Prove that $(A^{-1} + B^{-1})^{-1}=A(A+B)^{-1}B$ I have this equality 
$$(A^{-1} + B^{-1})^{-1}=A(A+B)^{-1}B$$ where  $A$ and $B$ are square symmetric matrices.
I have done many test of R and Matlab that show that this holds, however I do not know how to prove it.
 A: Note that
$$ \mathbf{A} \left(\mathbf{A} + \mathbf{B} \right)^{-1} \mathbf{B}$$
is the inverse of 
$$\left(\mathbf{A}^{-1} + \mathbf{B}^{-1} \right) $$
if and only if 
$$ \mathbf{A} \left(\mathbf{A} + \mathbf{B} \right)^{-1} \mathbf{B} \left(\mathbf{A}^{-1} + \mathbf{B}^{-1} \right) = \mathbf{I} $$
and 
$$ \left(\mathbf{A}^{-1} + \mathbf{B}^{-1} \right) \mathbf{A} \left(\mathbf{A} + \mathbf{B} \right)^{-1} \mathbf{B} = \mathbf{I} $$
so that the left and right inverses coincide. Let's prove the first statement. We can see that 
$$\begin{align} \mathbf{A} \left(\mathbf{A} + \mathbf{B} \right)^{-1} \mathbf{B} \left(\mathbf{A}^{-1} + \mathbf{B}^{-1} \right) & = \mathbf{A} \left(\mathbf{A} + \mathbf{B} \right)^{-1} \left(\mathbf{B} \mathbf{A}^{-1} + \mathbf{I} \right) \\ &=  \mathbf{A} \left(\mathbf{A} + \mathbf{B} \right)^{-1} \left( \mathbf{A} + \mathbf{B} \right) \mathbf{A}^{-1} \\ & = \mathbf{I} \end{align} $$
as desired. A similar trick will prove the second statement as well. Thus $ \mathbf{A} \left(\mathbf{A} + \mathbf{B} \right)^{-1} \mathbf{B}$ is indeed the inverse of $\left(\mathbf{A}^{-1} + \mathbf{B}^{-1} \right) $.
A: Assuming $A$, $B$, $A+B$, and $A^{-1}+B^{-1}$ are all invertible, note that
$$A^{-1} + B^{-1} = B^{-1} + A^{-1} = B^{-1}(A+B)A^{-1}$$
and then invert both sides, QED.
Symmetry is unnecessary for this to hold.
