The Symmetric Non-negative Inverse Eigenvalue Problem (SNIEP) asks when is a list $ \sigma = \left( \lambda_{1}, \lambda_{2}, \dots, \lambda_{n} \right) $ of real, monotonically decreasing numbers, the spectrum of an $n \times n$, symmetric, non-negative matrix $A$. In that case, we say $\sigma$ is realizable and $A$ is a realizing matrix. Here, we consider the case $n=5$, the lowest value of $n$ for which the problem is unsolved. Let $ s_{1}(\sigma) = \sum_{i=1}^5 \lambda_{i} $ and $ s_{3}(\sigma) = \sum_{i=1}^5 {\lambda_{i}}^3 $. It is known that to complete the solution for $n=5$, it remains to consider the case $\lambda_{3} > s_{1}(\sigma)$, so let $y=\lambda_{3}- s_{1}(\sigma)$ and assume $y \geq 0$. We prove that if $\sigma$ is realizable, then $s_{3}(\sigma) \geq s_{1}(\sigma)^3+6s_{1}(\sigma)y(s_{1}(\sigma)+y)$. This strengthens the inequality $s_{3}(\sigma) \geq s_{1}(\sigma)^3$ obtained by Loewy and Spector, which in turn strengthens the inequality $ 25s_{3}(\sigma) \geq s_{1}(\sigma)^3 $, one of the Johnson--Loewy--London inequalities. As an application of the new inequality, we show that certain lists previously unknown as far as their realizability is concerned are not realizable.