6. By Cauchy-Schwarz inequality, $\left(1+\frac{y}{x}+\frac{z}{x}\right)(1+x y+x z) \geqslant(1+y+z)^{2} \Rightarrow$ $\frac{1+x y+x z}{(1+y+z)^{2}} \geqslant \frac{x}{x+y+z}$. Similarly, $\frac{1+y z+y x}{(1+z+x)^{2}} \geqslant \frac{y}{x+y+z} \cdot \frac{1+z x+z y}{(1+x+y)^{2}} \geqslant$ $\frac{z}{x+y+z}$. Adding the above three inequalities yields $\frac{1+x y+x z}{(1+y+z)^{2}}+\frac{1+y z+y x}{(1+z+x)^{2}}+$ $\frac{1+z x+z y}{(1+x+y)^{2}} \geqslant 1$