I would like some feedback on my reasoning for the following problem:
Let $A$ be a $\mathbb{R}$-Banach algebra. Let $x \in A$. Show that there exists $a$,$b \in \mathbb{R}$ such that $(x + a1)^2 + b^2 1$ is not invertible.
Here $1$ is the multiplicative unit of $A$.
(Tentative) Proof:
Let $P(x) = (x + a1)^2 + b^2 1$.
We know that a polynomial expression in $x \in A$ is invertible iff $\textrm{Sp}_A(x) = \{ \lambda \in \mathbb{R}, (x-\lambda) \notin A^{-1}\}$ does not contain any roots of $P$. So, to show that $P(x)$ is not invertible, we must find conditions on $a$ and $b$ such that $P$ has at least one root in $\textrm{Sp}_A(x)$.
Suppose that $x$ is not invertible. Then it is sufficient to choose $a=b=0$.
Now suppose that $x$ is invertible, with spectral radius $\rho$. (most of the proofs I have seen of the spectral radius formula rely on complex Banach algebras but I believe it generalizes to the real case). The roots of $P(x)$ are $-a + ib$ and $-a - ib$. So we need to set $b=0$ and take $|a| < \rho$. Then either root is in $\textrm{Sp}_A(x)$ and the claim is proved.