2015/092) Solve $\dfrac{x-a}{b+c} + \dfrac{x-b}{a+c} + \dfrac{x-c}{a+b} = 3$ given a,b,c positive

$\dfrac{x-a}{b+c} + \dfrac{x-b}{a+c} + \dfrac{x-c}{a+b} = 3$
or $(\dfrac{x-a}{b+c}-1) + (\dfrac{x-b}{a+c}-1) + (\dfrac{x-c}{a+b}-1) =0$
or $\dfrac{x-a-b-c}{b+c} + \dfrac{x-b-a-c}{a+c} + \dfrac{x-c-a-b}{a+b} =0$
or  $(x-a-b-c)(\dfrac{1}{b+c} + \dfrac{1}{a+c} + \dfrac{1}{a+b}) =0$
so $x-a-b-c) =0$ or  $\dfrac{1}{b+c} + \dfrac{1}{a+c} + \dfrac{1}{a+b} =0$
as a,b,c are positive so 2nd expression >0 and hence x = a+b+c