Product of the differences of two pd matrices and their respective inverses is pd Given two $\textbf{positive definite (pd), Hermitian}$ matrices X and Y, I am trying to determine whether $(X-Y)(Y^{-1}-X^{-1})$ will always be pd as well, and how to prove this.
This formulation appears when trying to prove that the inverse function is convex on pd matrices: Is inverse matrix convex?.
It was trying to prove that if $(X-Y)\succ 0$, then $(Y^{-1}-X^{-1})\prec 0$ and vice versa, and subsequently $(X-Y)(Y^{-1}-X^{-1})\succ 0$ always. However I wasn't able to prove either of these two steps - nor disprove either of them.
 A: [Edited after After Michael's comment] In the general case, the OP's claim is not verified. For example if
$$
X =
\begin{bmatrix}
29& 11& 19\\
11& 10& 12\\
19& 12& 19
\end{bmatrix}
, \quad
Y =
\begin{bmatrix}
14& 16& 17\\
16& 21& 22\\
17& 22& 29
\end{bmatrix}
, \quad
w =
\begin{bmatrix}
1\\
2\\
3
\end{bmatrix}
$$
we have that
$$w^T(X-Y)(Y^{-1}-X^{-1})\,w = -21002/11025.$$
The claim is verified, instead, in the special case that $X$ and $Y$ are such that $XY^{-1}$ is symmetric but one needs to relax the claim from positive definite to positive semi-definite. In fact, if $X=Y=I$, the results is the matrix of all zeros.
We could rewrite $(X-Y)(Y^{-1}-X^{-1})$ as $XY^{-1}+YX^{-1}-2I$. Then, we observe that $XY^{-1} = (YX^{-1})^{-1}$, which implies that the two must have the same eigenvectors with reciprocal eigenvalues. Therefore, the eignevalues of $XY^{-1}+YX^{-1}$ will be $\lambda_i + 1/\lambda_i$, where $\lambda_i$ is the $i$-th eigenvalue of $XY^{-1}$. We observe that $x + 1/x \ge 2\; \forall x>0$, which implies that all eigenvalues of $XY^{-1}+YX^{-1}$ are at least 2. If $XY^{-1}$ is symmetric, this implies that $XY^{-1}+YX^{-1}-2I$ is at least positive semi-definite.
