Random problem from vector cal...

[Saint Medici]

Can anyone point me in the right direction for this problem? I can't seem to approach it from the right angle:


Let A, B, C be real numbers such that A>0, B>0, AC-B^2>0.

a. Prove that a number Q>0 exists such that Ax^2+2Bxy+Cy^2 \geq Q(x^2+y^2)

b. Find the largest possible Q.


I don't need answers (or rather, I do, but, you know what I mean; my main concern is figuring out how to approach this without my head exploding. Thanks in advance.

[NateTG]

I don't need answers (or rather, I do, but, you know what I mean; my main concern is figuring out how to approach this without my head exploding. Thanks in advance.

Are you familiar with the quadratic formula?

[Saint Medici]

Solving for x? Solving for y? Solving for Q? I don't understand how that solves anything. Not being a smart-alec; I just don't understand.

[NateTG]

I'm just trying to give you hints that you might be able to use to find an answer.

The LHS of the inequality looks a lot like:
(ax+cy)^2=a^2x^2+2acxy+c^2y^2

Similarly, the quadratic formula:
\frac{-b \pm \sqrt{b^2-4ac}}{2a}
has
b^2-4ac
which looks a lot like
B^2-AC

[HallsofIvy]

Not necessarily the best way but here's how I would do it:
You want
Ax^2+2Bxy+Cy^2 \geq Q(x^2+y^2)
which is the same as
(A-Q)x^2+ 2Bxy+ (C-Q)y^2\geq 0.
The will be true as long as the "discriminant" is not positive:
4B^2- 4(A-Q)(C-Q)\leq 0
or
4B^2- 4AC+ 4AQ+4CQ- 4Q^2\leq 0
That, of course, is the same as
Q^2+ (A+C)Q+ AC- B^2\geq 0

You can use the quadratic formula to find values of Q for which that is true.