Let $A$ be unital $C^*$-algebra, and $c$ be a positive element in $A$. Then $a$ is invertible if and only if $a \ge \epsilon 1$ for some $\epsilon > 0$.
I think I was able to prove the backward implication. Let $C^(a)$ be the commutative unital $C^*$-subalgebra generated by $a$. By the Gelfand-Naimark theorem, $C^*(a)$ is $*$-isomorphic to the space of continuous functions on some compact topological space; call the isomorphism $\phi$. If $a \ge \epsilon 1$, then $a - \epsilon 1 \ge 0$ and therefore $\phi (a) - \epsilon > 0$ or $\phi (a) \ge \epsilon$. But this means that $\phi(a)$ is an invertible function, and this inverse can be pulled-back via $\phi$ get furnish an inverse for $a$ itself.
Here is what I have for the forward implication. Suppose that $a$ is invertible. Then because $a$ is invertible and positive, and because the spectrum $\sigma (a)$ is nontrivial, there exists $\epsilon \in \sigma (a)$ such that $\epsilon > 0$ and $a - \epsilon 1$ is not invertible...I want to argue that $a - \epsilon 1 \ge 0$, but I don't see how to.
Let's do a forward implication instead of contradiction. Since $a$ is invertible then there is a neighborhood of $0\in U\subset\mathbb{C}$ such that $a-t1$ is invertible for all $t\in U$. Take $0<\epsilon\in U$. Then $a-\epsilon1$ is invertible. Now, the Gelfand transform $\hat{a}-\epsilon\hat{1}$ is continuous and takes all values $\sigma(a)-\epsilon$. Since $0\notin \sigma(a)-\epsilon$, $\sigma(a)\subset(0,+\infty)$ and $[\epsilon,0]\cap\sigma(a)=\emptyset$, then $\sigma(a)-\epsilon\subset (0,+\infty)$.
Thereofore, $a-\epsilon1\geq0$.