Very Challenging Math Problem on Discriminant!?

Question

Given that a, b, c, A, B, C are real numbers with a, A>0 and B^2-4AC≠0 such that /ax^2+bx+c/ ≤ /Ax^2+Bx+C/ for all real number x. Show that


/b^2-4ac/ ≤ /B^2-4AC/

Answer 1

Let a = 1, b = 5, c = 1, and A = 2, B = 7, C = 5, so that

b² - 4ac = 21
B² - 4AC = 9

Yet x² + 5x + 1 < 2x² + 7x + 5 for all x, since

(B-b)² < 4(A-a)(C-c), or
2² < 4(1)(4)
so that there are no real intersections

So, what's going on?

Addendum: novice's "proof" forgets that the height of the minumum is

-(1/4a)(b²-4ac), not √(b²-4ac)

It's the presence of the factor (1/4a) that complicates this proof, of a conjecture which turns out to be WRONG by the counterexample I've given.

Answer 2

well logically, the statements given tell us that the equation with the capitals must either have 2 roots or no roots.
the second poitn given tells us that the small letter equation's y value is smaller or equal to in magnitude with that of the capital letter equation.
both curves are minimas
the larger the magnitude of c is, the further away it is from the x axis and the smaller the discriminant.
the larger the magnitude of a, the smaller the discriminant will become
the larger the mag of b, the larger the discriminant. this is the main factor.

arghhh now im confused myself! i'll come back and write more when i've figured it out..

Answer 3

let y1 = |ax^2 + bx + Cc| and y2 be the other.
both of these are "upward facing U-shaped" parabolas(with the modulus sign), with a global minima.
It is given that y1<=y2. Therefore, y2 is "inside" y1 (if you think graphically). i.e. y1 curve never crosses/cuts y2. One U inside another U. Or in other words, for all x, y2 is above y1. Iam not clear but I hope you are getting it.
and also the discriminant is a measure of the height of the minima. (convince yourself of this). since y2 is inside y1, its minima is above the minima of y1.(graphically speaking).
hence the result.