= Solution
Set $u=A\psi$, $v=B\psi$ and $z=\langle u,v\rangle$. The <self-adjointness> of $A,B$ implies
$$
z=\langle\psi,AB\psi\rangle,\qquad \overline z=\langle\psi,BA\psi\rangle,\qquad \mathbb E_\psi(C)=-i(z-\overline z)=2\operatorname{Im}z.
$$
The <Cauchy-Schwarz inequality> followed by $|z|^2\geq(\operatorname{Im}z)^2$ gives
$$
\mathbb E_\psi(A^2)\mathbb E_\psi(B^2)=\|u\|^2\|v\|^2\geq|z|^2\geq(\operatorname{Im}z)^2=\boxed{\frac14\bigl(\mathbb E_\psi(C)\bigr)^2}.
$$
Both inequalities must be equalities. The exact conditions are \b[$A\psi$ and $B\psi$ are linearly dependent, and $\operatorname{Re}\langle A\psi,B\psi\rangle=0$]. Equivalently, equality holds if either vector is zero; when both are nonzero it holds precisely when
$$
\boxed{B\psi=i\lambda A\psi\quad\text{for some real }\lambda\ne0.}
$$
The zero-vector cases matter: they saturate the inequality but cannot all be expressed by solving for $B\psi$ as a multiple of $A\psi$. The real-part condition can also be written $\mathbb E_\psi(AB+BA)=0$, using the <anticommutator>. This is the <uncentered second-moment commutator inequality>.
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