If a, b, c are real, then both the roots of the equation(x−b)(x−c)+(x−c)(x−a)+(x−a)(x−b)=0                                  (1)are always

We can write (1) as3x2−2(a+b+c)x+bc+ca+ab=0                                    (2)The discriminant D of (2) is given byD=4(a+b+c)2−4×3(bc+ca+ab)=4a2+b2+c2+2bc+2ca+2ab−3bc−3ca−3ab=4a2+b2+c2−bc−ca−ab=2b2+c2−2bc+c2+a2−2ca+a2+b2−2ab=2(b−c)2+(c−a)2+(a−b)2As a, b, c are real, we get D ≥ 0. Thus, roots of (1) are real.