2004年亚太地区数学奥林匹克试题5为:证明:对任意正实数a,b,c,均有(a2+2)(b2+2)(c2+2)≥9(bc+ca+ab).杨志明老师将其加强为:若a,b,c为正整数,则(a2+2)(b2+2)(c2+2)≥3(a+b+c)2.文[1]对这一加强不等式的证明虽然精彩漂亮,但思路不易想到且变形较繁冗.这里,笔者试给出一个非常简 In the Mathematics Olympiad Question and Answer in the Asia-Pacific Region 2004, it was proved that for any positive real numbers a, b, c, there are (a2+2)(b2+2)(c2+2)≥9(bc+ca+ab). The teacher reinforces it as follows: If a, b, c are positive integers, then (a2+2)(b2+2)(c2+2)≥3(a+b+c)2. Although the proof of strengthening the inequality is wonderful and beautiful, but the ideas are not easy to think and deformation is more tedious. Here, I try to give a very simple