2015/081) If $a(y + z) = x, b(z + x) = y, c(x + y) = z$ show that $bc + ca + ab + 2abc = 1$

$a(y + z) = x$

Hence $\dfrac{1}{a} = \dfrac{y+z}{x}$

add 1 to both sides

$\dfrac{a+1}{a} = \dfrac{x + y +z}{x}$

so $\dfrac{a}{a+1} = \dfrac{x}{x+y+z}$

similarly

 

$\dfrac{b}{b+1} = \dfrac{y}{x+y+z}$

$\dfrac{c}{c+1} = \dfrac{z}{x+y+z}$

adding the above we get

$\dfrac{a}{a+1}+\dfrac{b}{b+1} +\dfrac{c}{c+1}= 1$

or $a(b+1)(c+1) + b(a+1)(c+1) + c(a+1)(b+1) = (1+a)(1+b)(1+c)$

or $(abc + ab + ac + a) + (abc + bc + ca + b) + (abc + ca + cb + c) = 1 + a + b+ c + ab + bc+ ca +abc$

hence $2abc + ab + bc + ca = 1$