<s>[INST] 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!
 [/INST] It seems your approach is the most straightforward, and you are almost done. Observe that WLOG we may assume $a, b \ge0$, and we, by AM-GM, have:
$$9a^4+10a^2b^2+b^4 \ge (6a^4+6a^2b^2)+(4a^2b^2+b^4) \ge  12a^3b +4ab^3.$$

EDIT: Another easy way is: $D=c^2(3a^4+6(a^2-ab)^2+(b^2-2ab)^2),$ not using the AM-GM inequality.
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