= Stability of a cubic-quartic scalar potential
{title2=$\mu^2\le3\lambda m^2$}
For $V(\phi)=m^2\phi^2/2-\mu\phi^3/6+\lambda\phi^4/24$, positive $\lambda$ makes the classical potential bounded below. For $m^2\ge0$, completing the square gives
$$
V=\phi^2\left[\frac{\lambda}{24}\left(\phi-\frac{2\mu}{\lambda}\right)^2+\frac{m^2}{2}-\frac{\mu^2}{6\lambda}\right].
$$
Thus zero is a global minimum precisely when $\mu^2\le3\lambda m^2$. At a nontrivial equality there are degenerate minima at zero and $2\mu/\lambda$. If $m^2>0$ and the inequality fails, zero is a <false vacuum>: its Hessian is positive but another minimum has lower energy. A negative quartic coupling, or a nonzero cubic term without a quartic term, is unbounded below. These are classical or tree-level criteria.
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