= Solution
Identify the nonrelativistic <mass> with the <rest mass>, $m=m_0$. Expanding the <relativistic kinetic energy> in $p^2/(m^2c^2)$ gives
$$
T=\frac{p^2}{2m}-\frac{p^4}{8m^3c^2}+O(c^{-4}),\qquad H'=-\frac{p^4}{8m^3c^2}.
$$
First-order <perturbation theory> requires $\Delta E_0=\langle0|H'|0\rangle$. Put $\alpha=\sqrt{m\hbar\omega/2}$, so $p=i\alpha(a^\dagger-a)$. The annihilation relations are $a|0\rangle=0$ and $\langle0|a^\dagger=0$. In particular, $a^\dagger|0\rangle$ does \b[not] vanish. A short calculation retaining the operator ordering gives
$$
p^2|0\rangle=\alpha^2\bigl(|0\rangle-\sqrt2|2\rangle\bigr),\qquad
\langle0|p^4|0\rangle=\|p^2|0\rangle\|^2=3\alpha^4.
$$
Equivalently, of the ordered words in $(a^\dagger-a)^4$, only $aa^\dagger aa^\dagger$ and $aa a^\dagger a^\dagger$ have nonzero ground-state expectation values, respectively $1$ and $2$. Words with a <creation operator> on the far left or an <annihilation operator> on the far right vanish; the remaining unbalanced words also vanish. Thus
$$
\boxed{\Delta E_0=-\frac{3\hbar^2\omega^2}{32mc^2},\qquad E_0=\frac12\hbar\omega-\frac{3\hbar^2\omega^2}{32mc^2}+O(c^{-4}).}
$$
The correction is negative because the <relativistic kinetic energy> is smaller than $p^2/(2m)$ at fixed nonzero <momentum>. This <relativistic oscillator ground-state kinetic correction> requires $\hbar\omega\ll mc^2$.
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