For invertible $A$, $C$, prove that: $(A^{−1} + B^TC^{−1}B)^{−1}B^TC^{−1} = AB^T(BAB^T + C)^{−1}$

Question

Here's what I have to prove:

For invertible $A$, $C$, prove that: $$ (A^{−1} + B^TC^{−1}B)^{−1}B^TC^{−1} = AB^T(BAB^T + C)^{−1}. $$

Here's what I have so far. I couldn't do the proof entirely because I got stuck at some point (marked in red) and I'm asking myself if the problem would have to be constrained more.

I couldn't really see a way to start with the left (or right) side and manipulate it such that I end up with the other side - so i proved it through equivalences (which is kind of ugly).

But maybe there is a way to prove it (and also a better style for proving it). Could you help me with this?

Here's what I've done so far:

Answer 1

Assuming that the parentheses are in fact invertible, you could multiply both sides with the parentheses to eliminate the inverted parentheses: $$\begin{aligned} (A^{−1} + B^TC^{−1}B)^{−1}B^TC^{−1} &= AB^T(BAB^T + C)^{−1} \\ B^TC^{-1}(BAB^T + C) &= (A^{-1} + B^TC^{-1}B)AB^T \\ B^TC^{-1}BAB^T + B^T &=B^T + B^TC^{-1}BAB^T \end{aligned}$$