I am working on problem 2.a in Murphy's $\textit{$C^\ast$-Algebras and Operator Theory}$, which asks to show that for positive elements $a, b$ of a unital $C^\ast$-algebra $A$, $\sigma(ab) \subset [0, \infty)$. By the definition given in this textbook, $a \in A$ is positive if $a$ is hermitian and $\sigma(a) \subset [0, \infty)$.
It's true that, if $a$ and $b$ commute, then $ab$ is positive, as: $$ab = (a^{1/2}b^{1/2})^\ast (a^{1/2}b^{1/2}),$$ from which it follows that $\sigma(ab) \subset [0, \infty)$.
Then, to solve the given problem, I invoke the following argument: $$\sigma(ab) \cup \{0\} = \sigma((a^{1/2}b^{1/2})^\ast (a^{1/2}b^{1/2})) \cup \{0\} \subset [0, \infty),$$
from which it follows that $\sigma(ab) \subset [0, \infty)$.
My question is this: we have that, for arbitrary positive $a, b \in A$, $a$ and $b$ are hermitian, from which it follows that $ab$ is hermitian. Furthermore, by the above argument, it follows that $\sigma(ab) \subset [0, \infty)$. Does it not follow from this that $ab$ is positive for arbitrary positive $a,b$?
It is not true that the product of hermitian operators is hermitian in general. For instance try $$ \begin{bmatrix} 1&0\\0&2\end{bmatrix} \begin{bmatrix} 1&2\\2&3\end{bmatrix} =\begin{bmatrix} 1&2\\4&6\end{bmatrix} $$ Actually, for $a,b$ hermitian we have that $ab$ is hermitian if and only if $ab=ba$.