If $a\ge 0$ and $b\ge 0$, then $\sigma(ab)\subset\mathbb{R}^+$. This is an exercise in Murphy's book:


  Let $A$ be a unital $C^*$-algebra and $a,b$ are positive elements in $A$. Then $\sigma(ab)\subset\mathbb{R}^+$.


The problem would be trivial if the algebra is abelian. On the other hand I do not have a clue for the non-abelian case. I guess one needs to use the fact $\sigma(ab)\cup\{0\}=\sigma(ba)\cup\{0\}$ and then maybe use some algebraic manipulation.

Anyway, I wonder whether someone has a hint on this. I guess I am missing a trick. 

Better though, maybe someone has some general insight on the many techniques concerning positive elements and approximate identities. For me they all seem very tricky and mysterious. For instance, how can they think of those strange functions when proving the existence of approximate identities and quasicentral approximate identities.

Thanks!
A common trick is to write a positive element of a C*-algebra as a square of a positive element:  $b=\sqrt b\sqrt b$.  Thus $\sigma(ab)\cup\{0\}=\sigma(a\sqrt b\sqrt b)\cup\{0\}=\sigma(\sqrt b a\sqrt b)\cup \{0\}$.  If you haven't already seen it, you can think about why if $a$ is positive, then $xax^*$ is positive for all $x$.
$$\sigma(ab)\subset \sigma(a)\sigma(b)\subset \mathbb R^+$$