Cho BT trên là S
Ta có: \(1+x^2=x^2+xy+yz+zx=\left(x+y\right)\left(x+z\right)\\ 1+y^2=\left(y+x\right)\left(y+z\right);1+z^2=\left(z+x\right)\left(z+y\right)\\ \Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+xz+yz\right)=2\)