= Solution
For a normalized state and a <self-adjoint operator> $Q$ with finite second moment, the <quantum uncertainty> is defined by $\Delta Q=\|(Q-\langle Q\rangle)\psi\|$. Since $\langle Q\rangle$ is real,
$$
(\Delta Q)^2=\langle(Q-\langle Q\rangle)^2\rangle
=\boxed{\langle Q^2\rangle-\langle Q\rangle^2}.
$$
In an energy <eigenstate> of the <quantum harmonic oscillator>,
$$
E=\langle H\rangle
=\frac{(\Delta p)^2+\langle p\rangle^2}{2m}
+\frac{m\omega^2}{2}\bigl((\Delta x)^2+\langle x\rangle^2\bigr).
$$
Discarding the nonnegative squared means proves the stated lower bound. The <Heisenberg uncertainty principle> is $\Delta x\,\Delta p\geq\hbar/2$, following from $[x,p]=i\hbar$. Applying the <arithmetic-geometric mean inequality> to the two variance contributions gives
$$
E\geq\frac{(\Delta p)^2}{2m}+\frac{m\omega^2(\Delta x)^2}{2}
\geq\omega\Delta p\Delta x\geq\boxed{\frac{\hbar\omega}{2}}.
$$
Here $\omega>0$ is the oscillator frequency. The bound is the ground-state energy; equality requires zero means and the appropriate minimum-uncertainty width.
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