 Solution

Use the Gaussian from part 1. Since $(-\Delta +|x{|}^2)\psi =2\psi$,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 ${\inf }_{\Sigma }J<0$.