If a, b, c are distinct positive numbers, then the nature of roots of the equation 1x−a+1x−b+1x−c=1x is

The equation on simplifying givesx(x−b)(x−c)+x(x−c)(x−a)+x(x−a)(x−b)−(x−a)(x−b)(x−c)=0         (1)Let f(x)=x(x−b)(x−c)+x(x−c)(x−a)+x(x−a)(x−b)−(x−a)(x−b)(x−c)We can assume without loss of generality that a < b < c. Now,f(a)=a(a−b)(a−c)>0f(b)=b(b−c)(b−a)<0f(c)=c(c−a)(c−b)>0So, one root of (1) lies in(a, b) and one root in (b, c). Obviously the third root must also be real.