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 ?
 A: At first sight one tends to believe the inequality because of the Mean Value Theorem,
$$
|u(x)|=|u(x)-u(y)|=|\nabla u(c)\cdot(x-y)|
$$
for all $y$ in the boundary. 
But there is an inner product involved, and so one cannot guarantee that the right-hand-side above is bounded below. 
Indeed, let $\Omega$ be the unit disc in $\mathbb R^2$ and 
$$
u(x)=2y(1-x^2-y^2).
$$
Being a polynomial, $u$ is smooth. It is also zero at the boundary. We have
$$\tag{1}
|\nabla u(x,y)|^2=16x^2y^2+4(1-x^2-3y^2)^2,
$$
which is never zero in the closed disk. As the expression in $(1)$ is continuous on a compact set, it achieves its minimum and so there exists $\alpha>0$ with $|\nabla u(x,y)|>\alpha$ for all $(x,y)\in\overline\Omega$.
Finally, $|u(1/2,0)|=0<\alpha\,|(1/2,0)-(1,0)|$
In hindsight, the assertion becomes suspicious when you notice that the condition on the gradient is only requiring for the gradient to be nonzero.
A: Let $x_0,y_0 \in \partial \Omega$. Let $x \in \Omega$. Then you have $$|u(x)| \ge \alpha |x - y_0|.$$ If $x \to x_0$ then by continuity you obtain $$|u(x_0)| \ge \alpha |x_0 - y_0|.$$ That is,
$$ 0 \ge \alpha |x_0 - y_0|.$$
The problem is misstated.
A: 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 $\nabla u(P)=0$. 
