Background (question below)
Let $A$ be a Banach algebra, i.e. $A$ is a complex vector space with norm $\left\Vert \cdot\right\Vert$ and multiplication satisfying $\left\Vert ab\right\Vert \leq\left\Vert a\right\Vert \left\Vert b\right\Vert$ such that $A$ is complete with respect to the norm. When $A$ is unital, i.e. there is a unit $1 \in A$ such that $1a = a1 = a$ for any $a \in A$, define the spectral radius of $a \in A$ as
$r\left(a\right)=\sup_{\lambda\in\sigma\left(a\right)}\left|\lambda\right|$
where
$\sigma\left(a\right)=\left\{ \lambda\in\mathbb{C}:\lambda1-a\text{ not invertible in }A\right\} $
is the spectrum of $a$ in $A$. When $A$ is not unital, that is $A$ has no unit, let $\widetilde{A}$ be the unitization of $A$, which is the unital Banach algebra $\left\{ a+\lambda:a\in A,\lambda\in\mathbb{C}\right\} $ with unit $0+1$, addition
$\left(a+\lambda\right)+\left(b+\mu\right)=\left(a+b\right)+\left(\lambda+\mu\right)$
and multiplication
$\left(a+\lambda\right)\left(b+\mu\right)=\left(ab+\mu a+\lambda b\right)+\lambda\mu$
and define $\sigma(a)$ and $r(a)$ the same except within $\widetilde{A}$ instead of $A$. $A$ is abelian if $ab=ba$ for any $a,b\in A$, and it is true that $\widetilde{A}$ is abelian when $A$ is.
Question
Let $A$ be a Banach algebra, and let $a,b\in A$ such that $ab=ba$. When $A$ is unital and abelian, it follows (e.g. Gelfand Representation) that
$r\left(a+b\right)\leq r\left(a\right)+r\left(b\right)$ and $r\left(ab\right)\leq r\left(a\right)r\left(b\right)$
If $A$ is not necessarily unital and abelian, one can consider the closed subalgebra $B$ of $\widetilde{A}$ generated by $\{1, a, b\}$, which is unital and abelian because $ab=ba$, and conclude that, restricted to $B$, the above inequalities hold (using the abelian case). Is this restriction the best you can say for these inequalities holding in this case?