已知a,b,c∈正实数,a+b+c=1.求证：1/(根号a+根号b)+1/(根号b+根号c)+1/(根号c+根号a)≥(3根号3)/2

题目

已知a,b,c∈正实数,a+b+c=1.求证：1/(根号a+根号b)+1/(根号b+根号c)+1/(根号c+根号a)≥(3根号3)/2

答案

首先,由Cauchy不等式,(√a+√b+√c)² ≤ (a+b+c)(1+1+1) = 3,得√a+√b+√c ≤ √3.
同样由Cauchy不等式,((√a+√b)+(√b+√c)+(√c+√a))(1/(√a+√b)+1/(√b+√c)+1/(√c+√a)) ≥ (1+1+1)².
即得1/(√a+√b)+1/(√b+√c)+1/(√c+√a) ≥ 9/(2(√a+√b+√c)) ≥ 3√3/2.