证明：∵a＞b＞c,∴a-b＞0,b-c＞0,a-c＞0,且a-c=(a-b)+(b-c).由柯西不等式可知,[(a-b)+(b-c)]×{[1/(a-b)]+[1/(b-c)]}≥(1+1)²=4.∴1/(a-b)+1/(b-c)≥4/(a-c).