If $A,B,C\in M_n(\mathbb{R})$ and $A+B+C=0$, then what are the possible values of the triple $\big(\mbox{rank}(A),\mbox{rank}(B),\mbox{rank}(C)\big)$?
I know that $\mbox{rank}(A)\leq \mbox{rank}(B)+\mbox{rank}(C), \mbox{rank}(B)\leq \mbox{rank}(A)+\mbox{rank}(C), \mbox{rank}(C)\leq \mbox{rank}(A)+\mbox{rank}(B)$.
Thus $\big(\mbox{rank}(A),\mbox{rank}(B),\mbox{rank}(C)\big)$ could be any natural number less than equal to $n$ such that the triple $\big(\mbox{rank}(A),\mbox{rank}(B),\mbox{rank}(C)\big)$ forms three sides of a triangle.
Is it fine or we can say more on the triple $\big(\mbox{rank}(A),\mbox{rank}(B),\mbox{rank}(C)\big)$?
The ranks simply have to follow the "Triangle Inequality" that you have already found. That is, for any integers $a$, $b$, and $c$ such that $0\leq a,b,c\leq n$, there are matrices $A$, $B$, and $C$ in $\text{Mat}_{n\times n}(\mathbb{R})$ of ranks $a$, $b$, and $c$, respectively, such that $$A+B+C=0$$ if and only if $$|a-b|\leq c\leq a+b\,.$$ All that is left to do is to find a specific example for each of such triples $(a,b,c)$. Your solution would not be complete without this step of finding examples.
If $c=a+b-2k$ for some integer $k$ such that $0\leq k\leq \min\{a,b\}$, then we can take $C:=-A-B$, where $$A:=\text{diag}_n\big(\underbrace{1,1,\ldots,1}_{a\text{ terms}},0,0,\ldots,0\big)$$ and $$B:=-\text{diag}_n\big(\underbrace{0,0,\ldots,0}_{a-k\text{ terms}},\underbrace{1,1,\ldots,1}_{b\text{ terms}},0,0,\ldots,0\big)\,.$$ If $c=a+b-2k+1$ for some integer $k$ such that $0\leq k\leq \min\{a,b\}$, then $a\geq 1$ and we can take $C:=-A-B$, where $$A:=\text{diag}_n\big(\underbrace{1,1,\ldots,1}_{a-1\text{ terms}},2,0,0,\ldots,0\big)$$ and $$B:=-\text{diag}_n\big(\underbrace{0,0,\ldots,0}_{a-k\text{ terms}},\underbrace{1,1,\ldots,1}_{b\text{ terms}},0,0,\ldots,0\big)\,.$$