The sum of three positive numbers is 1, and none of the numbers is greater than twice any other. What is the minimum of the product of the three numbers?
For any triplet of numbers $x, y, z$ that satisfy the conditions, let $t$ denote the middle number in size (equality is possible). The sum of the other two is then $1-t$, and since none of them is more than twice the other, the difference between them can be at most $\frac{1-t}{3}$. Using the identity $u \cdot v=\frac{(u+v)^{2}-(u-v)^{2}}{4}$, the product of the two "extreme" numbers is greater than or equal to

$$
\frac{(1-t)^{2}-\left(\frac{1-t}{3}\right)^{2}}{4}=\frac{2}{9}(1-t)^{2}
$$

Therefore, the product of the three numbers is certainly not less than $\frac{2}{9} t(1-t)^{2}$.

Let's examine the range within which $t$, the middle number among the three, can fall. Since the largest number is at most $2t$ and at least $t$, and the smallest is at most $t$ and at least $\frac{t}{2}$, we have

$$
\begin{gathered}
2 t+t+t \geq x+y+z \geq t+t+\frac{1}{2}, \quad \text { hence } \\
\frac{1}{4} \leq t \leq \frac{2}{5}
\end{gathered}
$$

We have found that for any triplet of numbers $x, y, z$ that satisfy the conditions, there is a $t \in\left[\frac{1}{4} ; \frac{2}{5}\right]$ such that $x y z \geq f(t)=$ $\frac{2}{9} \cdot t(1-t)^{2}$. Therefore, the product of the three numbers is certainly not less than the minimum of the function $f$ on the interval $\left[\frac{1}{4} ; \frac{2}{5}\right]$, which exists since $f$ is continuous. It is also true that for any $t \in\left[\frac{1}{4} ; \frac{2}{5}\right]$, the numbers $\frac{1-t}{3}, t, \frac{2}{3}(1-t)$ sum to 1, none of them is more than twice the other, and their product is exactly $f(t)$. Thus, the given products of $x y z$ also have a minimum, which is exactly the minimum of $f$ on the given interval.

$$
\text { 1986-12-436-1.eps }
$$

The function $f$ is concave on the interval $\left[\frac{1}{4} ; \frac{2}{5}\right]$, since its second derivative, $f^{\prime \prime}(t)=-\frac{2}{9}(4-6 t)$, is negative in this interval, so $f$ takes its minimum on $\left[\frac{1}{4} ; \frac{2}{5}\right]$ at one of the endpoints of the interval.

Substituting $f\left(\frac{1}{4}\right)=\frac{1}{32}<\frac{4}{125}=f\left(\frac{2}{5}\right)$, the minimum of the function is at $\frac{1}{4}$. Therefore, the product of the three numbers is at least $\frac{1}{32}$, and it is exactly this much when the numbers are $\frac{1}{4}, \frac{1}{4}$, and $\frac{1}{2}$.