Proof - a property in positive semi definite (PSD) cone? The clue for proving:

It can also be shown that for any $A,B\in\mathbb S^n_{++}$, $$A\preceq B\quad\Leftrightarrow\quad {\cal E}_A=\left\{u\mid u^TA^{-1}u\le1\right\}\subseteq {\cal E}_B.$$

 A: Prove first that ${\cal E}_A\subseteq {\cal E}_B$ $\Leftrightarrow$ $B^{-1}\preceq A^{-1}$.
"$\Leftarrow$" Easy.
"$\Rightarrow$" For any nonzero $u\in\Bbb{R}^n$ set $w=\frac{u}{\sqrt{u^TA^{-1}u}}$ and note that $w\in{\cal E}_A$. Then $w^TB^{-1}w\le 1$ $\Leftrightarrow$ $u^TB^{-1}u\le u^TA^{-1}u$.

Now the claim in question follows from this answer that proves that $B-A\succeq 0$ $\Leftrightarrow$ $A^{-1}-B^{-1}\succeq 0$. (It proves actually only $\Rightarrow$, but the equvalence follows from applying the implication to $A_1=B^{-1}$ and $B_1=A^{-1}$.) 
As an alternative proof, both statements "$B-A$ is PSD" and "$A^{-1}-B^{-1}$ is PSD" are equivalent (via Schur complement) to the matrix
$$
M=\begin{bmatrix}
B & A^{1/2}\\A^{1/2} & I
\end{bmatrix}
$$
being PSD. Here $A^{1/2}$ is the principal square root of $A$. Indeed, the Schur complements are
$$
M/I=B-A^{1/2}A^{1/2}=B-A,\quad M/B=I-A^{1/2}B^{-1}A^{1/2},
$$
and $B-A$ is PSD $\Leftrightarrow$ $M$ is PSD $\Leftrightarrow$ $I-A^{1/2}B^{-1}A^{1/2}$ is PSD $\Leftrightarrow$ $A^{-1}-B^{-1}$ is PSD (pre- and postmultiply the previous matrix by $A^{-1/2}$).
