Showing a trinomial is always positive I have a trinomial of the form $$ax^2+2bxy+cy^2$$ where $x,y$ are non-zero from $\mathbb{R}$. I want to show that $ax^2+2bxy+cy^2>0$ **if and only if** $a>0$ and $ac-b^2$<0

I will only show the if part.

Let $z = ax^2 + 2bxy + cy^2$.

Dividing all the terms by $y^2$, we get $\dfrac {z}{y^2} = a(\dfrac {x}{y})^2 + 2b(\dfrac xy) + c$. If we let $X = \dfrac {x}{y}$ and $Y = \dfrac {z}{y^2}$, then we are treating the curve of $Y = aX^2 + 2bX + c$.

$a>0$ means the curve is opening upward.

$\triangle = ac−b^2<0$ means the curve will never cut across the X-axis.

The two conditions together make the given expression positive definite.