= Solution
For any nonzero state, the <Rayleigh quotient> is
$$
R[\psi]=\frac{\langle\psi|\widehat H|\psi\rangle}
{\langle\psi|\psi\rangle}.
$$
Expand $|\psi\rangle=\sum_nc_n|\psi_n\rangle$ in normalized energy eigenstates. Then
$$
R[\psi]=\frac{\sum_n|c_n|^2E_n}{\sum_n|c_n|^2}
\geq E_0.
$$
Because the ground state is unique, equality holds exactly when all coefficients except $c_0$ vanish. Thus the <Rayleigh-Ritz variational principle> gives its minimum at the ray of $|\psi_0\rangle$.
For the proposed exact state, write $\widetilde\psi=e^{-\beta x^n}$. Its logarithmic derivatives give
$$
\frac{\widetilde\psi''}{\widetilde\psi}
=\beta^2n^2x^{2n-2}-\beta n(n-1)x^{n-2}.
$$
The stationary Schrödinger equation, after multiplication by $2m/\hbar^2$, is
$$
\left[-\frac{d^2}{dx^2}+x^6-3x^2+2\right]\widetilde\psi
=\varepsilon\widetilde\psi,
\qquad E=\frac{\hbar^2}{2m}\varepsilon.
$$
Matching the highest power requires $2n-2=6$, so $n=4$, and cancellation of $x^6$ requires $16\beta^2=1$. Normalizability selects $\beta=1/4$. Then $12\beta=3$ also cancels the quadratic term, leaving $\varepsilon=2$. Therefore
$$
\boxed{n=4,\qquad\beta=\frac14,\qquad E=\frac{\hbar^2}{m}.}
$$
This is the <exact ground state of a solvable sextic potential> candidate $e^{-x^4/4}$.
For the Gaussian trial state $\psi_\alpha=e^{-\alpha x^2/2}$, normalization cancels from the quotient. With respect to the probability density proportional to $e^{-\alpha x^2}$,
$$
\langle x^2\rangle=\frac1{2\alpha},
\qquad
\langle x^6\rangle=\frac{15}{8\alpha^3},
\qquad
\left\langle-\frac{d^2}{dx^2}\right\rangle=\frac\alpha2.
$$
Hence
$$
R[\psi_\alpha]
=\frac{\hbar^2}{2m}\varepsilon(\alpha),
\qquad
\varepsilon(\alpha)
=2+\frac\alpha2-\frac3{2\alpha}+\frac{15}{8\alpha^3}.
$$
The stationary equation is
$$
\varepsilon'(\alpha)=0
\quad\Longleftrightarrow\quad
4\alpha^4+12\alpha^2-45=0.
$$
Writing $z=\alpha^2$, the quadratic has exactly one positive root,
$$
z=\frac{-3+3\sqrt6}{2}
=\frac{-3+\sqrt{54}}2.
$$
Since the quotient tends to infinity as $\alpha\downarrow0$ or $\alpha\to\infty$, this is the unique global minimizer. Thus
$$
\boxed{p=-3,\qquad q=54,\qquad
\alpha_*=\sqrt{\frac{-3+\sqrt{54}}2}.}
$$
Using the stationary equation to replace $15/(8\alpha_*^3)$ by $\alpha_*/6+1/(2\alpha_*)$ gives the best estimate
$$
\boxed{
E_0^*=\frac{\hbar^2}{2m}
\left(2+\frac{2\alpha_*}{3}-\frac1{\alpha_*}\right).}
$$
This is the <Gaussian variational estimate for a solvable sextic potential>.
The exact eigenfunction $e^{-x^4/4}$ is positive and has no nodes. The <nodeless theorem for a one-dimensional ground state> therefore identifies it as the true ground state, with dimensionless energy $2$. The variational value is consistent: since $\alpha_*^2>3/2$,
$$
\frac{2\alpha_*}{3}-\frac1{\alpha_*}>0,
$$
so $E_0^*>\hbar^2/m=E_0$, as every trial-state upper bound must satisfy.
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