设数列{an}满足a1=2,an+1=an+(1/an),(n=1,2,3…).(1)证明:an>(2n+1)1/2(根号)对一切正整数n都成立

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设数列{an}满足a1=2,an+1=an+(1/an),(n=1,2,3…).(1)证明:an>(2n+1)1/2(根号)对一切正整数n都成立
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设数列{an}满足a1=2,an+1=an+(1/an),(n=1,2,3…).(1)证明:an>(2n+1)1/2(根号)对一切正整数n都成立
设数列{an}满足a1=2,an+1=an+(1/an),(n=1,2,3…).(1)证明:an>(2n+1)1/2(根号)对一切正整数n都成立

设数列{an}满足a1=2,an+1=an+(1/an),(n=1,2,3…).(1)证明:an>(2n+1)1/2(根号)对一切正整数n都成立
用数学归纳法证明,
当n=1时,a1=2>√(2*1+1)=√3,成立;
当n=2时,a2=2+1/2=5/2>√(2*2+1)=√5,成立;
设n=k时,原式成立,ak>√(2k+1),(ak)²>2k+1,∵a(k+1)=ak+1/(ak),∴ak*a(k+1)=(ak)²+1,∵ak>0,且ak≠a(k+1),∴(ak)²+[a(k+1)]²>2ak*a(k+1)=2(ak)²+2,[a(k+1)]²>(ak)²+2>2(k+1)+1,则a(k+1)>√[2(k+1)+1],当n=k+1时,原式成立,故an>根号下2n+1成立.

a[n+1]^2=a[n]^2+2+1/a[n]^2,
a[n+1]^2-a[n]^2=2+1/a[n]^2,...
累加,得a[n]^2-a[1]^2=2(n-1)+1/a[n-1]^2+1/a[n-2]^2+...+1/a[1]^2