13.38 Let $x, y$, and $z$ be three positive real numbers, their sum is 1. If none of them is more than twice the others, then the minimum value of the product $x y z$ is
(A) $\frac{1}{32}$.
(B) $\frac{1}{36}$.
(C) $\frac{4}{125}$.
(D) $\frac{1}{127}$.
(E) None of the above values.
(33rd American High School Mathematics Examination, 1982)
［Solution］Let $m=x_{0} y_{0} z_{0}$ be the minimum value of their product, and without loss of generality, assume
$$
x_{0} \leqslant y_{0} \leqslant z_{0} \text {. }
$$

Suppose $x_{1}=x_{0}-h, \quad z_{1}=z_{0}+h$, where $h>0$, and sufficiently small, so that $z_{1} \leqslant 2 x_{1}$, then $x_{1}, y_{0}, z_{1}$ still satisfy the original constraints, thus we have
$$
\begin{aligned}
x_{1} y_{0} z_{1} & =\left(x_{0}-h\right) y_{0}\left(z_{0}+h\right) \\
& =x_{0} y_{0} z_{0}+y_{0}\left[h\left(x_{0}-z_{0}\right)-h^{2}\right] \\
& <x_{0} y_{0} z_{0},
\end{aligned}
$$

This is because $\quad x_{0}-z_{0} \leqslant 0,-h^{2}<0$, so
$$
y_{0}\left[h\left(x_{0}-z_{0}\right)-h^{2}\right]<0 .
$$

Since $x_{0} y_{0} z_{0}$ is the minimum, the above inequality does not hold, thus $z_{0}=2 x_{0}$.
From the constraint $y_{0}=1-x_{0}-z_{0}=1-3 x_{0}$, so $m=2 x_{0}^{2}\left(1-3 x_{0}\right)$.
Also, by $x_{0} \leqslant 1-3 x_{0} \leqslant 2 x_{0}$, we have $\frac{1}{5} \leqslant x_{0} \leqslant \frac{1}{4}$.
Therefore, $m$ can be seen as the value of the function $f(x)=2 x^{2}(1-3 x)$ on the interval
$$
D=\left\{x \left\lvert\, \frac{1}{5} \leqslant x \leqslant \frac{1}{4}\right.\right\}
$$

In fact, $m$ is the minimum value of the function $f$ on $D$, because for any $x \in D$, taking $y=$ $1-3 x, z=2 x$, the resulting $x, y, z$ satisfy the original constraints, and we get $f(x)=x y z$.

To find the minimum value of $f(x)$ on $D$, we plot the graph of the function $f(x)$ (as shown in the figure).

Since $f(x)$ has a minimum at $x=0$ (from $f(x)=$ $2 x^{2}(1-3 x)$ we can see that $f(x)$ is positive or zero for $x<\frac{1}{3}$, and when $x=0$, $f(x)=0$), and a cubic curve can have at most one minimum, because if the function $f(x)$ has another minimum, say at $x=a$, then the equation $f(x)=$ $f(a)$ will have at least 4 roots. This is impossible, as a cubic equation can have at most three roots.
Therefore, the minimum value of the function $f(x)$ on $D$ must be at one of the endpoints, in fact
$$
f\left(\frac{1}{4}\right)=\frac{1}{32} \leqslant f\left(\frac{1}{5}\right)=\frac{4}{125} .
$$

Hence, the answer is $(A)$.