Does ${\rm rk} AB \le {\rm min} ({\rm rk} A, {\rm rk} B)$ hold over a sum?

Question

For matrices $A, B$, does the following rank (rk) inequality hold \begin{equation} {\rm rk} (AB + BA) \le {\rm rk}(AB) + {\rm rk}(BA) \le {\rm min}({\rm rk} A, {\rm rk} B) + {\rm min}({\rm rk} B, {\rm rk} A) = 2 ~{\rm min}({\rm rk} A, {\rm rk} B) ? \end{equation}

Answer 1

Yes, because the following two inequalities hold:

(1) The one in the title,

(2) ${\rm rk}(X+Y) \le {\rm rk}(X) + {\rm rk}(Y)$.

Apply the two inequalities and you can prove the one in your question.

Answer 2

For Rob's extra question.
Try these: $$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \qquad B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$ both of rank $1$. Then $$ AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \qquad BA = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $$ and $AB+BA$ has rank $2$.