y=(ax^2+x+1)/(x+1)
=ax^2/(x+1)+1
因为a>0且x>=3,所以y>1
y=ax^2/(x+1)+1
=a/(1/x+1/x^2)+1
可见x越大,y就越大,函数在范围内单调递增,x取最少值3时,y取得最少值
y=9a/4+1