u(P)=0$. 
