Banach algebra: for $\lambda$ not in the spectrum of $a$, $\Vert (\lambda - a)^{-1} \Vert \ge 1/\text{dist}(\lambda, \sigma(a))$ Here, $\mathcal{A}$ is a Banach algebra, $a \in \mathcal{A}$, and $\lambda \in \rho(a) = \mathbb{C} \setminus \sigma(a)$. Why does this inequality hold?
I can get as far as saying that $\Vert (\lambda - a)^{-1} \Vert \Vert \lambda - a \Vert \le \Vert (\lambda - a)^{-1} (\lambda - a )\Vert = \Vert 1_\mathcal{A} \Vert = 1$, since we assume that the norm on $\mathcal{A}$ is submultiplicative, and we also assume $\Vert 1_\mathcal{A} \Vert = 1.$
Thereby, $$\Vert (\lambda - a)^{-1} \Vert \ge \frac{1}{\Vert \lambda - a \Vert }.$$ But I can't seem to get from this to having $\text{dist}(\lambda, \sigma(a)) = \inf\{\Vert a - z \Vert: z \in \sigma(a)\}$ on the denominator of the RHS.
 A: Let $\phi \colon \mathcal{A} \to \mathbb{C}$ be a (unital) algebra homomorphism. Then we know that $\phi$ is a continuous linear form with $\lVert \phi\rVert \leqslant 1$, and $\phi(a) \in \sigma(a)$. Moreover, we have
$$\sigma(a) = \{ \phi(a) : \phi \text{ is a complex homomorphism}\}.$$
Since $\sigma(a)$ is compact, there is a complex homomorphism $\psi$ such that $\lvert \lambda - \psi(a)\rvert = \operatorname{dist}(\lambda,\sigma(a))$.
Then we have
$$1 = \psi(1) = \psi\bigl((\lambda - a)(\lambda - a)^{-1}\bigr) = \bigl(\lambda - \psi(a)\bigr)\psi\bigl((\lambda - a)^{-1}\bigr),$$
whence
$$\bigl\lVert (\lambda - a)^{-1}\bigr\rVert \geqslant \bigl\lvert \psi\bigl((\lambda - a)^{-1}\bigr)\bigr\rvert = \frac{1}{\lvert \lambda - \psi(a)\rvert} = \frac{1}{\operatorname{dist}(\lambda,\sigma(a))}.$$

A more elementary proof, using less machinery:
If $x \in \mathcal{A}$ is invertible, then $x - y$ is invertible for every $y\in \mathcal{a}$ with $\lVert y\rVert < \lVert x^{-1}\rVert^{-1}$, for then $\lVert x^{-1}y\rVert < 1$, and the Neumann series
$$(e - x^{-1}y)^{-1} = \sum_{k = 0}^{\infty} (x^{-1}y)^k$$
converges. Now let $\mu \in \sigma(a)$ with $\lvert \lambda - \mu\rvert = \operatorname{dist}(\lambda,\sigma(a))$. Then
$$(\lambda - a) - (\lambda - \mu) = \mu - a$$
is not invertible, hence
$$\operatorname{dist}(\lambda,\sigma(a)) = \lvert \lambda - \mu\rvert \geqslant \lVert (\lambda - a)^{-1}\rVert^{-1}.$$
