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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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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

1 Answer 1

up vote 4 down vote accepted

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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Brilliant. If you don't mind, how did you get to that so fast? I spent well over an hour and didn't see that. Is it possible to learn with practice or is it genuinely just something you looked at and saw? Edit: annoyingly, it does seem quite "obvious" now. –  Tim Green Mar 29 '11 at 12:36
    
@Tim Green: It is definitely possible to learn with practice, even though I don't think there's much to learn in this particular case, you just fill in the missing hole between what you know and what you want to proof... –  Myself Mar 29 '11 at 12:41
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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
    
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. –  Tim Green Mar 29 '11 at 12:53

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