2016/072) If $x=a(y+z), y= b(z+x), z=c(x+y)$ then prove that, $ab+bc+ca+2abc=1$

we have
$x+xa = x(1+a) = a(x+y+z)$ or $\frac{a}{1+a} = \frac{x}{x+y+z}$
similarly
$\frac{b}{1+b} = \frac{y}{x+y+z}$
$\frac{c}{1+c} = \frac{z}{x+y+z}$
add the above 3 to get
$\frac{a}{1+a} + \frac{b}{1+b}+ \frac{c}{1+c} = 1$
or $a(1+b)(1+c) + b(1+c)(1+a) + c(1+a)(1+b) = (1+a)(1+b)(1+c)$
or $a + ab + ac + abc + b + bc + ab + abc = (1+a)(1+b) = 1 + a + b + ab$
or $  ac + abc  + bc + ab + abc = 1$
or $ab + bc+ca + 2abc = 1$