y=(3x^2+3x+4)/(x^2+x+1),
y(x^2+x+1)=3x^2+3x+4,
(y-3)x^2+(y-3)x+(y-4)=0 ,
判别式=(y-3)^2-4(y-3)(y-4)>=0 ,
所以 (y-3)(y-3-4y+16)>=0 ,
即 (y-3)(3y-13)