TITLE: Frobenius Inequality Rank
QUESTION [15 upvotes]: I was looking for an answer for this problem in terms of matrices, but I really don't know how to prove this result. The proposition says that:
Let $A\in M_{m\times k}(\mathbb{C})$, $B\in M_{k\times p}(\mathbb{C})$ and $C\in M_{p\times n}(\mathbb{C})$, then $\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}(B)+\textrm{rank}(ABC)$

REPLY [6 votes]: If you know the Sylvester inequality then it's a two lines proof.
 Sylvester Inequality:

Consider $A_{m\times r}$ and $B_{r\times n}$. Then 
  $$r(AB)\geq r(A)+r(B)-r \tag{1}$$
  where $r(\cdot)$ is the rank.

Let  $B=U_{k\times r} V_{r\times p}$ be a full-rank factorization of $B$. 
Then by $(1)$, 
\begin{align}
r(ABC)&\geq r(AU)+r(VC)-r\\
     &=r(AB)+r(BC)-r(B).
\end{align}

For a proof of Sylvester inequality refer the following: