Inverse of $A^{-1}+B^{-1}$

If $A, B$ and $A + B$ are all $n × n$ invertible matrices. Prove that $A^{−1} + B^{−1}$ is invertible and the inverse is $A(A + B) ^{−1}B$.

I am afraid I am really stuck on this one, and I haven't really tried much because I don't know what to try.

Thanks for all the help guys, I understand now.

Here's a concise proof: $$ \begin{align} A^{-1} + B^{-1} &= B^{-1} + A^{-1} \\ &= B^{-1}AA^{-1} + B^{-1}BA^{-1}\\ &= B^{-1}(A + B)A^{-1} \\ \end{align} $$ So $A^{-1} + B^{-1}$ is the product of invertible matrices and thus is invertible, with inverse equal to $$ \begin{align} (A^{-1} + B^{-1})^{-1} &= (B^{-1}(A + B)A^{-1})^{-1} \\ &= A(A + B)^{-1}B \\ \end{align} $$

Here's half of the solution. $$A(A+B)^{-1}B=((A+B)A^{-1})^{-1}B=(I+BA^{-1})^{-1}B$$ (Note that inversion reverses the order of products.) What happens if you do the same thing with $B$? We then get $$(B^{-1}(I+BA^{-1}))^{-1}=(B^{-1}+A^{-1})^{-1}$$ We didn't even have to multiply anything out!