a, b, c, A, B, C are reals with a, A non-zero such that |ax^{2} + bx + c| ≤ |Ax^{2} + Bx + C| for all real x. Show that |b^{2} - 4ac| ≤ |B^{2} - 4AC|.

**Solution**

Put f(x) = ax^{2} + bx + c, F(x) = Ax^{2} + Bx + C, D = B^{2} - 4AC, d = b^{2} - 4ac. wlog a, A > 0. Considering large x, it is clear that a ≤ A.

We consider separately the three cases D > 0, D = 0 and D < 0.

__Case 1__. D > 0. Then F has two distinct roots α, β, so F(x) = A(x-α)(x-β). But then α and β must also be roots of f, so f(x) = a(x-α)(x-β). Hence 0 < d = a^{2}(α-β)^{2} ≤ A^{2}(α-β)^{2} = D. So |d| ≤ |D|.

__Case 2__. D = 0. So F(α) = F'(α) = 0 for some α. Hence also f(α) = 0. But F(α+ε) = O(&epsilon^{2}), so f(α+ε) = O(&epsilon^{2}) also. Hence f '(α) = 0 also. So f also has a repeated root, so d = 0.

__Case 3__. D < 0. Then F(x) = A(x + B/2A)^{2} + C-B^{2}/4A > 0 for all x. So we have (A-a)x^{2} + (B-b)x + (C-c) ≥ 0 for all x. If A = a, then it follows that B = b also. But now |d| = 4A|f(-B/2A)| ≤ 4A|F(-B/2A)| = |D|. So assume A > a. Similarly, we also have (A+a)x^{2} + (B+b)x + (C+c) ≥ 0 for all x. Since these two quadratics are never negative we have (B-b)^{2} ≤ 4(A-a)(C-c) and (B+b)^{2} ≤ 4(A+a)(C+c). Adding, B^{2} + b^{2} ≤ 4AC + 4ac. Hence d ≤ -D = |D|. As before, taking x = -B/2A we get -d ≤ 4af(-B/2A) ≤ 4Af(-B/2A) ≤ 4A|F(-B/2A)| = |D|.

© John Scholes

jscholes@kalva.demon.co.uk

8 Dec 2003

Last corrected/updated 8 Dec 03