题目若a,b,c是满足a+b+c=1的正数,则(a+1/b)~(1/2)+(b+1/c)~(1/2)+(c+1/a)~(1/2)≥(30)~(1/2).文[1]用很巧妙的办法证明了该题,我觉得还可以用柯西不等式结合均值不等式加以证明,证法如下.根据极端原理,不等式取等号的条件是a=b=c=1/3,那么,不等式左边每个式子乘以常数 (A + 1 / b) ~ (1/2) + (b + 1 / c) ~ (1/2) + (b + 1 / c) if a, b and c are positive numbers satisfying a + b + c + 1 / a) ~ (1/2) ≥ (30) ~ (1/2). The text [1] proves the problem in a clever way, and I think it can be proved by Cauchy inequality and mean inequality , The proof is as follows: According to the extreme principle, the inequality takes the equals sign if a = b = c = 1/3, then each equation on the left side of the inequality is multiplied by a constant