在各级各类数学竞赛中,经常涉及在a+b+c=1条件下的不等式问题,经探索,此类问题有统一的简单证法,其思路是构造最简单的不等式(x-y)2≥0.一、证明一类整式不等式例1(前苏联奥尔德荣尼基市第三届数学竞赛试题)设a、b、c∈R+,且a+b+c=1, In various types of math competitions of various types, the inequality problems under the condition of a+b+c=1 are often involved. After exploration, such problems have a simple and simple proof method. The idea is to construct the simplest inequality (xy)2. ≥ 0. I. Prove one type of integer inequality Example 1 (test question of the third mathematic competition of the former Soviet Union, Alde Jonigi City) Let a, b, c∈R+, and a+b+c=1,