Frobenius Inequality Rank 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 $\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}(B)+\textrm{rank}(ABC)$
 A: 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:

A: Notation: 


*

*$\rho(\cdot)$ stands for rank, 

*$\ker(\cdot)$ for null space (aka kernel),

*$\text{im}(\cdot)$ for column space (aka image).


All we need is the following well known identity (see this answer for a proof): 
$$\rho(AB)=\rho(B)−\dim(\text{im}(B) \cap \ker(A))
\tag{1}$$
and the following observation: $$\text{im}(BC) \cap \ker(A) \subseteq \text{im}(B)\cap \ker(A)\tag{2}$$ which holds since $\text{im}(BC)\subseteq \text{im}(B)$.
Now we want to write $\rho(ABC)$ in such a way that $\text{im}(BC)\cap \ker(A)$ pops up, so we could make use of $(2)$. Analogously to $(1)$:
$$\rho(ABC)=\rho(BC)−\dim(\text{im}(BC) \cap \ker(A))
\tag{3}.$$
From $(1)$ and $(3)$:
$$\rho(AB)+ \rho(BC) = \rho(B) + \rho(ABC)  + \underbrace{\dim(\text{im}(BC) \cap \ker(A))−\dim(\text{im}(B) \cap \ker(A))}_{\leq 0 \text{ due to } (2)}$$ which implies the desired inequality.
