If $\,\,\frac{x^2}{cy+bz}=\frac{y^2}{az+cx}=\frac{z^2}{bx+ay}=1,$ then show that .... I am stuck on the following problem that one of my friends gave me: 
If $\,\,\frac{x^2}{cy+bz}=\frac{y^2}{az+cx}=\frac{z^2}{bx+ay}=1,$ then show that $$\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}=1$$. I did a problem which was similar to this one but I could not tackle this particular one. Can someone help? 

It's wrong. Put $x=y=1, z=-1$.  Then from the first condition  we get the system $$ -a+c=1, \\b+a=1,\\ -b+c=1. $$ It follows $ a= \frac 12,b= \frac 12,c= \frac 32.  $
Then 
$$
{\frac {a}{a+x}}+{\frac {b}{b+y}}+{\frac {c}{c+z}}={\frac {a}{a+1}}+{\frac {b}{b+1}}+{\frac {c}{c-1}}=\frac{11}{3}\neq 1.
$$
