Range.
Solution: Let F(x)=f(x)-(a+2)x=x? +alnx-(a+2)x≧0 in x？ [1, +∞) holds.
Let F'(x)=2x+(a/x)-(a+2)=[2x? -(a+2) x+a]/x = (2x-a) (x-1)/x = 0, then the stagnation point x? = 1，x？ = a/2； x？ [ 1,+∞)。
When a=2, F'(x)=2(x- 1)? /x≥0 is constant at [1, +∞), so F(x) monotonically increases at [1, +∞); At this time, x= 1 is the minimum point,
But because f (1) = 1-4 =-3.
When a>2, namely a/2 >; In 1, x? = 1 is the maximum point, x? =a/2 is the minimum point. Let F(x)⊙0 be in x? [1, +∞) holds, it must be
Make its minimum value F(a/2)=(a? /4)+aln(a/2)-(a+2)a/2=-a？ /4-a+AlN(a/2)≧0；
Therefore, ln (a/2) ≧ (a/4)+1; This inequality has no solution, that is, the a>2 point problem has no solution.
When a<2, that is, a/2 < At 1, x? = 1 is the minimum point, x? =a/2 is the maximum point; Therefore, from f (1) =1-(a+2) =-1-a ≧ 0, a≦- 1 is obtained.
Conclusion: -∞ < A≦- 1 is the range of a.