Given some pure state $|\psi\rangle$ we have the following bound on the uncertainty for two non-commuting operators $A$ and $B$ \begin{equation} \sigma_A\sigma_B\geq\left|\frac 1{2i}\langle[A,B]\rangle\right|^2. \end{equation}
My question is this. Given any pure state, can we always find two operators which saturate this bound? Bonus question, if we have a pure state and an operator $A$, can we always find some operator $B$ which saturates this bound?
The answer is yes, to both questions.
For the first we can choose \begin{align} A = |\psi\rangle\langle\psi|, \end{align} and let $B$ be anything we like and it'll saturate the bound (because the left and right hand sides are both $0$). There are some obvious choices for what to choose $B$ to be, for example a projector onto a state which is neither orthogonal to $|\psi\rangle$ nor proportional to it works.
For the bonus question for any state $\psi$ and self-adjoint operator $A$ you can choose $B=I$ as Tobias Fünke mentioned and it saturates the bound. If you want $B$ to not commute with $A$ then you can make some less trivial examples. For example if $\psi$ is not an eigenstate of $A$ then choose $B=|\psi\rangle\langle\psi|$ and it works, on the other hand if $\psi$ is an eigenstate of $A$ then again any self-adjoint operator works, and you can cook up examples which do or do not commute with $A$ as you choose.
It might be worth noting that in all these examples the "bound is saturated" because the left and right hand sides are both zero. To come up with less trivial examples you can note that if you have any state and observables $(\psi, A,B)$ such that the bound is saturated, and you want an example for some different state $\phi$, then you can pick any unitary $U$ which maps $\psi$ to $\phi$ and the operators $UAU^\dagger$ and $U BU^\dagger $ are observables for which the bound is saturated by $\phi = U\psi$
