= Solution
Although $ab$ need not be self-adjoint, it is similar to a <Positive element of a C-star algebra>. Namely,
$$
a^{-1/2}(ab)a^{1/2}=a^{1/2}ba^{1/2}=:d.
$$
By <congruence preserves positivity in a C-star algebra>, $d\geq0$. In fact $b\geq1$ gives
$$
d-a=a^{1/2}(b-1)a^{1/2}\geq0,
$$
so $d\geq a\geq1$. Similar elements have the same <spectrum of an element>, because $s^{-1}(x-\lambda1)s$ is invertible exactly when $x-\lambda1$ is. Therefore the stronger conclusion is
$$
\boxed{\sigma(ab)=\sigma(d)\subseteq[1,\infty)\subseteq\mathbb R^+.}
$$
Positivity of the spectrum is being asserted, not positivity of the generally non-self-adjoint product itself.
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