An inequality on the rank of a block matrix Let $\mathbb F$ be a field, and let $r_1, r_2, s_1, s_2$ be positive integers. Consider the matrix
$$X:=\left[\begin{array}{cc}
 A & B \\
 C & D  \end{array}
 \right],$$
where $A \in \mathbb F^{r_1 \times s_1}$, $B\in \mathbb F^{r_1\times s_2}$, $C \in \mathbb F^{r_2\times s_1}$ and $D \in \mathbb F^{r_2\times s_2}$.
Q1: Is the following inequality true? 
  $$rank [X] \geq rank \left[\begin{array}{c}
 B \\
 D  \end{array}
 \right]+\max \left\{rank[A\;\; B]- rank[B], rank[C\;\;D]-rank[D] \right\}$$
 If so, when does the equality hold?
Q2: Is there some nice formula for $rank[X]$ depending only on the blocks?
Thanks.
Alessandro

Edit: The answer to the first question is YES. Is there an elementary proof of it? I can prove it using using results on completions of partial matrices, but I believe there should be an easy way to prove it.
For the equality I suspect that it holds if and only if
$$rank \left[\begin{array}{c}
 A \\
 C  \end{array}
 \right]=\max \left\{rank[A],rank[C] \right\},$$
but I'm not sure.
 A: Here is a slicker proof for Question 1:
The Frobenius rank inequality says that if $m,k,p,n$ are four nonnegative
integers, and if $U\in\mathbb{F}^{m\times k}$, $V\in\mathbb{F}^{k\times p}$
and $W\in\mathbb{F}^{p\times n}$ are three matrices, then
$\operatorname*{rank}\left(  UV\right)  +\operatorname*{rank}\left(
VW\right)  \leq\operatorname*{rank}V+\operatorname*{rank}\left(  UVW\right)  $.
(This is proven in the answer to question #497830 in the case when
$\mathbb{F}=\mathbb{C}$ (notice that the matrices $U$, $V$ and $W$ are denoted
by $A$, $B$ and $C$ in said question); the same proof applies for arbitrary
$\mathbb{F}$.)
For any $n\in\mathbb{N}$, we let $I_{n}$ denote the $n\times n$ identity
matrix. For any $n\in\mathbb{N}$ and $m\in\mathbb{N}$, we let $0_{n\times m}$
denote the $n\times m$ zero matrix.
Now, if we set $m=r_{1}$, $k=r_{1}+r_{2}$, $p=s_{1}+s_{2}$, $n=s_{2}$,
$U=\left[
\begin{array}
[c]{cc}
I_{r_{1}} & 0_{r_{1}\times r_{2}}
\end{array}
\right]  $, $V=X$ and $W=\left[
\begin{array}
[c]{c}
0_{s_{1}\times s_{2}}\\
I_{s_{2}}
\end{array}
\right]  $, then it is easy to see that $UV=\left[
\begin{array}
[c]{cc}
A & B
\end{array}
\right]  $, $VW=\left[
\begin{array}
[c]{c}
B\\
D
\end{array}
\right]  $ and $UVW=B$; therefore, the Frobenius rank inequality (applied to
these $m,k,p,n,U,V,W$) yields
$\operatorname*{rank}\left[
\begin{array}
[c]{cc}
A & B
\end{array}
\right]  +\operatorname*{rank}\left[
\begin{array}
[c]{c}
B\\
D
\end{array}
\right]  \leq\operatorname*{rank}X+\operatorname*{rank}B$.
Hence,
(1) $\operatorname*{rank}X-\operatorname*{rank}\left[
\begin{array}
[c]{c}
B\\
D
\end{array}
\right]  \geq\operatorname*{rank}\left[
\begin{array}
[c]{cc}
A & B
\end{array}
\right]  -\operatorname*{rank}B$.
On the other hand, if we set $m=r_{2}$, $k=r_{1}+r_{2}$, $p=s_{1}+s_{2}$,
$n=s_{2}$, $U=\left[
\begin{array}
[c]{cc}
0_{r_{2}\times r_{1}} & I_{r_{2}}
\end{array}
\right]  $, $V=X$ and $W=\left[
\begin{array}
[c]{c}
0_{s_{1}\times s_{2}}\\
I_{s_{2}}
\end{array}
\right]  $, then it is easy to see that $UV=\left[
\begin{array}
[c]{cc}
C & D
\end{array}
\right]  $, $VW=\left[
\begin{array}
[c]{c}
B\\
D
\end{array}
\right]  $ and $UVW=D$; therefore, the Frobenius rank inequality (applied to
these $m,k,p,n,U,V,W$) yields
$\operatorname*{rank}\left[
\begin{array}
[c]{cc}
C & D
\end{array}
\right]  +\operatorname*{rank}\left[
\begin{array}
[c]{c}
B\\
D
\end{array}
\right]  \leq\operatorname*{rank}X+\operatorname*{rank}D$.
Hence,
(2) $\operatorname*{rank}X-\operatorname*{rank}\left[
\begin{array}
[c]{c}
B\\
D
\end{array}
\right]  \geq\operatorname*{rank}\left[
\begin{array}
[c]{cc}
C & D
\end{array}
\right]  -\operatorname*{rank}D$.
Combining (1) with (2), we obtain
$\operatorname*{rank}X-\operatorname*{rank}\left[
\begin{array}
[c]{c}
B\\
D
\end{array}
\right]  $
$\geq\max\left\{  \operatorname*{rank}\left[
\begin{array}
[c]{cc}
A & B
\end{array}
\right]  -\operatorname*{rank}B,\operatorname*{rank}\left[
\begin{array}
[c]{cc}
C & D
\end{array}
\right]  -\operatorname*{rank}D\right\}  $.
This answers your Q1.
As for Question 2, here is a pretty comprehensive negative answer: In general, the rank of the block matrix $X$ is not uniquely determined from the ranks of $A$, $B$, $C$, $D$ and of the $1\times 2$ and $2\times 1$-block submatrices of $X$. This is probably easiest to check in the case of $r_1 = s_1 = r_2 = s_2 = 1$ and $A, B, C, D \neq 0$ (in this case, all nontrivial submatrices of $X$ have rank $1$, but $X$ itself can have either rank $1$ or rank $2$).
A: You can arrange row reduction of a block matrix $\left[ \matrix{A\cr B\cr} \right]$ so that you get
a set of $\text{rank} \left[\matrix{A\cr B\cr}\right] - \text{rank}(A)$ 
rows of $B$ that are linearly independent of the rows of $A$, i.e. the only linear combination of these rows of $B$ that is in the row space of $A$ has
all coefficients $0$. 
Taking transposes and interchanging $A$ and $B$, we have a similar result for columns: there is a set of $\text{rank}[A\ B] - \text{rank}(B)$ columns of
$A$ that are linearly independent of the columns of $B$.  Now for any $C$ and $D$, the corresponding columns of $\left[\matrix{A\cr C\cr}\right]$ are 
linearly independent of the columns of $\left[\matrix{B\cr D\cr}\right]$.
Thus  $\text{rank}\left[\matrix{B\cr D\cr}\right]$ linearly independent columns of $\left[\matrix{B\cr D\cr}\right]$ together with those 
$\text{rank}[A\ B] - \text{rank}(B)$ columns of $\left[\matrix{A\cr C\cr}\right]$ form a linearly independent set, which says 
$$\text{rank}(X) \ge \text{rank}\left[\matrix{B\cr D\cr}\right] + 
\text{rank}[A\ B] - \text{rank}(B)$$
Similarly for $\text{rank}\left[\matrix{B\cr D\cr}\right] + 
\text{rank}[C\ D] - \text{rank}(D)$.
EDIT: Your suspicion is incorrect in both directions.  Try
$$ \eqalign{A = \pmatrix{0 & 1\cr 0 & 1\cr},\ & B = \pmatrix{0 & 1\cr 1 & 1\cr}\cr
C = \pmatrix{0 & 1\cr 1 & 1\cr},\ & D = \pmatrix{0 & 0\cr 1 & 1\cr}}$$
Then $X$ has rank $4$, while $\text{rank}\left[\matrix{B\cr D\cr}\right] = 2$,  $\max(\text{rank}[A\ B] - \text{rank}(B), \text{rank}[C\ D] - \text{rank}(D)) = 1$,
and $\text{rank} \left[ \matrix{A\cr C\cr} \right] = \max\{\text{rank}[A],\text{rank}[C]\} = 2$.
In the other direction, try 
$$ \eqalign{A = \pmatrix{0 & 0\cr 1 & 1\cr}, \ &B = \pmatrix{1 & 1\cr 0 & 0\cr}\cr
C = \pmatrix{0 & 1\cr 0 & 1\cr}, \ &D = \pmatrix{1 & 0\cr 1 & 0\cr}}$$
Here $\text{rank}(X) = 3$, $\text{rank}\left[\matrix{B\cr D\cr}\right] = 2$, 
$\max(\text{rank}[A\ B] - \text{rank}(B), \text{rank}[C\ D] - \text{rank}(D)) = 1$, but $\text{rank} \left[ \matrix{A\cr C\cr} \right] =2 \ne \max\{\text{rank}[A],\text{rank}[C]\} = 1$.
