[proofplan]
Let $\hbar>0$, let $Q:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ be the position operator defined by $(Qf)(x)=xf(x)$, and let $P:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ be the momentum operator defined by $(Pf)(x)=-i\hbar f'(x)$. Center these operators by subtracting their expectations in the state $\psi$. Cauchy-Schwarz bounds the product of the centered norms below by the absolute value of their inner product, and the imaginary part of that inner product is controlled by the commutator $[Q,P]=i\hbar I$.
[/proofplan]
[step:Center the two observables and apply Cauchy-Schwarz]
Let $Q:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ denote the position operator defined by $(Qf)(x)=xf(x)$, and let $P:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ denote the momentum operator defined by $(Pf)(x)=-i\hbar f'(x)$, where $f\in\mathcal S(\mathbb R)$. Let $I:L^2(\mathbb R)\to L^2(\mathbb R)$ denote the identity operator.
Let
\begin{align*}
\mu_Q=(Q\psi,\psi)_{L^2}
\end{align*}
and
\begin{align*}
\mu_P=(P\psi,\psi)_{L^2}.
\end{align*}
Since $Q$ and $P$ are symmetric on $\mathcal S(\mathbb R)$, the scalars $\mu_Q$ and $\mu_P$ are real. Define
\begin{align*}
u=(Q-\mu_Q I)\psi
\end{align*}
and
\begin{align*}
v=(P-\mu_P I)\psi.
\end{align*}
Then $\|u\|_{L^2}=\Delta_\psi Q$ and $\|v\|_{L^2}=\Delta_\psi P$. By the Cauchy-Schwarz inequality in $L^2(\mathbb R)$,
\begin{align*}
\Delta_\psi Q\,\Delta_\psi P=\|u\|_{L^2}\|v\|_{L^2}\geq |(u,v)_{L^2}|.
\end{align*}
Since the absolute value of a complex number dominates the absolute value of its imaginary part,
\begin{align*}
\Delta_\psi Q\,\Delta_\psi P\geq |\operatorname{Im}(u,v)_{L^2}|.
\end{align*}
[guided]
Let $Q:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ be defined by $(Qf)(x)=xf(x)$, let $P:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ be defined by $(Pf)(x)=-i\hbar f'(x)$, and let $I:L^2(\mathbb R)\to L^2(\mathbb R)$ be the identity operator. The uncertainty product is the product of the norms of the centered position and momentum vectors. Define
\begin{align*}
\mu_Q=(Q\psi,\psi)_{L^2}
\end{align*}
and
\begin{align*}
\mu_P=(P\psi,\psi)_{L^2}.
\end{align*}
Because $Q$ and $P$ are symmetric on $\mathcal S(\mathbb R)$, both expectation values are real. Set
\begin{align*}
u=(Q-\mu_Q I)\psi
\end{align*}
and
\begin{align*}
v=(P-\mu_P I)\psi.
\end{align*}
By definition, $\|u\|_{L^2}=\Delta_\psi Q$ and $\|v\|_{L^2}=\Delta_\psi P$. The Cauchy-Schwarz inequality gives
\begin{align*}
\Delta_\psi Q\,\Delta_\psi P=\|u\|_{L^2}\|v\|_{L^2}\geq |(u,v)_{L^2}|.
\end{align*}
For any complex number $z$, one has $|z|\geq |\operatorname{Im}z|$. Applying this to $z=(u,v)_{L^2}$ gives
\begin{align*}
\Delta_\psi Q\,\Delta_\psi P\geq |\operatorname{Im}(u,v)_{L^2}|.
\end{align*}
Thus it remains only to compute the imaginary part of the centered inner product.
[/guided]
[/step]
[step:Compute the commutator contribution]
The real shifts $\mu_Q I$ and $\mu_P I$ commute with all operators, so
\begin{align*}
[(Q-\mu_Q I),(P-\mu_P I)]=[Q,P].
\end{align*}
For $\psi\in\mathcal S(\mathbb R)$,
\begin{align*}
(QP\psi)(x)=-i\hbar x\psi'(x)
\end{align*}
and
\begin{align*}
(PQ\psi)(x)=-i\hbar(\psi(x)+x\psi'(x)).
\end{align*}
Therefore
\begin{align*}
([Q,P]\psi)(x)=i\hbar\psi(x).
\end{align*}
Equivalently, $[Q,P]\psi=i\hbar\psi$. Since $\|\psi\|_{L^2}=1$,
\begin{align*}
([Q,P]\psi,\psi)_{L^2}=i\hbar.
\end{align*}
The space $\mathcal S(\mathbb R)$ is invariant under $Q$, $P$, $Q-\mu_Q I$, and $P-\mu_P I$, so all compositions below are defined on $\psi$. For symmetric operators $S$ and $T$ on a common invariant domain and a vector in that domain, with the $L^2$ inner product linear in the first variable,
\begin{align*}
2i\,\operatorname{Im}(S\psi,T\psi)_{L^2}=-(([S,T]\psi,\psi)_{L^2}).
\end{align*}
Applying this identity with $S=Q-\mu_Q I$ and $T=P-\mu_P I$ yields
\begin{align*}
|\operatorname{Im}(u,v)_{L^2}|=\frac{\hbar}{2}.
\end{align*}
Combining this with the previous step gives
\begin{align*}
\Delta_\psi Q\,\Delta_\psi P\geq \frac{\hbar}{2}.
\end{align*}
[guided]
Because $\mu_Q$ and $\mu_P$ are real scalars, the scalar multiples $\mu_Q I$ and $\mu_P I$ commute with $Q$ and $P$. Therefore the centered commutator is the same as the original commutator:
\begin{align*}
[(Q-\mu_Q I),(P-\mu_P I)]=[Q,P].
\end{align*}
We now compute $[Q,P]$ on the Schwartz function $\psi$. The position-after-momentum expression is
\begin{align*}
(QP\psi)(x)=-i\hbar x\psi'(x).
\end{align*}
The momentum-after-position expression uses the product rule:
\begin{align*}
(PQ\psi)(x)=-i\hbar(\psi(x)+x\psi'(x)).
\end{align*}
Subtracting these expressions gives
\begin{align*}
([Q,P]\psi)(x)=i\hbar\psi(x).
\end{align*}
Thus $[Q,P]\psi=i\hbar\psi$. Since $\|\psi\|_{L^2}=1$, this implies
\begin{align*}
([Q,P]\psi,\psi)_{L^2}=i\hbar.
\end{align*}
The space $\mathcal S(\mathbb R)$ is invariant under multiplication by $x$, differentiation, and scalar shifts. Therefore it is invariant under $Q$, $P$, $Q-\mu_Q I$, and $P-\mu_P I$, and all compositions used in the commutator are defined on $\psi$. For symmetric operators $S$ and $T$ on a common invariant domain, with the $L^2$ inner product linear in the first variable, the identity
\begin{align*}
2i\,\operatorname{Im}(S\psi,T\psi)_{L^2}=-(([S,T]\psi,\psi)_{L^2})
\end{align*}
follows by expanding the commutator and using symmetry. With $S=Q-\mu_Q I$ and $T=P-\mu_P I$, we obtain
\begin{align*}
|\operatorname{Im}(u,v)_{L^2}|=\frac{\hbar}{2}.
\end{align*}
The previous step already proved
\begin{align*}
\Delta_\psi Q\,\Delta_\psi P\geq |\operatorname{Im}(u,v)_{L^2}|.
\end{align*}
Combining the two displays gives
\begin{align*}
\Delta_\psi Q\,\Delta_\psi P\geq \frac{\hbar}{2}.
\end{align*}
[/guided]
[/step]
