Consider the classical Boundary Point Lemma:
Let $L$ be an elliptic operator.
Boundary Point Lemma Suppose $\Omega$ has the interior sphere property and that $u\in C^2(\Omega)\cap C^1(\bar\Omega)$ satisfies $Lu \geq 0$ in $\Omega$. Assume $u$ attains its non negative maximum at $x_0 \in \bar \Omega$. Then $x_0 \in \partial \Omega$ and for any outward direction $v$ at $x_0$ to $\partial \Omega$ $$\dfrac{\partial u}{\partial v}(x_0) > 0$$ unless $u$ is constant in $\bar\Omega$.
Now my question is: could I say that the gradient $Du(x_0)$ has the opposite direction of inner normal to $\partial\Omega$ at $x_0$?
I thought this from the fact that $(Du(x_0),v)=\dfrac{\partial u}{\partial v}(x_0)$. Now, if what I am said before is not true, one can find an outward direction such that $0>(Du(x_0),v)=\dfrac{\partial u}{\partial v}(x_0)$, so one have a contraddiction.
Is this true or I am wrong in some steps?
Hopf Lemma tells you that (under the required assumptions) the maximum of the solution $u$ is attained at a point $x_0$ on the boundary $\partial\Omega$ but also that the derivative of the función $u$ in the outward direction near $x_0$ has to be positive, i.e.
$$ \frac{\partial u}{\partial \mathbf{n}}(x_0):= \langle\nabla u(x_0),\mathbf {n}\rangle > 0. $$
The intuition one gets from here is that as you approach $x_0\in\partial\Omega$ from the interior of $\Omega$ the derivative of $u$ has to be positive since otherwise it is imposible for the maximum of $u$ to be in $\partial\Omega$. Or either $\nabla u(x_0)=0$ and then $u$ is constant in all the domain (this is one case in the Maximum Principle), or either $u$ has its maximum at $x_0\in\partial\Omega$ and hence if you move from $x_0$ towards the interior of $\Omega$ (i.e. direction $-\mathbf{n}$), $u$ has to decrease and thus $\langle\nabla u(x_0),-\mathbf{n}\rangle < 0$.
So, if I understood well your question, the answer is yes.
Hope it helps!