思路分析:
证明:设a1≥a2≥…≥an>0,
可知a12≥a22≥…≥an2,an-1≥an-1-1≥…≥a1-1.
由排序原理,得
a12b1-1+a22b2-1+…+an2bn-1≥a12a1-1+a22a2-1+an2an-1
即a12b1-1+a22b2-1+…+an2bn-1≥a1+a2+…+an.