= Solution
For the <quantum harmonic oscillator>, split the two <expectation values> into their <variances> and squared means:
$$
E=\frac{(\Delta p)^2}{2m}+\frac{m\omega^2(\Delta x)^2}{2}+\frac{\langle p\rangle^2}{2m}+\frac{m\omega^2\langle x\rangle^2}{2}.
$$
The last two terms are nonnegative, so
$$
\boxed{E\geq\frac{(\Delta p)^2}{2m}+\frac{m\omega^2(\Delta x)^2}{2}.}
$$
Apply the <arithmetic-geometric mean inequality> to the two remaining nonnegative terms, then the <Heisenberg uncertainty principle>:
$$
E\geq2\sqrt{\frac{(\Delta p)^2}{2m}\frac{m\omega^2(\Delta x)^2}{2}}=\omega\Delta p\Delta x\geq\boxed{\frac{\hbar\omega}{2}}.
$$
The argument applies to any normalized state with finite oscillator energy; the stated <stationary state> is a particular case. Equality requires zero position and momentum means, minimum <quantum uncertainty>, and balanced kinetic and potential contributions, as in the oscillator <ground state>.
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