a,b,c,A,B,C are real numbers such that a ≠ 0 , A≠0
It is given that |ax^2+bx+c| ≤|Ax^2+Bx+C| for all real number x
prove that |b^2-4ac|≤|B^2-4AC|
Great thanks to the one who can help me to solve the problem.
更新1:
I want to know why can you simipify it to |Ax^2+Bx+C|≥|x^2+d| =〉|B^2-4AC|≥|4d|, where B≥0