46. By the AM-GM inequality, we have $\frac{x y}{z}+\frac{y z}{x} \geqslant 2 y, \frac{y z}{x}+\frac{z x}{y} \geqslant 2 z, \frac{x y}{z}+\frac{z x}{y} \geqslant 2 x$. Adding these three inequalities, we get $\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} \geqslant x+y+z=1$. Also, by $1=x+y+z \geqslant 3 \sqrt[3]{x y z}$, we have $x y z \leqslant \frac{1}{27}$. Therefore,
$$\begin{array}{l} \left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right)\left(\frac{1}{z}-z\right)= \\ \frac{1}{x y z}[(1+x)(1+y)(1+z)][(1-x)(1-y)(1-z)]= \\ \frac{1}{x y z}(2+x y+y z+z x+x y z)(x y+y z+z x-x y z)= \\ 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-2+\frac{(x y+y z+z x)^{2}}{x y z}-x y z= \\ 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-2+\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}+2(x+y+z)-x y z \geqslant \\ \frac{6}{\sqrt[3]{x y z}}-2+1+2-x y z \geqslant 18+1-\frac{1}{27}=\left(\frac{8}{3}\right)^{3} \end{array}$$

Equality holds if and only if $x=y=z=\frac{1}{3}$.