Given a + B + C = 0, prove AB + BC + AC = 1

Given a + B + C = 0, prove AB + BC + AC = 1

This problem should have no solution in the range of real number
The simplest verification is: let a = 1; b = 1; C = - 2, and substitute it into the formula to be proved, then 1-2-2 = 1
It doesn't hold water
Complex point verification: multiply both sides of the formula by 2 to get (AB + BC) + (BC + AC) + (AB + AC) = 2;
That is: B (a + C) + C (a + b) + a (B + C) = 2, because a + B + C = 0, there are: a + C = - B, a + B = - C, B + C = - A, substituting the former formula, we get
-B ^ 2-C ^ 2-A ^ 2 = 2, that is, B ^ 2 + C ^ 2 + A ^ 2 = - 2