2003 Putnam A4

Suppose that $a,b,c,A,B,C$ are real numbers, $a\ne 0$ and $A \ne 0$, such that

$$

| a x^2 + b x + c | \leq | A x^2 + B x + C |

$$

for all real numbers $x$. Show that

$$

| b^2 - 4 a c | \leq | B^2 - 4 A C |.

$$

假设 $a,b,c,A,B,C$ 是实数，$a\ne 0$ 和 $A \ne 0$，这样

$$

| a x^2 + b x + c | \leq | A x^2 + B x + C |

$$

对于所有实数 $x$。表明

$$

| b^2 - 4 a c | \leq | B^2 - 4 A C |。

$$