（1）用数学归纳法.
A（n+1）=An^2-nAn+1=An(An-n)+1>=An*2+1>=(n+2)*2+1=2n+5>n+1+2
（2）因为an>=n+2,所以an-n>=2
A(n+1)=An(An-n)+1>=2An+1
A(n+1)+1>=2(An+1)
1/(A(n+1)+1)
设数列{An}满足An+1=An^2-nAn+1,n为正整数,当A1>=3时,证明对所有的n>=1,有
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