For $n \in \mathbb{N}$, denote by $\mathbb{P}^n$ complex projective space, just as a set.
Suppose we are given a map $\mathbb{P}^{r_1} \times \mathbb{P}^{r_2} \rightarrow \mathbb{P}^{r_3}$ that has finitely many fibers for each point in the image $\mathbb{P}^{r_3}$, and suppose moreover that the map is of the following form $$([x_0: ... :x_{r_1}], [y_0 : .. :y_{r_2}]) \mapsto [f_1(x_0,..., x_{r_1},y_0, ..., y_{r_2}): .. : f_{r_3}(x_0, .., x_{r_1}, y_0, .., y_{r_2})] $$ where the $f_i$ are polynomials only having terms of the form $x_iy_j$ (you could say they come from bilinear forms). Can we conclude that $r_1+r_2 \leq r_3$?
There's two things I tried up till now. First I tried to prove something more general: if $f: M \rightarrow N$ is a holomorphic map of complex manifolds with finite fibers, I tried to prove that it's always the case that $\dim(M) \leq \dim(N)$ (maybe not true but something close to it?)? What I did is the following: suppose $\dim(M) > \dim(N)$, then if $f$ has a regular value we have a point in $N$ whose inverse image is a submanifold of dimension $\dim(M)-\dim(N)$ of $N$, i.e. an infinite fiber, contradiction. So we may assume $f$ has no regular values. In that case, by Sard's theorem the image of $f$ has measure $0$ in $N$ and then I could not work it out any further.
The second thing I tried is actually translating the problem into the more "algebraic" form in the first paragraph of this question.
Thanks in advance for any help.
Here is an algebraic-geometry way to do this. Your map is a morphism of schemes. It is quasi-finite by assumption, and if $X\to Y$ is a morphism of schemes with $X$ proper and $Y$ separated, then this morphism is proper (ref). Proper + quasi-finite = finite (ref), so your morphism is finite, and any finite morphism $X\to Y$ necessarily has $\dim X\leq\dim Y$ (ref).
If you want to use the approach from your post ("I tried to prove something more general"), then one option if you have a map of varieties $f:X\to Y$ in characteristic zero is to restrict to the closure of the image of your map and think about $f:X\to \overline{Im(f)}$. Then the image of your map is constructible by Chevalley's theorem, dense in $\overline{Im(f)}$, and thus contains a nontrivial open subset. Restricting to the smooth locus of $\overline{Im(f)}$ and applying generic smoothness, one may finish. Here's a reference for generic smoothness: ref, starting on page 7.