Positive numbers a, B, C satisfy a + B + C-2 = 0, and prove that (2-A) (2-B) (2-C) is greater than or equal to 8abc
From a + B + C - 2 = 0, we can get: 2 - a = B + C; 2 - B = a + C; 2 - C = a + B; so: (2-A) (2-B) (2-C) = (B + C) (a + C) (a + B). And a, B, C are all positive numbers, so: a + b > = 2 radical (AB) a + C > = 2 radical (AC) C + b > = 2 radical (CB) triple multiplication can get: (a + b) (B + C) (C
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