Showing a trinomial is always positive

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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

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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.