已知三次函数f(x)=a/3x^3+b/2x^2+cx+d(a

R上单调递增,则f’(x)=ax^2+bx+c恒为非负数,
所以a > 0 且 b^2 - 4ac = (a+b+ b^2/(4a))/(b-a)
= (2a+b)^2/(4a(b-a)) =[(b-a)+3a]^2/(4a(b-a))
>=[ 4(b-a)* 3a]/(4a(b-a)) = 3.