Question

试确定实数m的取值范围,使[(a+1)x2+ax+a]/[x2+x+1]>m为绝对不等式

Answer 1

[(a+1)x2+ax+a]/[x2+x+1]>m

a+x^2/(x^2+x+1)>m

x^2/(x^2+x+1)>m-a

由于x^2>=0,x^2+x+1=(x+1/2)^2+3/4>0

所以x^2/(x^2+x+1)>0

要使上面不等式为绝对不等式,只需m-a