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: $$
A+B=B(B^{-1}+A^{-1})A,
$$
So
$$
(A+B)^{-1}=A^{-1}(A^{-1}+B^{-1})^{-1}B^{-1}.
$$
Finally,$$
A(A+B)^{-1}B=A(A^{-1}(A^{-1}+B^{-1})^{-1}B^{-1})B=(A^{-1}+B^{-1})^{-1}.
$$
A: $$(A(A+B)^{-1}B)^{-1}=B^{-1}(A+B)A^{-1}=A^{-1}+B^{-1}.$$
A: Hint: To show that $A^{-1}=C$, compute $AC$ and show that $AC=I$.
In your case, you just need to show that $(A^{-1} + B^{-1})\cdot(A(A + B)^{-1}B)=I$, where $I$ is the identity matrix.
A: You can do it the way as other users have mentioned or observe that $M(N^{-1})=(NM^{-1})^{-1}$. See how the matrix $M$ came inside the inverse. Also $M(N^{-1})O=(O^{-1}NM^{-1})^{-1}$. It is just a generalization of the first statement. Now put $M=A$, $N=A+B$ and $O=B$. You will have your answer. 
A: How about a purposedly complicated proof, but without matrix inversion?
Let $X = A(A + B)^{-1}B$. By writing the rightmost $B$ as $(A+B)-B$, we get
$$X = A(A + B)^{-1}[(A+B)-A] = \underbrace{A - A(A + B)^{-1}A}_{X_1}.$$
Similarly, if we write the leftmost $A$ as $(A+B)-B$ instead, we get
$$X = [(A+B)-B](A + B)^{-1}B = \underbrace{B - B(A + B)^{-1}B}_{X_2}.$$
Let $Y=A^{-1} + B^{-1}$. Then
\begin{align}
XY &= X(A^{-1} + B^{-1})\\
&= X_1 A^{-1} + X_2B^{-1}\\
&= [I - A(A + B)^{-1}] + [I - B(A + B)^{-1}]\\
&= 2I - (A+B)(A + B)^{-1} = I.
\end{align}
So $XY=I$ and we are done.
