Let $K(x,y)$ be the Poisson kernel for the Dirichlet problem of Laplace equation on a bounded domain $D$ with smooth boundary, i.e. for a harmonic function $u$ on $\bar D$ with $u|_{\partial D}=g$, we have $$u(x)=\int_{\partial D}K(x,y)g(y)\mathrm{d}y:=\langle K(x,\cdot),g\rangle$$
We know that if $x\in D^\circ$ and $y\in \partial D$, then $K(x,y)\geq 0$. The reason is that for arbitrary continuous function $g\geq 0$ on $\partial D$ that does not identically vanish, $u(x)>0$ by maximal principle, so $\langle K(x,\cdot),g\rangle>0$. The above argument actually shows something stronger: $K(x,\cdot)$ cannot vanish on an open subset of $\partial D$.
My question is the following: can we show that $K(x,y)$ must be positive everywhere if $x\in D^\circ$? We know this is true for the half plane and the unit ball, and is it true in general? If not, what condition on the domain $D$ should be imposed?
This fact is essentially a very well-known and important result called Hopf's lemma:
If $\Omega$ is a smooth domain, $u\in C^1(\overline{\Omega})\cap C^2(\Omega)$ is a harmonic function, and $u(x)< u(y)$ for every $x\in \Omega$ and some $y\in \partial \Omega$, then $$ \frac{\partial u}{\partial \nu} (y) >0, $$ where $\nu$ denotes the outer unit normal.
This is obviously a local result, so that the condition $u(x)< u(y)$ for all $x\in \Omega$ could be relaxed to hold only in a neighborhood of $y$. There is a neat proof of this result on Wikipedia (search for Hopf's lemma).
As to why this implies you result, we note that on smooth domains, $K(x,y)=\partial_{\nu_y} G(x,y)$, where $G$ is the Green function for the domain, with the convention that $G\geq 0$. Since $G(x,y)=0$ for $y\in \partial \Omega$ and $x\in \Omega$ by construction, the hypothesis are satisfied and so $K(x,y)>0$.