运用柯西不等式：[a(b+2c)+b(c+2a)+c(a+2b)]×[a/(b+2c)+b/(c+2a)+c/(a+2b)]≥(a+b+c)^2 (其实就是用3ab+3bc+3ca去乘)a,b,c都是正数,故a/(b+2c)+b/(c+2a)+c/(a+2b) ≥ (a+b+c)^2 / (3ab+3bc+3ca)∵(a-b)^2 + (a-b)^2...