[(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<=0
即m<=a