I have been asked to prove that a Voronoi region $V=\{x \in \mathbb{R}^n : \|x-x_0\| \le \|x-x_i\|, i=1,\dots,k\}$ around $x_0$ with respect to $x_1,\dots,x_k$ is a polyhedron.
My idea is to find the set of hyperplanes $h_i : a_i^Tx=b_i$ defined by the pairs $(x_0,x_i)$, such as they divide the entire space in halfspaces that fulfill $a_i^Tx \le b_i \Leftrightarrow \|x-x_0\| \le \|x-x_i\|$.
Intuitively, I can see that the hyperplanes I am looking for are symmetric with respect to each pair of points, so $a_i = x_i-x_0$ and $b_i = (x_i-x_0)^T (\frac{x_i+x_0}{2})$ could be a valid solution. As always, I am able to imagine it in a visual way, but I think I have no idea of how could I try to prove it in a more formal way, that is, to derive the valid $a_i$ and $b_i$ values without drawing.
Added: Following the Boyd and Vanderberghe book on convex optimization, the definition of polyhedron I am using is that of a intersection of a finite number of halfspaces and hyperplanes.
Could anybody please give me a hint on whether this is possible or not? Could a graphical proof like this just do its job?
Thanks in advance.