= Solution
The <expectation value> of an <observable> is the mean outcome of repeated measurements on identically prepared systems in the normalized state. For a <self-adjoint> $Q$ it is real and equals $\langle\psi,Q\psi\rangle$. Assume the state belongs to the domains needed for the second moments and the commutator below, so all these quantities exist.
For real $\lambda$, positivity of the squared norm gives
$$
0\leq\|(Q+i\lambda P)\psi\|^2=\langle Q^2\rangle+\lambda^2\langle P^2\rangle+i\lambda\langle[Q,P]\rangle.
$$
The commutator expectation is purely imaginary, so this is a real quadratic in $\lambda$. If $\langle P^2\rangle>0$, its minimum is nonnegative exactly when its discriminant is nonpositive. If $\langle P^2\rangle=0$, nonnegativity for all real $\lambda$ forces the linear coefficient to vanish, yielding the same conclusion. Thus
$$
\boxed{\langle Q^2\rangle\langle P^2\rangle\geq\frac14|\langle[Q,P]\rangle|^2.}
$$
Replace the observables by their centred versions $Q-\langle Q\rangle I$, $P-\langle P\rangle I$. Their commutator is unchanged, and their second moments are the <variances> $(\Delta Q)^2,(\Delta P)^2$. Taking square roots gives the <Robertson uncertainty principle>
$$
\boxed{\Delta Q\,\Delta P\geq\frac12|\langle[Q,P]\rangle|.}
$$
For the <quantum harmonic oscillator>, $p=-i\hbar\partial_x$ gives $[x,p]\psi=i\hbar\psi$. The uncentred inequality above consequently gives $\langle x^2\rangle\langle p^2\rangle\geq\hbar^2/4$. The <arithmetic-geometric mean inequality> now yields
$$
\langle H\rangle=\frac{\langle p^2\rangle}{2m}+\frac{m\omega^2\langle x^2\rangle}{2}\geq\omega\sqrt{\langle x^2\rangle\langle p^2\rangle}\geq\boxed{\frac12\hbar\omega}.
$$
This is a bound for every normalized state of finite oscillator energy, not just an eigenstate. It is attained by the centred Gaussian ground state, for which $\langle p^2\rangle=m^2\omega^2\langle x^2\rangle$ and the uncertainty product equals $\hbar/2$.
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