|=a(bc−a2)−b(b2−ac)+c(ab−c2) =abc−a3−b3+abc+abc−c3 =3abc−(a3+b3+c3) As we know, AM ≥ GM ⇒
a3+b3+c3
3
≥3√a3×b3×c3 ⇒a3+b3+c3>3abc Now it is given that a, b and c are distinct positive real numbers. Therefore, the sum of the cube a3+b3+c3>abc, it implies that the value of the determinant is always gonna be negative