I'm trying to prove that the following potential is stable at its critical point:
$$ F_{\textbf{n}}(x) = x - \sum_{\ell=1}^{r} \ln G_{n_{\ell}}(x), $$
where $\textbf{n} = (n_1, n_2, \ldots, n_r)$ and
$$ G_{n}(x) = \sum_{j=0}^{n} \binom{n}{j}\frac{x^{j}}{j!} \equiv L_{n}(-x)$$
with $L_{n}(-x)$ $n$th Laguerre polynomial. From plotting $F_{\textbf{n}}(x)$ as a function of $x$ for a few sets of $\textbf{n}$ values, it appears that the potential is always stable, but I am seeking an analytical proof.
Solution Attempt
The critical point $x = \bar{x}$ is defined by
\begin{equation} 0 = 1 - \sum_{\ell=1}^r \frac{G_{n_{\ell}}'(\bar{x})}{G_{n_{\ell}}(\bar{x})} = 1 - \frac{1}{\bar{x}} \sum_{\ell=1}^r n_{\ell} \left(1- \frac{G_{n_{\ell}-1}(\bar{x})}{G_{n_{\ell}}(\bar{x})}\right) \end{equation} Where the final expression comes from the Laguerre Polynomial identity $x G_n'(x) = n(G_n(x) - G_{n-1}(x))$.
We can show that the second derivative of the potential is given by \begin{align} F_{\textbf{n}}''(\bar{x}) & = \sum_{\ell=1}^r \left[ \left(\frac{G_{n_{\ell}}'(\bar{x})}{G_{n_{\ell}}(\bar{x})}\right)^2 - \frac{G_{n_{\ell}}''(\bar{x})}{G_{n_{\ell}}(\bar{x})}\right]\\[.75em] & =\sum_{\ell=1}^r \left[ \left(\frac{G_{n_{\ell}}'(\bar{x})}{G_{n_{\ell}}(\bar{x})}\right)^2 - \frac{1}{\bar{x}}\frac{-(1+ \bar{x})G_{n_{\ell}}'(\bar{x})+ n_{\ell}G_{n_{\ell}}(\bar{x})}{G_{n_{\ell}}(\bar{x})}\right] \\[.75em] & =\sum_{\ell=1}^r \left(\frac{G_{n_{\ell}}'(\bar{x})}{G_{n_{\ell}}(\bar{x})}\right)^2 + \frac{1+ \bar{x}}{\bar{x}}\sum_{\ell=1}^r\frac{G_{n_{\ell}}'(\bar{x})}{G_{n_{\ell}}(\bar{x})} - \frac{1}{\bar{x}}\sum_{\ell=1}^rn_{\ell} \\[.75em] & = \frac{1}{\bar{x}^2}\sum_{\ell=1}^r n_{\ell}^2 \left(1- \frac{G_{n_{\ell}-1}(\bar{x})}{G_{n_{\ell}}(\bar{x})} \right)^2 + 1 + \frac{1-N}{\bar{x}} \end{align} In the second equality we used the identity $x G''_n(x) = n G_{n}(x) - (1+x) G_n'(x)$ which can be established by differentiating both sides of $x G'_n(x) = n(G_n(x) - G_{n-1}(x)$ (1st Laguerre Polynomial identity) and applying $G'_n(x) = G_{n-1}'(x) + G_{n-1}(x)$ (intentionally a sum and not a difference). In the final equality we used the critical point condition and defined $N \equiv \sum_{\ell=1}^{r} n_{\ell}$.
From the final line above, it is not clear to me how to show that $F_{\textbf{n}}''(\bar{x})$ is always positive. Any help, or alternative approaches would be appreciated.