Question

若f(x)=x2+ax+b,求证2分之f(x1+x2)小于等于二分之〔f(x1)+f(x2)〕

Answer 1

f(x)=x2+ax+bf(x1)=x1^2+ax1+bf(x2)=x2^2+ax2+bf((x1+x2)/2)=((x1+x2)/2)^2+a((x1+x2)/2)+bf((x1+x2)/2)-(f(x1)+f(x2))/2=-(x1-x2)^2/4-b.(1)而x1+x2=-a,x1x2=b上式=-a^2小于等于0