Problem: Let $A$ and $B$ be $n\times n$ matrices such that their sum $A+B$ is invertible. Then show that $$A(A+B)^{-1}B=B(A+B)^{-1}A$$ (Hat tip: Sam Walters)
Solution. \begin{aligned}I&=(A+B)(A+B)^{-1}\\&=A(A+B)^{-1}+B(A+B)^{-1}\end{aligned}\label{eq:matrix} Multiply \eqref{eq:matrix} by $B$ from the right $$\label{eq:matrix2}B=A(A+B)^{-1}B+B(A+B)^{-1}B$$ Also multiply \eqref{eq:matrix} by $A$ from the left $$\label{eq:matrix3}A=A(A+B)^{-1}A+B(A+B)^{-1}A$$ Subtract \eqref{eq:matrix3} from \eqref{eq:matrix2}. $$\label{eq:matrix4}B-A=A(A+B)^{-1}B-B(A+B)^{-1}A+B(A+B)^{-1}B-A(A+B)^{-1}A$$ In a similar manner from $I=(A+B)^{-1}(A+B)$, we obtain $$\label{eq:matrix5}A-B=A(A+B)^{-1}B-B(A+B)^{-1}A+A(A+B)^{-1}A-B(A+B)^{-1}B$$ \eqref{eq:matrix4}+\eqref{eq:matrix5} results $$A(A+B)^{-1}B=B(A+B)^{-1}A$$