I'm looking at Takesaki's Theory of Operator Algebras I, chapter I.4 exercise 3. The following statement is implicitly used:
Let $A$ be a unital $C^*$-algebra and $a, b\in A$ selfadjoint. Then $\Vert a+ib\Vert\geq\Vert a\Vert$.
I know how to prove this by taking for granted that for selfadjoint $a$, the following formula holds: $$ \Vert a\Vert = \sup_{\omega\text{ state}}\vert \omega(a)\vert. $$ Then simply $\Vert a + ib\Vert\geq \vert\omega(a+ib)\vert = \sqrt{\omega(a)^2+\omega(b)^2}\geq \vert\omega(a)\vert$ and take $\sup$ over $\omega$. However, the exercise appears very early on in the book before states on $C^*$-algebras are even mentioned, so I'm assuming there is a more elementary proof. By elementary here I'm admitting the use of various facts about the spectral radius in $C^*$-algebras and the continuous functional calculus (and of course standard Banach space results). Does anyone know how to do this? I've been trying to directly show that $\rho(a+ib)\geq\rho(a)$ ($\rho$ spectral radius), but haven't gotten anywhere.
The inequality looks scary when written that way. But it's just the inequality $$ \|\operatorname{Re}x\|=\frac12\|x+x^*\|\leq\frac12(\|x\|+\|x^*\|)=\|x\|. $$