Let $X$ and $Y$ be symmetric positive definite matrices. Suppose that $X-Y$ is positive semidefinite. Does it follow that $Y^{-1}- X^{-1}$ is also positive semidefinite?
If $X$ and $Y$ are simultaneously diagonalizable, then we may diagonalize both of them to get the result.
After playing around with this a bit, I am starting to think that the answer is "no". But I didn't immediately see how to construct a counterexample. For example, it seems like there is no $2\times 2$ counterexample.
(1) If $A >0$ and $A$ is symmetric, there is some symmetric $S>0$ such that $A = S^2$ (square root).
(2) It is straightforward to show that if $C$ is symmetric and $A \ge B$, then $CAC \ge CBC$.
(3) It is straightforward using the above to show that if $A \ge I$ (with $A$ symmetric), then $I \ge A^{-1}$.
Suppose $S$ is a square root of $Y$, that is $S>0$ and $Y=S^2$.
Suppose $X \ge Y$, with $X>0,Y>0$.
Then (2) with $C=S^{-1} $ gives $S^{-1} X S^{-1} \ge I$. Hence (3) gives $I \ge S X^{-1} S$ and so applying (2) again with $C=S^{-1}$ gives $Y^{-1} \ge X^{-1}$.