Solution
= Solution
Use the Gaussian from part 1. Since $(-\Delta+|x|^2)\psi=2\psi$,
$$
J(a\psi)=\frac{a^2}{2}(2-\omega)\|\psi\|_2^2
+\frac{a^4}{4}\|\psi\|_4^4.
$$
The quadratic coefficient is negative because $\omega>2$. For every sufficiently small nonzero $a$, the negative quadratic term dominates the quartic term, so $J(a\psi)<0$ and therefore $\inf_\Sigma J<0$.