已知x/y+z=a,y/x+z+b,z/x+y=c,求:a/1+a+b/1+b+c/1+c=1

Question

已知x/y+z=a,y/x+z+b,z/x+y=c,求:a/1+a+b/1+b+c/1+c=1

Answer 1

取倒数得：
x/A=(B+C)/BC=1/C+1/B
X/A+1/A=(X+1)/A=1/C+1/B+1/A
1/(X+1)=1/[A(1/A+1/B+1/C)]
Y/B=1/C+1/A
(Y+1)/B=1/C+1/A+1/B
Z/C=1/A+1/B
(Z+1)/C=1/A+1/B+1/C
1/(X+1)+1/(Y+1)+1/(Z+1)=1/[A(1/A+1/B+1/C)]+1/[B(1/A+1/B+1/C)]+1/[C(1/A+1/B+1/C)]
=(BC+AC+AB)/[ABC(1/A+1/B+1/C)]
=(BC+AC+AB)/(BC+AC+AB)
=1