proof that rank$(A+B+AB) \le $rank$(A) + $rank$(B)$ when $A,B$ are square matrices

Question

I'm trying to figure out how to proof this. I know how to prove that rank$(A+B+AB) \le $ rank$(A) + \;$ rank$(B)$ but can't see how $AB$ connects to that.

Answer 1

Solution $$\operatorname{rank}(A)+\operatorname{rank}(B)=\operatorname{rank}\left( \begin{array}{cc} A & 0 \\ 0 & B \end{array} \right)\\ =\operatorname{rank}\left( \begin{array}{cc} A & -AB-B \\ 0 & B \end{array} \right)\\ =\operatorname{rank}\left( \begin{array}{cc} AB+A+B & -AB-B \\ -B& B \end{array} \right)\\ \ge\operatorname{rank}(AB+A+B) $$

Answer 2

\begin{align*} \operatorname{rank}(A+B+AB) &\leq \operatorname{rank}(A(B+I)+B)\\ &\leq \operatorname{rank}(A(B+I)) + \operatorname{rank}(B) \qquad \text{triangle inequality of rank}\\ &\leq \operatorname{rank}(A) + \operatorname{rank}(B) \qquad\qquad\quad\,\, \text{since}\ \operatorname{rank}(XY) \leq \operatorname{rank}(X) \end{align*}