There's a little dettail that I can't get in this proof. It all comes from Evans' PDEs book
Here is the statement:
Assume $u \in C^2(U) \cap C^1(\bar{U})$ (where U is open and bounded in $\mathbb{R}^n$)
Let L be an elliptic operator with $c \equiv$ in U (c(x) is L's coefficent of 0-degree term)
Suppose that $Lu \leq 0$ and there exists a point $x_0 \in \partial U$ s.t. $u(x) < u(x_0) \space \forall x \in U$
Assume finally that U statisfies the interior ball condition at $x_0$ (that is, there exists an open ball $B \subset U$ with $x_0 \in \partial B)$
Then $\frac{ \partial u}{ \partial \nu}(x_0) > 0$
Proof sketch:
Since $x_0$ satisfies interior ball condition, let $B_r$ be the ball. We want to apply weak-maximum principle in $\Omega := B_r \setminus B_{\frac{r}{2}}$ to the function $$ u + \epsilon v_{\lambda} $$ where $$v_{\lambda} := e^{-\lambda |x|^2} - e^{-\lambda r^2}$$
so that $u(x) + \epsilon v_{\lambda}(x) < \max\limits_{\partial \Omega} (u + \epsilon v_{\lambda}) = u(x_0) \forall x \in \Omega$
My doubt is on showing $\max\limits_{\partial \Omega} = u(x_0)$ (1): since $$ v_{\lambda} \equiv 0 \text{in} \partial B_r$$
$$u(x) < u(x_0) \forall x \in \partial B_{\frac{r}{2}} \subset U$$
$$\lambda >> 0 \implies v_{\lambda}(x) = k_{\lambda} \in \mathbb{R} \text{,} k_{\lambda} > 0 \forall x \in \partial B_{\frac{r}{2}}$$
then by choosing $\epsilon$ small enough, we have that $\max\limits_{\partial B_{\frac{r}{2}}} u(x) + \epsilon v_{\lambda} < u(x_0)$.
It is written that's enough for the thesis (1), but where it comes from that $u(x_0) = \max\limits_{\partial B_r} u(x)$ ? I just can't see: couldn't $x_0$ be a generic point of the border that satisfie statement's conditions, but isn't the maximum value on the ball's border?
I tried to look at other questions here, but I found nothing. Hope I didn't miss something.
Since $B_r$ is an interior ball, the answer to your question follows from the hypothesis that $u(x_0)>u(z)$ for $z \in U$. For $x \in \partial B_r$, either $x \in U$, or $x \in \partial U$. The former case is clear. In the latter case, there exists $z_n$ in $U$ tending to $x$, so continuity of $u$ in $\bar U$ and the hypothesis imply that $u(x_0) \ge u(x)$.