不等式的證明,14

1/a + 1/b + 1/c= (1/a + 1/b + 1/c) (a + b + c)= a/a + a/b + a/c + b/a + b/b + b/c + c/a + c/b + c/c= 3 + (a/b + b/a) + (b/c + c/b) + (a/c + c/a)≥ 3 + 2 + 2 + 2= 9;a + b + c = 1a² + b² + c² + 2ab + 2bc + 2ca = 1a² + b² + c² = 1 - (2ab + 2bc + 2ca)a² + b² + c² ≥ 1 - (a²+b² + b²+c² + c²+a²)3(a² + b² + c²) ≥ 1 a² + b² + c² ≥ 1/3;(1/a - 1) (1/b - 1) (1/c - 1)= (1 - a) (1 - b) (1 - c) / (abc)= (b + c) (a + c) (a + b) / (abc)≥ 2√(bc) 2√(ac) 2√(ab) / (abc)= 8

還有:a,b,c是正數

2011-04-30 19:29:19 補充：
1. 1/a + 1/b + 1/c ≥ 3 x 3/(a+b+c) = 9 / 1 = 9 (A.M.≥H.M.)