Matrix inequality after taking inverse

Consider $B-A\geq 0$ using Schur complement, this is equivalent to $$\begin{bmatrix}B&I\\I&A^{-1}\end{bmatrix}\geq 0, \quad B>0$$ Since $A^{-1}>0$, now apply the Schur complement one more time to obtain $$A^{-1}-I(B)^{-1}I=A^{-1}-B^{-1}\geq 0$$ therefore we have $A^{-1}\geq B^{-1}$.

I will show that (for $A > 0$ and $B > 0$) we have that $A\le B$ is equivalent to $\sigma(A^{-1}B)\subset [1,\infty)$, where $\sigma(T)$ denotes the spectrum of a matrix $T$ (i.e., the set of eigenvalues). Then the claim immediately follows since $\sigma(BA^{-1}) = \sigma(A^{-1}BA^{-1}A) = \sigma(A^{-1}B)$.

For the equivalence, note that $T:=A^{-1}B - I$ is selfadjoint with respect to the (positive definite) inner product $\langle A\cdot,\cdot\rangle$. Now, $A\le B$ is equivalent to $T$ being positive semi-definite with respect to this inner product. But this is itself equivalent to $\sigma(T)\subset [0,\infty)$, i.e., $\sigma(A^{-1}B) = \sigma(T+I)\subset [1,\infty)$.