因为 (a + 1)(b + 1)(c + 1) > 0以及 (a - 1)(b - 1)(c - 1) < 0故 (a + 1)(b + 1)(c + 1) - (a - 1)(b - 1)(c - 1) > 0即 2(ab + bc + ca + 1) > 0因此 ab + bc + ca + 1 > 0.