# Given $a>0$ and $ac-b^2>0$ show

Given $a>0$ and $ac-b^2>0$ show

$cy^2+a[(x+\frac{by}{a})^2-(\frac{by}{a})^2] > 0$

I'm completely confused about this, I've tried a few approaches. I end up getting stuck saying that I know $cy^2>0$ using the 2nd of the given inequalities, but I can't show the $a[...]$ part is >0 since all I know is x and y are non-zero.

Any guidance? Comes from a larger question about showing a symmetric matrix [a, b, b, c] is positive definite if that helps.

Thanks for any nudges :)

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If you are workking with $\begin{pmatrix}a&b\\b&c\end{pmatrix}$ and the corresponding quadratic form $ax^2+2bxy+cy^2$, you should have there $-\frac{b^2y^2}{a^2}$ instead of $-\frac{by^2}{a}$ -- just as Myself writes in the answer bellow. – Martin Sleziak Mar 29 '11 at 12:41
@Martin: good remark, I actually checked the latexcode to see what exactly was meant by the last square :-) Since it says \frac{by}{a}^2, I interpreted it as I did – Myself Mar 29 '11 at 12:45
Sorry about that - fixed the question now. – Tim Green Mar 29 '11 at 12:52

Rewrite it as $cy^2 + a\left((x+\tfrac{by}{a})^2 - \frac{b^2y^2}{a^2}\right) = a(x+\tfrac{by}{a})^2 + y^2 (\frac{ac}{a} - \frac{b^2}{a})$.
Completing the square is a standard way of proving the quadratic formula or finding where a quadratic is $0$. By the way, you need to exclude $x=0,y=0$ if you want a strict inequality. – Douglas Zare Mar 29 '11 at 12:46