证明：若a1>a2>……>an,则1^2/(a1-a2)+2^2/(a2-a3)+……+(n-1)^2/(an-1-an
2个回答

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

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