I believe this inequality is true, but I am not sure how to show it rigorously.
For Matrices $A^{-1}$ and $B^{-1}$ where $A^{-1}, B^{-1} \preceq I$ where I is a matrix of all 1's and this inequality (I believe) represents that each element is less than 1, I want to show the following $$ ||A^{-1} - B^{-1}|| = ||A^{-1}A^{-1}A - B^{-1}B^{-1}B|| \le ||A-B|| $$ My reasoning behind me doing this is that I know, $|1/5-1/3| \le |5-3|$ and we can replace 5 or 3 with any natural numbers and this will work.
So how can I rigorously show this for my matrix inequality above? Or is it actually fine as is, as long as I state $A^{-1}, B^{-1} \preceq I$ ?
Thank you for taking the time!
Use identity $A^{-1}-B^{-1}=A^{-1}(B-A)B^{-1}$ and the fact that $\|A^{-1}\|\leq 1$ and $\|B^{-1}\|<1$.