Let $M_x = (I+xS)^{1/2}$ be a real $n \times n$ matrix with $S$ symmetric and positive semi-definite, and $x$ a real parameter. Let $H$ be another $n \times n$ matrix such that $$ u^T H u > 0 $$ for all $u \in U$, where $U$ is some compact subspace of the unit real vectors $S^{n^2-1}$. I would like to prove that there exists $\epsilon > 0$ such that $$ u^T M_x^{-1} H M_x u > 0 $$ for all $0 \leq x < \epsilon$ and $u \in U$. Notice that a Taylor expansion in $x$ yields $M_x = I + O(x)$ and $M_x^{-1} = I + O(x)$, so $$ u^T M_x^{-1} H M_x u = u^T H u + O(x) > 0 $$ for small enough $x$. More precisely, for each $u \in U$ there exists $\epsilon(u) > 0$ such that $$ u^T M_x^{-1} H M_x u > 0 $$ for all $0 \leq x < \epsilon(u)$. How do I extend this to produce a uniform bound for all $u \in U$?
My attempt: I tried to use Taylor's Theorem to produce an explicit and continuous function $\epsilon(u)$, whose infimum will be reached since $U$ is compact, and we would be done. My attempt is to write $F_x = M_x^{-1} H M_x$ and $$ F_x = H + xF_x'(\xi_x) $$ for some $\xi_x \in (0, x)$ by Lagrange's form of Taylor's Theorem, from which $$ u^T F_x u = u^T H u + xu^TF_x'(\xi_x)u > 0$$ iff $$ -xu^TF_x'(\xi_x)u < u^T H u \, .$$ So we might want to take $$ \epsilon(u) = \sup \{ \delta > 0 \mid -\delta u^TF_\delta'(\xi_\delta)u < u^T H u \} $$ or something of the kind. I have no idea how to show this is continuous, or whether there is an easier way to prove the claim.
There is no need to use Taylor expansion:
As $U$ is compact and $u \mapsto u^T H u$ continuous from your first inequality we can deduce that there exist $\delta >0$ such that: $$\forall u \in U, \, u^T H u \geq \delta$$
Then as the application: \begin{align*} \phi: &U \times [0,1] \to \mathbb R\\ &(u,x) \mapsto u^T M_x^{-1} H M_x u \end{align*} is continuous on the compact $U \times[0,1]$, by Heine it is uniformly continuous. So there exist $\epsilon >0$ such that: $$ |u_1-u_2|+|x_1-x_2| \leq \epsilon \implies |\phi(u_1,x_1)-\phi(u_2,x_2)| \leq \frac{\delta}{2}$$ and in particular if $0 \leq x < \epsilon$: $$|u-u|+|x-0| < \epsilon$$ so: $$\phi(u,x) \geq u^T H u -\frac{\delta}{2} \geq \frac{\delta}{2} >0$$