题1若a,b,c为任意正实数,且满足a+b+c=3,求证:a~2+b~2+c~2+abc≥4.解决这个问题有着众多方法,这里给出一种在笔者看来最为巧妙的证法.证明不难发现取等条件为a=b=c=1.由抽屉原理,a,b,c这3个数中必有两个同时不大于1或者同时不小于1,不妨设为a,b,则(a-1)(b-1)≥0,即ab≥a+b-1. Problem 1 If a, b, c are any positive real numbers and satisfy a + b + c = 3, verify that: a ~ 2 + b ~ 2 + c ~ 2 + abc ≥ 4. There are many ways to solve this problem, Out of a kind of I think in the most ingenious card law proved not difficult to find such conditions as a = b = c = 1. By the drawer principle, a, b, c of these three numbers must be no more than two at the same time 1 or at the same time not less than 1, may be set to a, b, (a-1) (b-1) ≥ 0, that is ab ≧ a + b-1.