Let $A : \mathcal{H} \to \mathcal{H}$ and $B : \mathcal{H} \to \mathcal{H}$ be trace-class (hence compact) Hermitian operators on a separable Hilbert space. Assume that $A$ is strictly positive and that $B$ is positive and rank-1. I'm interested in conditions when $A - \epsilon B \ge 0$ for some strictly positive $\epsilon \in \mathbb{R}$ (as usual, $>$ and $\ge$ for operators refers to being positive (semi)definite).

If $A$ is finite dimensional, then $A - \epsilon B \ge 0$ for some $\epsilon > 0$ always. This is because the smallest eigenvalue of $A - \epsilon B$ obeys $\lambda_\min >0$ for $\epsilon=0$ and varies continuously with $\epsilon$.

If $A$ is infinite dimensional and $B=\vert \phi\rangle\langle\phi\vert$ for some eigenvector $\vert \phi\rangle$ of $A$ (with corresponding eigenvalue $\lambda>0$), then it is clear that $A - \epsilon B\ge 0$ for $\epsilon\le \lambda$.

What about when $A$ is infinite dimensional and $B$ does not have the form of $\vert \phi\rangle\langle\phi\vert$?