How to prove that (and why is it that) if a state space realization (A,B,C, D) is a minimal realization, then $(A^{-1},B^{-1},C^{-1},D^{-1})$ will also be a minimal realization?
One way to define a minimal realization would be:
I should also add that for the above proof, we know that a state space realization $(\hat{A}, \hat{B}, \hat{C}, \hat{D})$ of the inverse transfer function is given by: $$ \begin{align} \hat{A} &= A-B\,D^{-1}C\\ \hat{B} &= B\,D^{-1}\\ \hat{C} &= -D^{-1}C\\ \hat{D} &= D^{-1} \end{align} $$
The "inverse" realization as you have defined it need not even exist for some minimal realizations.
For example, if $H(s)$ is any proper rational transfer function (MIMO or SISO--it doesn't matter) then it has a minimal realization with $D = 0$. In this case $D^{-1}$ does not exist and as a result all of the $A^{-1}, B^{-1}, C^{-1}$ as you have put them are not defined.
Furthermore, since $D$ is $\dim Y \times \dim U$ it isn't square in general, so it doesn't even make sense to talk about a matrix inverse of $D$.