Proof of $\text{rank}(A+B)\leq \text{rank}(A)+\text{rank}(B)$ by another way. Here's the problem:

Let $A,B\in Mat_{n}(\mathbb R)$. Use $
\begin{pmatrix}
A & A \\
A & A+B \\
\end{pmatrix}
$ to show that $$\text{rank}(A+B)\le \text{rank}(A)+\text{rank}(B)$$

By performing elementary operations, I've got that $\text{rank}\begin{pmatrix}
A & A \\
A & A+B \\
\end{pmatrix}=\text{rank}\begin{pmatrix}
A & O \\
O & B \\
\end{pmatrix}=\text{rank}(A)+\text{rank}(B)$. My intuitive approach is say that since $\begin{pmatrix}
A & A \\
A & A+B \\
\end{pmatrix}$ contains $\begin{pmatrix}A+B \\
\end{pmatrix}$, $$\text{rank}(A)+\text{rank}(B)=\text{rank}\begin{pmatrix}
A & A \\
A & A+B \\
\end{pmatrix}\geq \text{rank}\begin{pmatrix}A+B \\
\end{pmatrix}=\text{rank}(A+B)$$
but I can't prove this claim well. Could someone give me some hints? Thanks in advance!
 A: $$A + B = \begin{bmatrix}I&I\end{bmatrix} \begin{bmatrix} A\\B\end{bmatrix}$$
Therefore, $\operatorname{rank} (A+B) \leq \operatorname{rank} \begin{bmatrix} A\\B\end{bmatrix}$. Since $\operatorname{rank} \begin{bmatrix} A\\B\end{bmatrix} \leq \operatorname{rank} (A) + \operatorname{rank} (B)$, we have the desired inequality.
A: Your intuitive approach is correct. There are a few approaches that we could apply to get your inequality; here's one approach. In general, it holds that $\operatorname{rank}(PQ) \leq \min\{\operatorname{rank}(P),\operatorname{rank}(Q)\}$. Thus, if we take
$$
P = \pmatrix{A & A\\A&A+B}, \quad Q = \pmatrix{0 & I}
$$
(where $I$ denotes the identity matrix), then we have
$$
\operatorname{rank}(P)\geq \operatorname{rank}(QP) \geq \operatorname{rank}(QPQ^T) = \operatorname{rank}(A + B).
$$
A: Or like this.
For any matrix $A$
$$
\operatorname{rank}\begin{pmatrix} A & *\\* & *\end{pmatrix}\geq \operatorname{rank}(A).
$$
An outline of the proof. Choose a nonzero minor $M$ of order $r=\operatorname{rank}(A)$ in the matrix $A$. Then $M$ is a nonzero minor in the large matrix as well.
