如果知道Cauchy不等式,直接1/a+1/b+1/c = (a+b+c)(1/a+1/b+1/c) ≥ (1+1+1)² = 9.
如果只会均值不等式,就展开1/a+1/b+1/c = (a+b+c)(1/a+1/b+1/c)
= 3+(a/b+b/a)+(b/c+c/b)+(c/a+a/c) ≥ 3+2+2+2 = 9.