= Solution
Introduce the <dimensionless> width $s=\hbar\alpha/(m\omega)>0$. Multiplying the <variances> gives
$$
\boxed{(\Delta x')_t^2(\Delta p')_t^2
=\frac{\hbar^2}{4}\left[
1+\frac14(s-s^{-1})^2\sin^2(2\omega t)\right]
\ge\frac{\hbar^2}{4}.}
$$
The minimum holds \b[at all times precisely when $\alpha=m\omega/\hbar$], the oscillator ground-state width. For any other $\alpha$, it is still attained at the isolated quarter-period times $t=n\pi/(2\omega)$, including zero. At a fixed generic time, minimizing over $\alpha$ again selects $s=1$.
The excess product has a <covariance> interpretation:
$$
C_{xp}(t)=\frac12\langle x'p'+p'x'\rangle_t
=\frac{\hbar}{4}(s-s^{-1})\sin2\omega t,\qquad
(\Delta x')^2(\Delta p')^2-C_{xp}^2=\frac{\hbar^2}{4}.
$$
Thus the <breathing Gaussian state in a harmonic oscillator> saturates the <Robertson-Schrödinger uncertainty relation> throughout, even when its position-momentum product exceeds the lower bound in the <Heisenberg uncertainty principle>.
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