Proving an equation with invertible matrices I need to prove that if $A$, $B$ and $(A + B)$ are invertible then $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$
I'm a bit lost with this one,
I can't find a way to make any assumptions about $(A^{-1} + B^{-1})$,
Neither by using $A^{-1}$, $B^{-1}$ and $(A + B)^{-1}$.
If someone could clue me in I'll be grateful.

UPDATE:
Thanks for all of your input - it really helped,
I got a solution, I'd like to know if my way of proving it is valid:
$(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$  

multiplying both sides by $A^{-1}$ and $B^{-1}$  
$A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1} = (A+B)^{-1}$

now multiplying both sides by $(A + B)$, I get to
$(A+B)(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1}) = I$
Lets call $(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})$ -> $C$  
Now I determine that $C$ is $(A+B)^{-1}$,
because $(A+B)C$ equals the Identity Matrix.

So lastly to verify that I place $C$ in the original equation:
$(A^{-1} + B^{-1})^{-1} = A(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})B$
and from this I get:
$(A^{-1} + B^{-1})^{-1} = (A^{-1} + B^{-1})^{-1}$

$$
A+B=B(B^{-1}+A^{-1})A,
$$
So
$$
(A+B)^{-1}=A^{-1}(A^{-1}+B^{-1})^{-1}B^{-1}.
$$
Finally,$$
A(A+B)^{-1}B=A(A^{-1}(A^{-1}+B^{-1})^{-1}B^{-1})B=(A^{-1}+B^{-1})^{-1}.
$$
