= Solution
Use the normalized <Rayleigh quotient> $R(\psi)=\langle\psi,H\psi\rangle/\langle\psi,\psi\rangle$. Its real directional <derivative> in an arbitrary complex direction $\eta$ is
$$
\delta R=\frac{2\operatorname{Re}\langle\eta,(H-R)\psi\rangle}{\langle\psi,\psi\rangle}.
$$
It vanishes for every direction precisely when $(H-R)\psi=0$, since testing $\eta=(H-R)\psi$ gives its squared norm. Conversely an <eigenvector> makes every <derivative> zero. Stationarity over all states is stronger than stationarity in a restricted trial family.
Normalize the <Gaussian> as $\psi_\alpha=(\alpha/\pi)^{1/4}e^{-\alpha x^2/2}$, with $\alpha>0$. Its moments are $\langle X^2\rangle=1/(2\alpha)$, $\langle X^4\rangle=3/(4\alpha^2)$ and $\langle P^2\rangle=\hbar^2\alpha/2$. The printed <potential energy> consequently gives
$$
\boxed{\mathcal E(\alpha)=\frac{\hbar^2\alpha}{4m}+\frac{m\omega^2}{4\alpha}+
\frac{15m^2\omega^3}{256\hbar\alpha^2}.}
$$
Put $a=\alpha\hbar/(m\omega)$. Then $\mathcal E/(\hbar\omega)=a/4+1/(4a)+15/(256a^2)$, whose second <derivative> is $1/(2a^3)+45/(128a^4)>0$. It diverges at both positive endpoints, so its sole <stationary point> is the unique <global minimum>. Its equation is
$$
\boxed{a^3-a-\frac{15}{32}=0,\qquad a_*\simeq1.18179610.}
$$
The <variational principle> therefore yields the optimized upper bound
$$
\boxed{E_0\leq\mathcal E(\alpha_*)\simeq0.548944745\,\hbar\omega,\qquad
\alpha_*\simeq1.18179610\,\frac{m\omega}\hbar.}
$$
\b[The printed minimizer is inconsistent with the printed quartic coefficient.] At $a=3/2$ the <derivative> is $5/48>0$, so this is not the stationary <Gaussian> for that <potential energy>. It is nevertheless a legitimate trial vector and gives the simple, weaker bound $\boxed{E_0\leq109\hbar\omega/192}$. If the quartic coefficient were instead $5m^2\omega^3/(16\hbar)$, four times the printed value, then $a=3/2$ would indeed minimize the family and its <energy> would be $31\hbar\omega/48$. This possible coefficient correction is identified explicitly rather than silently used. The reusable calculation is <Gaussian trial for a quadratic-plus-quartic oscillator>.
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