(1/(a-b)+1/(b-c)+1/(c-d))*(a-d)
=[1/(a-b)+1/(b-c)+1/(c-d)]*[(a-b)+(b-c)+(c-d)]
=1+(b-c)/(a-b)+(c-d)/(a-b)+(a-b)/(b-c)+1+(c-d)/(b-c)+(a-b)/(c-d)+(b-c)/(c-d)+1
=3+[(b-c)/(a-b)+(a-b)/(b-c)]+[(c-d)/(a-b)+(a-b)/(c-d)]+[(c-d)/(b-c)+(b-c)/(c-d)]≥3+2+2+2=9.
當且僅當a-b=b-c=c-d,即a、b、c、d成等差數(shù)列時取“=”號.
所以,（1/(a-b)+1/(b-c)+1/(c-d)）*(a-d)的最小值為9.
因為不相等的正數(shù)a、b滿足a^3-b^3=a^2-b^2,
所以,a²+ab+b²=a+b,
即b²-(a+b)b+(a+b)²-(a+b)=0,
因為b有值使上式成立,所以,
(a+b)²-4(a+b)²+4(a+b)≥0,
即3(a+b)²-4(a+b) ≤0,
得0＜(a+b)≤4/3.