Show that $A^{-1}+B^{-1}$ is also invertible 
Let $A$ and $B$ be two invertible $n \times n$ real matrices. Assume that $A+B$ is invertible. Show that $A^{-1} + B^{-1}$ is also invertible.


My approach
\begin{aligned}
&|\mathrm{A}|\left|A^{-1}+B^{-1}\right||\mathrm{B}|=|\mathrm{B}+\mathrm{A}| \neq 0 \\
\Rightarrow &\left|A^{-1}+B^{-1}\right| \neq 0 \text { as }|\mathrm{A}|,|\mathrm{B}| \neq 0 \\
\Rightarrow & A^{-1}+B^{-1} \text {is invertible }
\end{aligned}
Am I correct? Any other method or hint would be greatly appreciated!
 A: Just for the sake of a slightly different approach. Let $C$ be the inverse of $A+B$ (same as $B+A)$. Then we can show that $ACB$ is the inverse of $A^{-1}+B^{-1}$.
\begin{align*}
(A^{-1}+B^{-1}) (ACB)&=CB+B^{-1}ACB\\
&=B^{-1}B (CB) +B^{-1}ACB\\
&=B^{-1}\underbrace{(B+A)C}_{=I}B\\
&=B^{-1}B\\
&=I.
\end{align*}
A: As @Michael has mentioned in his comment, you are like "using the result to prove the result", or at least you could show the proof for your first line to have your solution verified without any problems...
Anyway, this answer will probably deal with this problem:
Let's assume the matrix $X$ is inverse for $A^{-1}+B^{-1}$. Then we must have
$$(A^{-1}+B^{-1})X=I \implies A^{-1}X+B^{-1}X=I$$ Now denote $Y=B^{-1}X$ to get
$$A^{-1}BY+Y=I \implies BY+AY=A \implies (A+B)Y=A$$ Since $A+B$ is invertible
$$Y=(A+B)^{-1}A$$ By substituting back we get
$$B^{-1}X=(A+B)^{-1}A \implies \boxed{X=B(A+B)^{-1}A}$$ So, we got no contradiction with this assumption.

Indeed, one can check that this $X$ satisfies the conditions of the inverse matrix. Let's check!
We need to show that
$$(A^{-1}+B^{-1})(B(A+B)^{-1}A)=I \ \ \ \ (1)$$
Now we use the following trick: Note that
$$B(A+B)^{-1}A + A(A+B)^{-1}A=(A+B)(A+B)^{-1}A=A \implies$$$$\implies B(A+B)^{-1}A = A -  A(A+B)^{-1}A$$ and similarly
$$B(A+B)^{-1}A + B(A+B)^{-1}B=B(A+B)^{-1}(A+B)=B \implies$$$$\implies B(A+B)^{-1}A = B -  B(A+B)^{-1}B$$
Now, back to $(1)$,
$$(A^{-1}+B^{-1})(B(A+B)^{-1}A)=(A^{-1})(A -  A(A+B)^{-1}A) + (B^{-1})(B -  B(A+B)^{-1}B)=$$$$=I - (A+B)^{-1}A + I - (A+B)^{-1}B = 2I - (A+B)^{-1}(A+B)=2I-I=I$$ And we are done!
