 Solution

Since $a-1$ and $b-1=(b-a)+(a-1)$ are positive, their spectra are nonnegative. Translation of the spectrum of an element gives $\sigma (a),\sigma (b)\subseteq [1,\infty )$, so both elements are invertible. Their inverses and inverse square roots are positive by the continuous functional calculus. Also $1-a^{-1}$ is positive because the scalar function $1-t^{-1}$ is nonnegative on $\sigma (a)$.

For the order reversal, setThe inequality uses congruence preserves positivity in a C-star algebra. continuous functional calculus gives $k^{-1}\le 1$. Sinceanother congruence showsConsequentlyHere $b^{-1}$ is strictly positive in the stronger invertible sense: $b^{-1}\ge \Vert b{\Vert }^{-1}1>0$, because its spectrum is the set of reciprocals of the positive spectral values of $b$. This is inversion reverses the order of strictly positive elements; no commutativity of $a,b$ is required.