an+1^2=an^2+2+1/an^2
a(n+1)=an+1/an
用归纳法证明
n=1时,a1^2>(2*1+1)
n=2时,a2=5/2 a2^2=a1^2+1/a1^2+2=25/4>2*2+1
假设n=k时,ak^2>2n+1
则n=k+1时,
a(k+1)^2=ak^2+2+1/ak^2>2k+1+2+1/ak^2=2(k+1)+1+1/ak^2>2(k+1)+1
a(k+1)>根号下[2(k+1)+1]
所以 an>根号下2n+1恒成立
设数列{an}满足a1=2,an+1^2=an^2+2+1/an^2(n=1,2,3.)求证：an>根号下2n+1.
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