Lời giải:
Ta có: Thay \(xyz=1\)
\(S=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}\)
\(S=\frac{z}{z+xz+xyz}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}\)
\(S=\frac{z}{z+xz+1}+\frac{xz}{xz+xyz+xz.yz}+\frac{1}{1+z+xz}\)
\(S=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{1+z+xz}\)
\(S=\frac{z+xz+1}{xz+z+1}=1\)
Vậy \(S=1\)