How to prove $(b-2)^2 > 12a(5c + 2)$ provided $(3a + b + 5c)(5c + 2) < 0$? $a$, $b$, $c$ are rational numbers. It is known that $(3a + b + 5c)(5c + 2) < 0$.
How do I prove that $(b-2)^2 > 12a(5c + 2)$?
According to Bernoulli's inequality, I got this:
$(1 + b - 3)^2 \ge 1 + 2(b - 3) > 12a(5c + 2)$
$2b - 5 > 12a(5c + 2)$
What should I do next?

It seems the following.
Put $x=3a$, $y=b-2$, and $z=5c+2$. Given $(x+y+z)z<0$ we need to show that $y^2>4xz.$ Assume the converse. Then $(x+z)^2\ge 4xz>y^2$. So $|x+z|>|y|$ and the sign of $x+y+z$ is equal to the sign of $x+z$. Then $(x+z)z<0$, but $xz>y^2/4\ge 0$ and $z^2\ge 0$, a contradiction.
