Putnam 2003, Problem A-4

Suppose that \( a, b, c, A, B, C \) are real numbers, \( a \neq 0 \) and \( A \neq 0 \), such that
\[
|ax^2 + bx + c| \leq |Ax^2 + Bx + C|
\]
for all real numbers \( x \). Show that
\[
|b^2 - 4ac| \leq |B^2 - 4AC|.
\]

Solution:
We split into three cases. Note first that \( |A| \geq |a| \), by applying the condition for large \( x \).  

### Case 1: \( B^2 - 4AC > 0 \)  
In this case \( Ax^2 + Bx + C \) has two distinct real roots \( r_1 \) and \( r_2 \). The condition implies that \( ax^2 + bx + c \) also vanishes at \( r_1 \) and \( r_2 \), so \( b^2 - 4ac > 0 \). Now  
\[
B^2 - 4AC = A^2 (r_1 - r_2)^2 \geq a^2 (r_1 - r_2)^2 = b^2 - 4ac.
\]  


### Case 2: \( B^2 - 4AC \leq 0 \) and \( b^2 - 4ac \leq 0 \)  
Assume without loss of generality that \( A \geq a > 0 \), and that \( B = 0 \) (by shifting \( x \)). Then \( Ax^2 + Bx + C \geq ax^2 + bx + c \geq 0 \) for all \( x \); in particular, \( C \geq c \geq 0 \). Thus  
\[
4AC - B^2 = 4AC \geq 4ac \geq 4ac - b^2.
\]  


### Alternate derivation (due to Robin Chapman):  
The ellipse \( Ax^2 + Bxy + Cy^2 = 1 \) is contained within the ellipse \( ax^2 + bxy + cy^2 = 1 \), and their respective enclosed areas are \( \pi/(4AC - B^2) \) and \( \pi/(4ac - b^2) \).  


### Case 3: \( B^2 - 4AC \leq 0 \) and \( b^2 - 4ac > 0 \)  
Since \( Ax^2 + Bx + C \) has a graph not crossing the \( x \)-axis, so do \( (Ax^2 + Bx + C) \pm (ax^2 + bx + c) \). Thus  
\[
(B - b)^2 - 4(A - a)(C - c) \leq 0, \quad (B + b)^2 - 4(A + a)(C + c) \leq 0
\]  
and adding these together yields  
\[
2(B^2 - 4AC) + 2(b^2 - 4ac) \leq 0.
\]  
Hence \( b^2 - 4ac \leq 4AC - B^2 \), as desired.