Let $a$ be a real number less than zero and $w,x,y,z$ real numbers with $w\geq 1, x\geq 2, y\geq 3, z\geq 4$ that have pairwise distances of at least $1$. Given that $$15a^2+6a\sigma _1+4\sigma _2=\sigma _1^2,$$ where $\sigma _1=w+x+y+z$ and $\sigma _2=wx+wy+wz+xy+xz+yz$ are the first two elementary symmetric polynomials in $w,x,y,z$, prove that $$(a+2)^2\geq \frac{2}{3}.$$ As of now, I could not succeed in proving any upper bound on this, partly due to the fact that I only have lower bounds on $w,x,y,z$.
Interpreting the given equation as a quadratic in $a$ or $\sigma _1$ and then deriving inequalities from the determinant did not give any sufficient results either.
Any help or hint is much appreciated.
NOTE: An earlier version of this question asked to prove that $4\sigma _2-\sigma _1^2\leq 50$. However, this was just an attempt by me to prove the inequality above since it is a sufficient condition. Unfortunately, it does not hold in general, as proven by Chris Custer.
A little algebra gives that you need to prove $$-6a\sigma_1\le50+15a^2$$. So we need $\sigma_1\le -\frac {50+15a^2}{6a}$.
But there's no upper bound on $\sigma _1$. So the inequality doesn't hold, in general...
For instance if $a=-1$ and $\sigma _1=20$, the inequality fails.