upper bounds on a certain matrix norm Is there some simple upper bound on $||(B^{-1}+A^{-1})^{-1}||$, where $A,B$ are $n \times n$ symmetric matrices?
 A: You can use the surprising identity $(A^{-1}+B^{-1})^{-1}=A(A+B)^{-1}B$, and take the norms of the three factors separately.
A: Probably not unless $A$ and $B$ are positive-definite, since if $B$ is very close to $-A$ then $B^{-1}+A^{-1}$ is very small and so its inverse is very large. In fact, depending on the norm, they probably need to be close only on one shared or almost-shared eigenvector.
For spectral norm of positive-definite matrices, we have a nice answer. The highest eigenvalue of $(A^{-1}+B^{-1})^{-1}$ is the lowest eigenvalue of $A^{-1}+B^{-1}$, which one can find by minimizing $x^T(A^{-1}+B^{-1})x$ with respect to $x^Tx=1$. But the minimum for $A^{-1}$ is its lowest eigenvalue, $1/||A||$, and the minimum for $B^{-1}$ is similarly $1/||B||$. Thus:
$x^T( A^{-1}+B^{-1})^{-1} x= x^T A^{-1} x+ x^T B^{-1} x\geq 1/||A||+1/||B||$
So the spectral norm of the harmonic sum is bounded by the harmonic sum of the spectral norms!
A: To expand on my first comment, if $A, B > 0$ are symmetric positive definite matrices. Then, it is known that
$$\left(\frac{A^{-1}+B^{-1}}{2}\right)^{-1} \le A\sharp B \le \frac{A+B}{2},$$
where the inequalities are in the Löwner partial order, and $A\sharp B := A^{1/2}(A^{1/2}BA^{1/2})^{-1/2}A^{1/2}$ denotes the matrix geometric mean. 
These operator inequalities are of course, stronger than corresponding norm inequalities (based on unitarily invariant norms).
For the case where you don't have positive matrices, I think the conjecture mentioned in my second argument can be expanded into a proof --- maybe if I get time, I'll try to expand that.
