Verify that $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots As the title states, the goal is to verify that the quadratic equation: $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots. This problem comes from an interschool mathematics contest for High-schoolers, here's my (brute force) attempt:
Since we need to verify that the roots are rational, we need to prove that the discriminant of this equation (which I'll denote with $D$) is greater than $0$. In other words:
$D=B^2-4AC>0$ where $B=c(3a^2+b^2)$, $A=abc^2$ and $C=3a^2-ab+b^2$ Therefore:
$D=[c(3a^2+b^2)]^2-4(abc^2)(3a^2-ab+b^2)$
$D=[c^2(9a^4+6a^2b^2+b^4)]+[c^2(-12a^3b+4a^2b^2-4ab^3)]$
$D=c^2(9a^4+10a^2b^2+b^4-12a^3b-4ab^3)$
I'm not quite sure where to proceed from here. And this certainly does not prove that the roots are real as this expression can be negative. Is there a way to proceed from here? Are there any better or alternative ways to answer this? Please share your approaches!

$$
p(x) = ab{c}^{2}{x}^{2}+c \left( 3\,{a}^{2}+{b}^{2} \right) x+3\,{a}^{2}-ab+{b}^{2} .
$$
Let's try to factor this with the idea of finding a factor $Ax+B$ where $A$ is a divisor of $abc^2$ and $B$ is a divisor of $3a^2-ab+b^2$.  Since $3a^2-ab+b^2$ cannot be factored over the rationals, we try without factoring it.  (If this doesn't work, we have to do something harder.)  The possible factors of $p(x)$ are $Ax+B$ where
$$
A \in \big\{1, a, b, ab, c, ac, bc, abc, c^2, ac^2, bc^2, abc^2\big\} ,
\\
B \in \big\{1, -1, (3a^2-ab+b^2), -(3a^2-ab+b^2)\big\} .
$$
Trying these four candidates for $B$ until one works, we arrive at:
$$
p(x)= \left( cx+1 \right)  \left( abcx+(3\,{a}^{2}-ab+{b}^{2})\right)
$$
