We are given that system of equations has non-trivial solution. So
⇒ 1( 1 - a2) + c(-c - ab) - b(ac + b) = 0
⇒ 1 - 2abc - a2 - b2 - c2 = 0
or a2 + b2 + c2 + 2abc = 1
⇒ a2 + b2c2 + 2abc = 1 - b2 - c2 + b2c2 = (1 - b2)(1 - c2)
⇒ (a + bc)2 = (1 - b2)(1 - c2)
Similarly, (b + ac)2 = (1 - a2)(1 - b2) and (c + ab)2 = (1 - a2)(1 - b2)
Hence, (1 - a2), (1 - b2) and (1 - c2) all have same sign. Since at least one of them is proper fraction, it implies all of them are positive.
So 1 - a2 > 0, 1 - b2 > 0, 1 - c2 > 0
⇒ a2 + b2 + c2 < 3
⇒ 1 - 2abc < 3
⇒ abc > -1