# Random problem from vector cal

1. Oct 12, 2004

### 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.

2. Oct 12, 2004

### NateTG

Are you familiar with the quadratic formula?

3. Oct 12, 2004

### 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.

4. Oct 13, 2004

### 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:
$$(ax+cy)^2=a^2x^2+2acxy+c^2y^2$$

$$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
has
$$b^2-4ac$$
which looks a lot like
$$B^2-AC$$

5. Oct 13, 2004

### HallsofIvy

Staff Emeritus
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.