jmc

Putting everything over a common denominator, we get $$\begin{array}{rl}\frac{1}{(1-x)(1-y)(1-z)}+\frac{1}{(1+x)(1+y)(1+z)}& =\frac{(1+x)(1+y)(1+z)+(1-x)(1-y)(1-z)}{(1-x)(1-y)(1-z)(1+x)(1+y)(1+z)}\\ & =\frac{2+2(xy+xz+yz)}{(1-x^2)(1-y^2)(1-z^2)}.\end{array}$$Note that $2+2(xy+xz+yz)\ge 2$ and $(1-x^2)(1-y^2)(1-z^2)\le 1,$ so $$\frac{2+2(xy+xz+yz)}{(1-x^2)(1-y^2)(1-z^2)}\ge 2.$$Equality occurs when $x=y=z=0,$ so the minimum value is $\boxed{2}.$