若n=2
左邊=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
即證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)
原不等式得證