Greatest open ball of invertible elements in a Banach algebra Let $a$ be an invertible element of a Banach algebra $A$. Then we know that also each $a+b$ with $b\in A$ and $||b||<||a^{-1}||^{-1}$ is invertible. Now my question is whether $B_{||a^{-1}||^{-1}}(a)$ is already the greatest open ball of invertible elements around $a$?
Or, in other words, is the following true:
For all invertible $a\in A$ there is a $b\in A$ such that $||b||=||a^{-1}||^{-1}$ and $a+b$ is not invertible.
 A: We'll analyse some cases where we have equality.
Consider $A = M_{n}(\mathbb{C})$ with the operator norm induced by the $L^2$ norm on $\mathbb{C}^n$. The norm of a matrix $a$ is $\max ( \sigma_l)$ the largest of the singular values of $a$. Assume that $a$ is invertible. Then the norm of $a^{-1}$ is $\max( \sigma_l^{-1})$. Therefore
$$||a|| = \max (\sigma_l)$$
$$||a^{-1}||^{-1} = \min (\sigma_l)$$
Consider the singular value decomposition of $a$. It is clear now that we can substract from $a$ $ \min (\sigma_l) \cdot$ operator of norm $1$ and obtain a noninvertible element.
It is not hard to show that the property holds for finite dimensional $C^*$ algebras by considering an imbedding in some matrix algebra. In fact, it holds for all $C^*$ algebras.
Consider $ A$ a $C^*$ algebra and $a \in A$.
Assume first that $a$ is hermitian that is $a = a^*$. Then $||a|| = \rho(a)$ the spectral radius of $a$. If $a$ is also invertible then 
$$||a^{-1}||^{-1} = \rho (a^{-1})^{-1} = (\max \{ |\lambda \| | \lambda \in \sigma(a^{-1})\})^{-1}= \min\{ |\lambda| \ | \lambda \in \sigma(a) \}$$
Take $\lambda_0 \in \sigma(a)$ of minimal absolute value. By the above $|\lambda_0| = ||a^{-1}||^{-1}$. Observe that $a - \lambda_0 \cdot 1$ is not invertible.
Let $a\in A$ invertible, not necessarily hermitian. Consider the polar decomposition
$a = u \cdot h$ with $u$ unitary, $h$ hermitian. We have 
$||a|| = ||h||$ and $||a^{-1}|| = ||h^{-1}||$. By the above there exists $\lambda_0$ of absolute value $||h^{-1}||^{-1}$ so that $h - \lambda_0 \cdot 1$ is not invertible. If follows that $a - \lambda_0 \cdot u$ is not invertible. 
Let $X$ be a Banach space and $A$ the Banach algebra of linear bounded operators on $X$ with the operator norm. Let $a\in A$ invertible. For any $\epsilon > 0$ there exists $x \in X$ so that $||y|| \le ||a^{-1}||^{-1} + \epsilon$ and $x \colon = a^{-1} y $ is of norm $1$.
If $X$ is finite dimensional we can even take $\epsilon =0$. Let now $\phi \colon X \to \mathbb{K}$ a linear functional of norm $1$ so that $\phi(x) = 1$. Then the linear operator $b \colon = \phi \otimes y$ is of norm $||y||$ and $b x = y$. Therefore $x = a^{-1} b x$ and so 
$$ax - b x = (a-b)x=0$$ 
 We have exhibited a  $b$ of norm $\le ||a^{-1}||^{-1}$ so that $a-b$ is not invertible. 
Observation: Your question seem to be this: For a given invertible element of a Banach algebra $A$ does the distance from $a$ to the set of noninvertible elements equal $||a||^{-1}$ and is the distance achieved. By the above, both have answer yes for $C^*$ algebras and for the algebra of operators on a (finite dimensional) Banach space. 
Here is an example of an algebra where in general we do not have equality. Let 
$A$ be the Wiener algebra, the algebra of complex valued functions with period $2\pi$ and with absolute convergent Fourier series. $A$ consists of functions $f$ of form
$$f(\theta) = \sum_{n \in \mathbb{Z}} \alpha_n e^{i n \theta} $$
such that $\sum_{n \in \mathbb{Z}} |\alpha_n| < \infty$. It turns out that if a function $f$ in $A$ has no zeroes then the Fourier series of its inverse has again coefficients in $l^1(\mathbb{Z})$ and hence is invertible in $A$. ( this is not a tautology). It follows that the spectrum of an element equals the set of values of that element considered as a function of $\theta$. Let now $a \in A$ a function, $a$ invertible. It follows that the distance from $a$ to the set of noninvertible elements ( i.e. the elements which have zero as a value) equals $\min_{\theta} a(\theta)$ in other words, equals $\rho(a^{-1})^{-1}$ where $\rho(\cdot)$ is the spectral radius. Now it's enough to exhibit an invertible $a$ such that $\rho(a^{-1}) < ||a^{-1}||$. This happens generically when $a^{-1}$ has more than two terms in its Fourier expansion. Example: $a = (3 + e^{i \theta} - e^{2 i \theta})^{-1}$.
We see that $\rho(a^{-1})  = \max_{\theta} |3 + e^{i \theta} - e^{2 i \theta}| < 3 + 1 + 1 = ||a^{-1}||$. ( the maximum is $2 \sqrt{\frac{13}{3}} = 4.16...$ but that is not very important at this point ) 
$\bf{Added:}$ From the above reasoning for the Wiener algebra it should be become clear that for a commutative Banach algebra (assume complex for simplicity) the distance from a given invertible element $a$ to the noninvertibles is $\rho(a^{-1})^{-1}$. In order to get an  algebra $A$ and an invertible $a$ in $A$ for which we have this distance strictly between $||a^{-1}||$ and $\rho(a^{-1})^{-1}$ we stay away from commutative ones. But now we remember that for any group $G$ (with some measure) we can consider the algebra of $L^1$ functions on $G$. Let's consider a finite and noncommutative $G$. The simplest example is $S_3$. It turns out that here for a generic invertible this distance is strictly in between the above stated bounds.Note: In order to understand what the invertibles are one should look at the so called "group determinant". So now we have a family of Banach algebras (finite dimensional) for which the inequality is strict.
