Let $A,B$ be two $(m,n)$ matrices. Prove that $$\def\rk{\operatorname{rank}}\rk{(A+B)}\leq \rk A+\rk B$$ using only the definition of $\rk$ involving non-zero minors.
I expressed every minor $\Delta_r$ of $A+B$ as a linear combination of minors of $A$ and $B$, each with order not less than $\lceil \frac{r}{2} \rceil$, but this approach didn't get me to the result.
I am a beginner in linear algebra and that is the only definition of $\rk$ I have learnt, so I would highly appreciate any solution using it.
EDIT: I think I managed to find a solution. Let $A=(a_{ij}), B=(b_{ij})$ with $i\leq m$ and $j\leq n$. Also, $\rk A=r$ and $\rk B=s$. Then $A+B=(a_{ij}+b_{ij})$ and let $\Delta_{r+s+1}$ be a minor of $A+B$ of order $(r+s+1)$. Let $k=r+s+1$. $$\Delta_{k}=\det \begin{pmatrix} a_{i_1j_1}+b_{i_1j_1} & a_{i_1j_2}+b_{i_1j_2} &... &a_{i_1j_{k}}b_{i_1j_{k}} \\ ...&...&...&...\\ a_{i_{k}j_1}+b_{i_{k}j_1}& a_{i_kj_2}+b_{i_kj_2}& ...& a_{i_kj_k}+b_{i_kj_k}\end{pmatrix}$$ Using the properties of determinants, we can develop the one above as a linear combination of determinants with the form $$\det \begin{pmatrix} a_{i_1p_1} & a_{i_1p_2}&... &a_{i_1p_{x}} & b_{i_1q_1} &b_{i_1q_2} &... &b_{i_1q_y} \\ a_{i_2p_1} & a_{i_2p_2}&... &a_{i_2p_{x}} & b_{i_2q_1} &b_{i_2q_2} &... &b_{i_2q_y} \\ ...&...&...&...&...&...&...&...\\ a_{i_kp_1}& a_{i_kp_2}& ...& a_{i_kp_x} & b_{i_kq_1} & b_{i_kq_2} & ... &b_{i_kq_y}\end{pmatrix}$$ where $x+y=k=r+s+1$ and $ \{ p_1,p_2,...p_x,q_1,q_2,...q_y \} = \{j_1,j_2,...j_k \}$
If $x \leq r$ then $y \geq s+1$ and we can write the last determinant as a linear combination of $(k-x)$order determinants consisting only of elements of $B$. Since $k-x=y$, these will be minors of $B$ with order $y \geq s+1$ and since $\rk B=s$ it means that all these minors are $0$.
The same holds true if $x\geq r+1$, when we write the determinant as a linear combination of minors of $A$ with order $x$, thus all equal to $0$.
In conclusion, all the determinants with the form as above are $0$ and so $\Delta_k=0$, which means that $\rk(A+B) \leq r+s=\rk A+\rk B$