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I know that for any nonzero $x,y\in\mathbf{R}$,

$$ax^2+by^2+cxy > 0,$$

where $a,b,c\in\mathbf{R}$. What can I deduce about $a$, $b$, and $c$?


For example, letting $x=1$ and $y=0$, I know that $\boxed{a>0}$. Letting $x=0$ and $y=1$, I know that $\boxed{b>0}$. Letting $x=y=1$, I know that $\boxed{c>-(a+b)}$.

What else can I deduce about $a$, $b$, and $c$? How will I know when it's time to stop looking? (Is that last question even answerable?)

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You can "complete the square" as you would for solving a quadratic equation. –  Mark Bennet Oct 24 '12 at 3:25
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2 Answers

up vote 4 down vote accepted

You can write $$ax^2+by^2+cxy=a\left(x+\frac{c}{2a}\,y\right)^2+\left(b-\frac{c^2}{4a}\right)y^2$$ by completing the square. This is a sum of two squares. In order to ensure that it's always positive, the coefficients both have to be positive, so $$a>0$$ and $$c^2<4ab.$$

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Dividing the whole thing by $y^2$ , we get $a(x/y)^2+c(x/y)+b>0$ (Note: we can divide by $y^2$ since it is non zero) Since it is positive for all values of $(x/y)$, the discriminant must be negative (since $a$ is given to be positive) Therefore, $c^2 - 4 ab < 0$

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Jonathan's method is more explanatory. –  RahulHP Oct 24 '12 at 3:25
    
Do you mean the discriminant? –  Ron Oct 24 '12 at 3:32
    
Yes, sorry about that and thanks for pointing it out :) –  RahulHP Oct 25 '12 at 17:59
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