Let $A$ be a $C^{\ast}$-algebra and $c$ be an invertible positive element in $A.$ Then there exists $\varepsilon \gt 0$ such that $c \geq \varepsilon \mathbf 1.$
My Attempt $:$ I guess that $0 \lt \varepsilon \leq \text {inf} \left (\sigma(c) \right ).$ For that what I need to show is that $$c - \text {inf} \left (\sigma (c) \right ) \mathbf 1 \geq 0.$$ But It is obvious since $c - \text {inf} \left (\sigma (c) \right ) \mathbf 1$ is clearly self-adjoint and $$\sigma \left (c - \text {inf} \left (\sigma (c) \right ) \mathbf 1 \right ) = \left \{\lambda - \text {inf}\ \left (\sigma (c) \right )\ \bigg |\ \lambda \in \sigma(c) \right \} \subseteq [0, \infty).$$
Does it make sense what I argued above? Could anyone please check it?
Thanks for your time.
The argument is fine. The main point, though, that you shouldn't omit, is that because $c$ is invertible you have that $0\not\in\sigma(c)$, and then because $\sigma(c)$ is compact, $\inf\sigma(c)>0$.