= 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, set
$$
k=a^{-1/2}ba^{-1/2}=1+a^{-1/2}(b-a)a^{-1/2}\geq1.
$$
The inequality uses <congruence preserves positivity in a C-star algebra>. <continuous functional calculus> gives $k^{-1}\leq1$. Since
$$
b^{-1}=a^{-1/2}k^{-1}a^{-1/2},
$$
another congruence shows
$$
a^{-1}-b^{-1}=a^{-1/2}(1-k^{-1})a^{-1/2}\geq0.
$$
Consequently
$$
\boxed{0<b^{-1}\leq a^{-1}\leq1.}
$$
Here $b^{-1}$ is strictly positive in the stronger invertible sense: $b^{-1}\geq\|b\|^{-1}1>0$, 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.
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