For some $A,B \in \mathbb{S}_{++}^n$, let
$$\mathcal{E}_A := \{x\in\mathbb{R}^n \mid x^T A^{-1}x \le 1\}$$
$$\mathcal{E}_B := \{x\in\mathbb{R}^n \mid x^T B^{-1}x \le 1\}$$
I'm trying to prove that $\mathcal{E}_A \subseteq \mathcal{E}_B \implies B \succeq A$.
I've proven the other direction rather easily, but I'm stuck on the forward direction. I think the LHS has the interpretation that the ellipsoid represened by $A$ is fully contained in the one represented by $B$, but don't know how to go from there.
With the left side, we have $x^TA^{-1}x \leq 1 \implies x^TB^{-1}x \leq 1$. From this, we deduce that $x^TA^{-1}x \geq x^TB^{-1}x$ for all $x$.
Indeed: for any $x$, define $\alpha = \frac 1{\sqrt{x^TA^{-1}x}}$. We note that $y = \alpha x$ satisfies $y^TA^{-1}y = 1$, from which it follows that $y^TB^{-1}y \leq 1$, which we can rewrite as $$ \begin{align} y^TB^{-1}y \leq 1 &\implies (\alpha x)^TB^{-1}(\alpha x) \leq 1 \\ & \implies \frac{x^TB^{-1}x}{[\sqrt{x^TA^{-1}x}]^2} \leq 1 \\ & \implies \frac{x^TB^{-1}x}{x^TA^{-1}x} \leq 1 \implies x^T B^{-1}x \leq x^TA^{-1}x. \end{align} $$ Now, we can say that $x^T(A^{-1} - B^{-1})x \geq 0$ for all $x$, which is to say that $B^{-1} \preceq A^{-1}$. It follows that $A \preceq B$.