linear algebra - Sylvester rank inequality: $operatorname{rank} A + operatorname{rank}B leq operatorname{rank} AB + n$

If $A$ and $B$ are two matrices of the same order $n$, then
$$ \operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n. $$

I don't know how to start proving this inequality. I would be very pleased if someone helps me. Thanks!

Edit I. Rank of $A$ is the same of the equivalent matrix $A' =\begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix}$. Analogously for $B$, ranks of $A$ and $B$ are $r,s\leq n$. Hence, since $\operatorname{rank}AB = \min\{r,s\}$, then $r+s\leq \min\{r,s\} + n$. (This is not correct since $\operatorname{rank} AB \leq \min\{r,s\}$.

Edit II. A discussion on the rank of a product of $H_f(A)$ and $H_c(B)$ would correct this, but I don't know how to formalize that $\operatorname{rank}H_f(A) +\operatorname{rank}H_c(B) - n \leq \operatorname{rank}[H_f(A)H_c(B)]$.