Rank of the concatenation of three matrices

Question

We all know that the sum of two matrix ranks is greater than or equal to the rank of the concatenation of two matrices, so can this property be extended to $3$ or even $4$ matrices?$$\operatorname{rank}(A)\leq \operatorname{rank}(B)+\operatorname{rank}(C)+\operatorname{rank}(D),$$where $A=(B\mid C\mid D)$.

Answer 1

To clarify the comments a bit, here's what "applying the rule inductively" might look like.

Claim: $\operatorname{rank}[A_1 \mid \cdots \mid A_{n-1}\mid A_n] \leq \sum_{i=1}^n \operatorname{rank}(A_i)$.

Proof: Denote $M_n = [A_1 \mid \cdots \mid A_{n-1}\mid A_n]$. The statement trivially holds for $n=1$. If the statement holds for $n \geq 1$, then it follows that $$ \operatorname{rank}(M_{n+1}) = \operatorname{rank}([M_n \mid A_{n+1}]) \leq \operatorname{rank}(M_n) + \operatorname{rank}(A_{n+1}) \\ = \left[\sum_{i=1}^n \operatorname{rank}(A_i)\right] + \operatorname{rank}(A_{n+1}) = \sum_{i=1}^{n+1} \operatorname{rank}(A_i). $$ By induction, we therefore conclude that $\operatorname{rank}(M_n) \leq \sum_{i=1}^n \operatorname{rank}(A_i)$ for all $n$.