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.

an+1^2=an^2+2+1/an^2a(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)+1a(k+...