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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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
（1）an>=n+2
(2)1/(1+a1) + 1/(1+a2) + ……+1/(1+an) =< 1/2
同志们谢谢了特别是第二问,步骤谢谢万分感谢

（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)

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