5. Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=1$. If none of $x, y, z$ is more than twice another, then the minimum value of the product $xyz$ is ( ).
(A) $\frac{1}{127}$
(B) $\frac{4}{125}$
(C) $\frac{1}{36}$
(D) $\frac{1}{32}$
5.D.

Let $m=x_{0} y_{0} z_{0}$ be the minimum value of the product, and assume $x_{0} \leqslant y_{0} \leqslant z_{0}$. Suppose $x_{1}=x_{0}-t, z_{1}=z_{0}+t$, where $t>0$, and sufficiently small so that $z_{1} \leqslant 2 x_{1}$.
Thus, $x_{1}, y_{0}, z_{1}$ still satisfy the original constraints, therefore,
$x_{1} y_{0} z_{1}=\left(x_{0}-t\right) y_{0}\left(z_{0}+t\right)$
$=x_{0} y_{0} z_{0}+y_{0}\left[t\left(x_{0}-z_{0}\right)-t^{2}\right]0$;
When $\frac{2}{9}<x \leqslant \frac{1}{4}$, $f^{\prime}(x)<0$.
Hence $f(x)_{\min }=\min \left\{f\left(\frac{1}{4}\right), f\left(\frac{1}{5}\right)\right\}$
$=\min \left\{\frac{1}{32}, \frac{4}{125}\right\}=\frac{1}{32}$.