f(x)=ax^2+bx+c
a+b+c=f(1)
a-b+c=f(-1)
c=f(0)
解得
a=[f(1)+f(-1)]/2-f(0)
b=[f(1)-f(-1)]/2
c=f(0)
|4a+2b+c|
=|2f(1)+2f(-1)-4f(0)+f(1)-f(-1)+f(0)|
=|3f(1)+f(-1)-3f(0)|
≤|3f(1)|+|f(-1)|+|3f(0)|
≤7