已知a,b,c > 0满足a+b+c ≥ 1/a+1/b+1/c,求证a+b+c ≥ 3/(abc).
∵a+b+c ≥ 1/a+1/b+1/c > 0.
∴(a+b+c)² ≥ (1/a+1/b+1/c)² = 1/a²+1/b²+1/c²+2/(ab)+2/(bc)+2/(ca)
≥ 3/(ab)+3/(bc)+3/(ca) = 3(a+b+c)/(abc).
故a+b+c ≥ 3/(abc).我还有问题，你可以帮我解答吗，私信联系吧.