The general formula to get the potential energy of any spherical distribution is this : \begin{equation}\tag{1} U = - \int_0^R \frac{GM(r)}{r} \, \rho(r) \, 4 \pi r^2 \, dr, \end{equation} where $M(r)$ is the mass inside a shell of radius $r < R$. It is easy to get the gravitational energy of a uniform sphere of mass $M$ and radius $R$ : \begin{equation}\tag{2} U = -\, \frac{3 G M^2}{5 R}. \end{equation} In general, for any spherical distribution of total mass $M$ and exterior radius $R$, we can write this : \begin{equation}\tag{3} U = -\, \frac{k \, G M^2}{R}, \end{equation} where $k > 0$ is a constant that depends on the internal distribution. $k = \frac{3}{5}$ for the uniform distribution. For a thin spherical shell of radius $R$ (all mass concentrated on its surface), we can get $k = \frac{1}{2}$.
Now, I suspect that for all cases : \begin{equation}\tag{4} \frac{1}{2} \le k < \infty. \end{equation}
Physically, this makes sense. But how to prove this from the general integral (1) ?
To simply things a bit, we may introduce the dimensionless variable $x = r/R \le 1$, and defines relative mass $\bar{M}(x) \equiv M(r)/M \le 1$ and relative density $\bar{\rho}(x) = \rho(r) / \rho_{\text{average}}$, where $\rho_{\text{average}} = 3 M/4 \pi R^3$. Thus, integral (1) takes the following form : \begin{equation}\tag{5} U = -\, \frac{3 G M^2}{R} \int_0^1 \bar{M}(x) \, \bar{\rho}(x) \, x \, dx. \end{equation} The last integral is $\frac{k}{3}$. I'm not sure this may help to prove (4).