= Solution
<Wavefunction normalization> requires $\alpha>0$. For the normalized <probability> weight proportional to $e^{-\alpha x^2}$, the <Gaussian integral> and its derivatives give
$$
\langle x^2\rangle=\frac1{2\alpha},\qquad
\langle x^4\rangle=\frac3{4\alpha^2}.
$$
Since $\widetilde\Psi''=(\alpha^2x^2-\alpha)\widetilde\Psi$, the <kinetic energy> and <potential energy> expectations are
$$
\langle T\rangle=\frac\alpha2-\frac{\alpha^2}{2}\langle x^2\rangle=\frac\alpha4,
\qquad
\langle V\rangle=\frac12\langle x^4\rangle=\frac3{8\alpha^2}.
$$
Thus
$$
E(\alpha)=\frac\alpha4+\frac3{8\alpha^2},\qquad
E'(\alpha)=\frac14-\frac3{4\alpha^3}=0
\ \Longrightarrow\ \alpha_*=3^{1/3}.
$$
The second derivative is positive, and the <energy> diverges at both ends of $\alpha>0$, so this is the global minimum within the <Gaussian function> family. \b[The best <Gaussian function> upper bound is]
$$
\boxed{E_0\leq E(\alpha_*)=\frac{3^{4/3}}8\simeq0.54084359.}
$$
This is an upper bound, not the exact quartic ground <energy>.
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