I am reading a proof of this theorem:
If $a,b$ are positive elements of a $C^\ast$ algebra and $a \le b$ then $a^{1/2}\le b^{1/2}$.
I don't understand one step in the proof. I understand this: Let $t > 0$ and $c,d$ be such that $c + i d = (t + b + a)(t + b - a)$. Then $c \ge t^2$ therefore $c$ is positive and invertible.
Given this, why is $1 + i c^{-1/2}dc^{-1/2}$ invertible?
Here is the argument:
Since $d$ is selfadjoint, so is $c^{-1/2}dc^{-1/2}$. Then the spectrum of $1+ic^{-1/2}dc^{-1/2}$ is of the form $$ \{1+i\lambda:\ \lambda\in\sigma(c^{-1/2}dc^{-1/2})\}. $$ The $\lambda$ are real, so the spectrum does not contain $0$, and thus $1+ic^{-1/2}dc^{-1/2}$ is invertible.
As this equals $c^{-1/2}(c+id)c^{-1/2}$ we conclude that $c+id$ is invertible. Now one can conclude that $t+b-a$ is invertible. As this happens for all $t>0$, $b-a\geq0$, i.e. $a\leq b$.