= Solution
For a normalized state, write
$$
A'=\widehat A-\langle\widehat A\rangle_\psi,
\qquad
B'=\widehat B-\langle\widehat B\rangle_\psi.
$$
Then $\Delta_\psi A=\|A'\psi\|$ and $\Delta_\psi B=\|B'\psi\|$. The Schwarz inequality gives
$$
(\Delta_\psi A)(\Delta_\psi B)
\geq |(A'\psi,B'\psi)|
\geq |\operatorname{Im}(A'\psi,B'\psi)|.
$$
Since $A'$ and $B'$ are Hermitian and their commutator equals $[\widehat A,\widehat B]$,
$$
2i\operatorname{Im}(A'\psi,B'\psi)
=(\psi,[\widehat A,\widehat B]\psi).
$$
Therefore
$$
(\Delta_\psi A)(\Delta_\psi B)
\geq\frac12\left|(\psi,[\widehat A,\widehat B]\psi)\right|,
$$
which is the <Robertson uncertainty principle>.
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