= Solution
For any real coupling, the same normalized Gaussian family has <energy>
$$
E_\epsilon(\alpha)=\frac\alpha2+\frac1{2\alpha}+\frac{3\epsilon}{4\alpha^2}.
$$
If $\epsilon<0$, then as $\alpha\downarrow0$ the negative $\alpha^{-2}$ term dominates the positive $\alpha^{-1}$ and $\alpha$ terms. Each trial still has finite <energy>, is normalized and decays at infinity, but its <energy> tends to $-\infty$. Therefore
$$
\boxed{\inf_{\|\psi\|_2=1}E[\psi]=-\infty;\qquad\text{there is no minimum energy.}}
$$
This variational sequence makes the instability of the negative quartic potential explicit, without mistaking any local well near the origin for a global ground-state bound.
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