设函数f(x)=ax2+bx+c且f(1)=-a/2,3a>2c>2b,求证

1.将f(1)=-a/2 代入有 a+b+c=-a/2 整理有 3a+2b+2c=0
因为3a>2c>2b,所以必有3a>0 2b0 b-3a-2b>2b 解得,-3-1/3b
f(2)=4a+2b+c=5/2a+b;f(0)=c=-3/2a-b
f(0)*f(2)=-15/4(a+8/15b)^2+1/15b^2