I am reading their book A course in minimal surfaces.
Suppose $D$ is the unit disc in $\mathbb{R}^2$. Lemma 6.26 says,
Lemma 6.26 Let $\Omega$ be a bounded mean convex region with smooth boundary, and suppose that $u:\bar D=\to \bar \Omega\subset \mathbb{R}^3$ is continuous, almost conformal, and harmonic. If $u(\partial D)\subset \partial \Omega$ and $u(D)\cap \Omega\neq \emptyset$, then $u(D)\subset\Omega$ and $|\nabla u|$ does not vanish on $\partial D$.
The idea to prove this is the following: Set $f(\rho)=2a\rho-\rho^2$ for some fixed constant $a$ small and $v(z)=f\circ \rho(u(z))$, $z\in D$, where $\rho=dist(x,\partial \Omega)$. Some calculation shows $$\Delta v(z)\leq 0,\quad \text{when }\rho(u(z))\text{ is small}$$ In other words, $v$ is superharmonic if $u(z)$ is in some tubular neighborhood of $\partial \Omega$. Next, the authors point out $v$ is nonnegative, vanishes on the boundary and $u(D)\cap \Omega\neq \emptyset$. Consquently $v$ is strictly positive on $D$ by maximum principle.
My question is how do we see that $v$ is nonnegative over $D$. If $a$ is very small and $u(z)$ is far from the boundary, wouldn't $v$ be negative?