设x/a=y/b=z/c=k
∴xyz/abc=k^3
∵x/a=y/b=z/c
∴(x+y)/(a+b)=(y+z)/(b+c)=(z+x)/(c+a)=k
∴(x+y)(y+z)(z+x)/(a+b)(b+c)(c+a)=k^3
∴(a+b)(b+c)(c+a)/(x+y)(y+z)(z+x)=1/k^3
∴xyz(a+b)(b+c)(c+a)/abc(x+y)(y+z)(z+x)=k^3×1/k^3=1
∴xyz(a+b)(b+c)(c+a)/abc(x+y)(y+z)(z+x)=1