Theorem: Suppose $\A$ and $\B$ are positive definite matrices satisfying $\A \ggt \B$. Then $\A^{-1} \glt \B^{-1}$.
Proof: Consider the partitioned matrix
\[\M = \begin{bmatrix} \A & \I \\ \I & \B^{-1} \end{bmatrix}.\]A symmetric partitioned matrix is positive definite if and only if one diagonal block is positive definite and its Schur complement is positive definite as well. Since $\B^{-1}$ is positive definite and its Schur complement is $\A - \I\B\I = \A - \B \gg \zero$, $\M$ is positive definite. Applying the same fact using the other diagonal block, the Schur complement of $\A$,
\[\B^{-1} - \I\A^{-1}\I = \B^{-1}-\A^{-1},\]must also be positive definite.