Deriving Uncertainty Relation for Canonical Commuting Observables from Schrödinger Uncertainty Relation In case the names are not standard:
\begin{equation}
 \sigma_{\hat{A}}^{2}\sigma_{\hat{B}}^{2}
  \geq \left\vert
    \frac{1}{2i} \left\langle \left[\hat{A},\hat{B}\right] \right\rangle
    \right\vert^{2}
  +  \left\vert
    \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle
  -  \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle
    \right\vert^{2}
\label{eq:1}
\tag{Scrhödinger's Uncertainty Relation}
\end{equation}
\begin{equation}
 \sigma_{\hat{A}}\sigma_{\hat{B}}
  \geq \frac{\left\vert
    \left\langle \left[\hat{A},\hat{B}\right]\right\rangle
    \right\vert}{2}
\label{eq:2}
\tag{Robertson's Uncertainty Relation}
\end{equation}
Given $\hat{Q},\hat{P}$ observables such that $\left[\hat{Q},\hat{P}\right] = i\hbar$
\begin{equation}
 \sigma_{\hat{Q}}\sigma_{\hat{P}}
  \geq \frac{\hbar}{2}
\label{eq:3}
\tag{Uncertainty Relation for Cannonical Commuting Observables}
\end{equation}
For \ref{eq:3}, we know that the equality is reached, meaning $\left[\hat{Q},\hat{P}\right] = i\hbar$ is a sufficient condition for
\begin{equation}
\left\vert
    \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle
  -  \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle
    \right\vert^{2}
        =       0
\end{equation}
on \ref{eq:1}.
This is what I'm not able to prove.
An ideal answer would contain the necessary conditions for this - that the \ref{eq:1} turns out to be \ref{eq:2} and the equality is reached - to happen.
PD:

*

*I'm aware of how to derive \ref{eq:3} from the commutation relation by other means. I'm interested in what is mentioned above.

*In case it sounds like homework, hints or references are highly appreciated.

 A: Robertson's Uncertainty Relation will hold even if
$$\begin{equation}
\left\vert
    \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle
  -  \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle
    \right\vert^{2}
        \neq       0
\end{equation}$$
Because
$$\sigma _{A}^{2}\sigma _{B}^{2}\geq \left|{\frac {1}{2}}\langle \{{\hat {A}},{\hat {B}}\}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle \right|^{2}+\left|{\frac {1}{2i}}\langle [{\hat {A}},{\hat {B}}]\rangle \right|^{2}$$
$$\Rightarrow\sigma _{A}^{2}\sigma _{B}^{2}\geq \left|{\frac {1}{2i}}\langle [{\hat {A}},{\hat {B}}]\rangle \right|^{2}$$
since both of them are positive and if something is bigger than the addition of two positive real numbers then it is also bigger than each positive number seperately.
$$\Rightarrow\sigma _{A}^{2}\sigma _{B}^{2}\geq \left|{\frac {1}{2}}\langle [{\hat {A}},{\hat {B}}]\rangle \right|^{2}$$
$$\Rightarrow\sigma _{A}\sigma _{B}\geq {\frac {1}{2}}\left|\langle [{\hat {A}},{\hat {B}}]\rangle \right|$$
Scrhödinger's Uncertainty Relation is a stronger condition and if it satisfied then Robertson's Uncertainty Relation directly follows from that.
But Scrhödinger's Uncertainty Relation is satisfied only if the vector $ {\hat {B}}|\Psi \rangle $ is in the domain of the unbounded operator $ {\hat {A}}$, which is not always the case. A counter example is given here in Wikipedia.
Also refer to chapter 12 of Quantum Theory for Mathematicians by Brian C. Hall 2013 for more information.
Edit: Based on the comment I understood that you are asking the condition for a state in which both inequalities will imply each other and both equalities are satisfied.
Let there be some states $|\Psi_i \rangle$ for which Scrhödinger's Uncertainty Relation satisfies the equality and let there be some states $|\psi_i \rangle$ for which Robertson's Uncertainty Relation satisfies the equality. If $|\Psi_i \rangle$ and $|\psi_i \rangle$ have a common state, then for that state both relations satisfy with equality.
Since such a state satisfies both equalities it also satisfies:
$\begin{equation}
\left\vert
    \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle
  -  \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle
    \right\vert^{2}
        =       0
\end{equation}$
But even if $\begin{equation}
\left\vert
    \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle
  -  \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle
    \right\vert^{2}
        =       0
\end{equation}$ for a state that doesn't imply that for that state both uncertainty relations are satisfied with equality.
