Given ABC = 1, a + B + C = 2, A2 + B2 + C2 = 3, find the sum of (AB + C-1), (BC + A-1), (AC + B-1) A2 + B2 + C2 = 3, 2 after a, B, C is the square,

The title is wrong
Because according to a + B + C = 2 and a ^ 2 + B ^ 2 + C ^ 2 = 3, AB + BC + AC = 1 / 2
However, AB + BC + Ca > = 3 (ABC) ^ (2 / 3) = 3 (mean inequality), which leads to a contradiction
So the proposition is wrong!

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