= Solution
Normalize the Gaussian trial first. The <Gaussian integral> gives
$$
1=C^2\sqrt{\frac\pi\alpha},\qquad\boxed{C=\pm\left(\frac\alpha\pi\right)^{1/4}.}
$$
Its second and fourth moments are $\langle x^2\rangle=1/(2\alpha)$ and $\langle x^4\rangle=3/(4\alpha^2)$. Since $\psi'=-\alpha x\psi$, the kinetic term is $\int(\psi')^2=\alpha^2\langle x^2\rangle=\alpha/2$. At the specified coupling, the <energy> is therefore
$$
E(\alpha)=\frac\alpha2+\frac1{2\alpha}+\frac3{2\alpha^2}.
$$
The stationary condition is
$$
E'(\alpha)=\frac{\alpha^3-\alpha-6}{2\alpha^3}=0,\qquad\alpha^3-\alpha-6=(\alpha-2)(\alpha^2+2\alpha+3).
$$
The quadratic factor is positive for real $\alpha$, so the unique positive stationary width is $\alpha=2$. Also $E''=\alpha^{-3}+9\alpha^{-4}>0$ and $E$ diverges at both ends of $\alpha>0$, proving a global minimum within this trial family. Its value is
$$
\boxed{E_{\rm trial,min}=\frac{13}{8},\qquad\psi_*(x)=\pm(2/\pi)^{1/4}e^{-x^2}.}
$$
The associated <Sturm-Liouville problem> is
$$
\boxed{-\psi''+(x^2+2x^4)\psi=E\psi,\qquad\psi\in L^2(\mathbb R),\quad\psi\to0\text{ at }\pm\infty.}
$$
Here $\rho=1$, $\sigma=x^2+2x^4$ and $w=1$. The <Rayleigh-Ritz variational principle> makes the exact lowest <eigenvalue> the minimum over all normalized admissible states. Restricting to Gaussians can only raise this minimum, hence $\boxed{E_0\leq13/8}$. The inequality is strict because $H\psi_*/\psi_*=2-3x^2+2x^4$ is not constant: the trial is not an exact <eigenfunction>. This is the <Gaussian trial for a quadratic-plus-quartic oscillator>.
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