证明:若a1>a2>……>an,则1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)+n^2/(an-a1)大于等于0

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证明:若a1>a2>……>an,则1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)+n^2/(an-a1)大于等于0

证明:若a1>a2>……>an,则1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)+n^2/(an-a1)大于等于0
证明:若a1>a2>……>an,则1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)+n^2/(an-a1)大于等于0

证明:若a1>a2>……>an,则1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)+n^2/(an-a1)大于等于0
若n=2
左边=1/(a1-a2)+4/(a2-a1)=3
1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)+n^2/(an-a1)>=0
即证1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)>=n^2/(a1-an)
由柯西不等式
(a1-a2+a2-a3+...+an-1-an)[1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)]>=(1+2+..+n-1)^2
即1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)>=(n(n-1)/2)^2/(a1-an)
(n(n-1))^2/[4(a1-an)]>=(2^2*n^2)/[4(a1-an)]=n^2/(a1-an)
所以1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an)>=n^2/(a1-an)
原不等式得证

毫无疑问啊,所有指数函数是不可能小于0的
任何一项都是大于0的,你的题目可能写错了

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