数学难题，f(x)=ax^2+ bx+ c,对任何x 都有|f(x)|小于等于1求证|a|小于等于2

f(x) = ax^2 + bx + c.

1 >= |f(1)| = |a + b + c|,
1 >= |f(0)| = |c|,
1 >= |f(-1)| = |a - b + c|.

|a + c| = |[a + b + c] + [a - b + c]|/2

<= [|a + b + c| + |a - b + c|]/2

<= [1 + 1]/2

= 1.

|a| = |a + c - c|

<= |a + c| + |c|

<= 1 + 1

= 2.

题哪里抄错了吧，没抄错的话a=0

a=0,b=0,|c|<=1

a=0,b=0,|c|<=1