Inequality with $ax^2+bx+c$

Question

Let $a,b,c,x,y > 0$ reals prove that: $$(ax^2+bx+c)(ay^2-by+c) \geq (4ac-b^2)xy$$ What I have done is this: $$ax^2+bx+c=a \left (x+\frac{b}{2a} \right)^2+\frac{4ac-b^2}{4a}$$ $$ay^2-by+c=a \left (y-\frac{b}{2a} \right)^2+\frac{4ac-b^2}{4a}$$ From here I have two cases:
$ 4ac-b^2 >0$ $$\left[ \sqrt{a}^2 \left (x+\frac{b}{2a} \right)^2+\sqrt \frac{4ac-b^2}{4a}^2\right ]\cdot \left[ \sqrt{a}^2 \left (y-\frac{b}{2a} \right)^2+\sqrt \frac{4ac-b^2}{4a}^2\right ]\geq$$ $$\left [ \sqrt a \left (x+\frac{b}{2a} \right)\cdot\sqrt \frac{4ac-b^2}{4a}+ \sqrt a \left (y-\frac{b}{2a} \right)\cdot\sqrt \frac{4ac-b^2}{4a}\right ]^2$$ This is just Cauchy. $$\left[ \sqrt{a}^2 \left (x+\frac{b}{2a} \right)^2+\sqrt \frac{4ac-b^2}{4a}^2\right ]\cdot \left[ \sqrt{a}^2 \left (y-\frac{b}{2a} \right)^2+\sqrt \frac{4ac-b^2}{4a}^2\right ]\geq$$ $$a \cdot \frac{4ac-b^2}{4a} \left ( x+\frac{b}{2a}+y-\frac{b}{2a} \right )^2=(4ac-b^2) \left ( \frac{x+y}{2}\right )^2\geq (4ac-b^2)xy. $$ Now the second case I have trouble proving : $4ac-b^2 < 0$. If someone can take a look and give me a solution or a hint I would much appreciate.

Answer 1

Hint: Use AM-GM inequality to obtain $$ ax+b+\frac{c}{x}\ge b+2\sqrt{ac}, $$ $$ ay-b+\frac{c}{y}\ge -b+2\sqrt{ac}. $$

Answer 2

It's wrong.

Try $b=c=1$, $x=1$, $y=3$ and $a=\frac{1}{3}.$