# Inequality with $ax^2+bx+c$

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.

Hint: Use AM-GM inequality to obtain $$ax+b+\frac{c}{x}\ge b+2\sqrt{ac},$$ $$ay-b+\frac{c}{y}\ge -b+2\sqrt{ac}.$$
• Yes. I have tried this too. But it's true only if $4ac-b^2>0$. And I don't know what to do if $4ac-b^2<0$ Commented Jan 23, 2019 at 17:31
• @ovetz13 Umm... the given inequality seems to fail if $4ac-b^2< 0$. Look at the example: $(a,b,c)=(1,3,1)$. Then for $(x,y)=(3,1)$, the LHS becomes $-19$ while the RHS is $-15$. It fails to hold $-19\geqslant -15$. Commented Jan 23, 2019 at 17:45
• So I can say that the given inequality is true only if $4ac-b^2>0$. Problem solved I guess. Commented Jan 23, 2019 at 17:47
Try $$b=c=1$$, $$x=1$$, $$y=3$$ and $$a=\frac{1}{3}.$$