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!

Since this problem was written for high-school contestants, I also have my doubts as to how many of them would spot the factorability of the discriminant.  I decided to take the approach of writing the quadratic polynomial in "vertex form", thus
$$ abc^2·x^2 \ + \ c·( \ 3a^2 + b^2 \ )·x \ + \ ( \ 3a^2 - ab + b^2 \ ) $$
$$ = \ \ abc^2 · \left[ \ x^2 \ + \ \frac{\overbrace{3a^2 \ + \ b^2}^{K}}{abc}·x \ \right] \ + \ ( \ K \ - \ ab \ ) $$
$$ = \ \ abc^2 · \left( \ x  \ + \ \frac{K}{2abc} \ \right)^2 \ + \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4a^2b^2c^2}·abc^2 \ \right) $$
$$ = \ \ abc^2 · \left( \ x  \ + \ \frac{K}{2abc} \ \right)^2 \ + \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ \right)  \ \ , $$
with $ \ K \ > \ 0 \ $ and (necessarily) $ \ ab \ \neq \ 0 \ \ . \ $  (I am presuming, as it is not stated, that $ \ a \ , \ b \ , \ c \ $ are all real.)
The only sign dependence of significance in this expression is that of the recurrent factor $ \ ab \ \ : \  $  if $ \ ab \ > \ 0 \ \  $ (the parabola "opens upward"), then this quadratic polynomial has real zeroes if $ \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ \right) \ \le \ 0 \ \ ; \ $ if $ \ ab \ < \ 0 \ \  $ (the parabola "opens downward"), then the direction of this "coefficients" inequality is reversed.  With $ \ ab \ > \ 0 \ , \ $ the term $ \  \frac{K^2}{4ab} \ $ is also positive, so it remains to determine whether $ \ K \ \le \ ab \ + \ \frac{K^2}{4ab} \ \ . \ $   Since both the numerator and denominator in these ratios are positive, we may apply the AM-GM inequality to show that indeed
$$  \frac{ ab  \ + \ \frac{K^2}{4ab}}{2} \ \ \ge \ \ \sqrt{ab · \frac{K^2}{4ab} } \ \ \Rightarrow \ \ ab  \ + \ \frac{K^2}{4ab} \ \ \ge \ \ 2·\sqrt{ \frac{K^2}{4} } \ = \ K \ \ . $$
Hence, $ \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ \right) \ \le \ 0 \ $ for this case.  [This portion ends up being similar to Reza Rajaei's argument.]  With $ \ ab \ < \ 0 \ \ , \ \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ $ is equivalent to a sum of three positive terms, hence $ \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ \right) \ \ge \ 0 \ \ . $
Therefore, the polynomial in question always has at least one real zero.  (Other than that we must have $ \ c \ \neq \ 0 \ \ , \ $ the value of $ \ c \ $ is of no consequence for this conclusion.)
