Question

x=by+cz,y=cz+ax,z=ax+by

若x=by+cz,y=cz+ax,z=ax+by.求证(a/1+a)+(b/1+b)+(c/1+c)=1

Answer 1

当x=by+cz①,y=cz+ax②,z=ax+by③时,
①-②得x-y=by-ax=z-ax-ax=z-2ax,得出a=(y+z-x)/2x,
①-③得x-z=cz-ax=cz-(y-cz)=2cz-y,得出c=(x+y-z)/2z,
②-③得y-z=cz-by=x-by-by=x-2by,得出b=x-y+z/2y,
由上得出1+a=x+y+z/2x
1+b=x+y+z/2y
1+c=x+y+z/2z
则1/(1+a)=2x/(x+y+z)
1/(1+b)=2y/(x+y+z)
1/(1+c)=2z/(x+y+z)
则a/(a+1)+b/(b+1)+c/(c+1) ==｛(y+z-x)/2x｝*｛2x/(x+y+z)｝+
｛(x-y+z)/2y｝*｛2y\(x+y+z)｝+｛(x+y-z)/2z｝*｛2z/(x+y+z)｝
=(y+z-x+x-y+z+x+y-z)/(x+y+z)=1