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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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    Sorry about that - fixed the question now.2011-03-29

1 Answers 1

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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}) $.

Now use what you know and use that squares are always positive.

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    Yep - one of the other features given was the vector x is non-zero so I can infer x and y cannot be zero at the same time.2011-03-29