Let $u:\Omega\subset\mathbb{R}^2\to\mathbb{R}$ such that $u$ satisfies the differential equation of minimal surfaces on the region $\Omega$. Let $p\in(\Omega\times\{0\})\subset\mathbb{R}^3$ and a plane, closed disk $D_r$ contained in $\Omega$ with center $p$ and radius $r>0$. Show that:
$$\text{Area}(B_r\cap\text{graph}(u))\leq 2\pi r^2$$
where $B_r\subset\mathbb{R}^3$ is a closed ball with radius $r$ and whose center is any point along the vertical line $\{p+(0,0,t)\mid t\in\mathbb{R}\}$.
As far as I've seen, problems envolving minimal surfaces usually have a nice, elegant geometric intuition, despite the technical details in a rigorous proof. But for this problem I have no clue what the intuition is supposed to be.
I've tried to actually calculate the area of the intersection, but it looked very complicated. I also don't know how to apply the maximim principle here, or any of the usual tricks I've learned (using catenoids, cones, convex hull property etc), so I'm stuck.
Any hints? Thanks!