已知a1+a2+…….+an=1求证：a1^2/(a1+a2) + a2^2/(a2+a3)…….+an-1^2/(an-1+an) +an^2/(an+a1)>1/2

来源：百度知道编辑：UC知道时间：2024/05/29 17:01:32

已知a1+a2+…….+an=1
求证：a1^2/(a1+a2) + a2^2/(a2+a3)……+an-1^2/(an-1+an) +an^2/(an+a1)>1/2

题目应该是求证大于等于0
用数学归纳法
当n=2时
若证a1^2/(a1+a2) + a2^2/(a2+a1)>1/2
即a1^2+ a2^2>1/2
a1^2+ a2^2>(a1+a2)^2/2
(a1-a2)^2>0
得证
假设n-1时成立
即a1^2/(a1+a2) + a2^2/(a2+a3)……+an-1^2/(an-1+a1)>1/2
即a1^2/(a1+a2) + a2^2/(a2+a3)……+an-1^2/(an-1+an) +an^2/(an+a1)>1/2+an^2/(an+a1)+an-1^2/(an-1+an)-an-1^2/(an-1+a1)
只需证明an^2/(an+a1)+an-1^2/(an-1+an)-an-1^2/(an-1+a1)>0即可
an^2/(an+a1)+an-1^2/(an-1+an)-an-1^2/(an-1+a1)>
an^2/(an+a1+an-1+an)-1^2/(an-1+an+a1)-an-1^2/(an-1+a1)+an=
an^2/(an+a1+an-1+an)>0
得证

题目应该是求证大于等于0
用数学归纳法
当n=2时
若证a1^2/(a1+a2) + a2^2/(a2+a1)>1/2
即a1^2+ a2^2>1/2
a1^2+ a2^2>(a1+a2)^2/2
(a1-a2)^2>0
得证

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