In a problem I am considering functions $T\geq \epsilon I$ where $T,I\in B(H)$, $\epsilon> 0$. What is the definition of this inequality? Is it saying $||T(x)||_{H}\geq ||\epsilon I(x)||_{H}$ $\forall x\in H$ or $||T||_{B(H)}\geq ||\epsilon I||_{B(H)}$?
EDIT: For context, below is the problem I am looking at.
Let $T\in B(H)$ be positive, that is, $T=T^{*}$ and $\langle T(x),x\rangle \geq 0$ $ \forall x\in H$. Prove that if there exists $\epsilon >0$ such that $T\geq\epsilon I$ then $T$ is invertible.
In FA we write $T\geq S$ to denote $\langle Tx, x \rangle \geq \langle Sx, x \rangle$ for all $x$. This notation is standard. So $T \geq \epsilon I$ means $\langle Tx, x \rangle \geq \epsilon \|x\|^{2}$ for all $x$.
Proof of the result you are trying to prove: From $\langle Tx, x \rangle \geq \epsilon \|x\|^{2}$ we see that $T$ is one-to-one. Since $T=T^{*}$ it follows that $T^{*}$ is one-to-one. But $(Ran(T))^{\perp}=\ker T^{*}$ so range of $T$ is dense. But $\langle Tx, x \rangle \geq \epsilon \|x\|^{2}$ also shows that the range is closed: if $Tx_n$ is Cauchy then $x_n$ is Cauchy so it converges to some $x$ and $Tx_n \to Tx$. It now follows that $T$ is also surjective. Now $\langle Tx, x \rangle \geq \epsilon \|x\|^{2}$ shows that $T^{-1}$ is bounded with its norm at most $\frac 1 {\epsilon}$.