Q.15see image...........................................................

a(y+z)=x ———————(I)

b(z+x)=y ———————(II)

c(x+y)=z ———————(III)

from eq (I) we get,

a=x/(y+z) ———————(IV)

from eq (II) we get,

b=y/(z+x) ———————(V)

from eq (III) we get,

c=z/(x+y) ———————(VI)

SUBSTITUTIN VALUE OF a,b and c in given eq

we get,

(1/1+(x/(y+z))+(1/1+(y/(z+x))+(1/1+(z/(x+y))

by solving above eq we get,

((y+z)/(x+y+z))+((z+x)/(x+y+z))+((x+y)/(x+y+z))

=(2x+2y+2z)/(x+y+z)

=2(x+y+z)/(x+y+z)

=2