jmc

Let the three numbers be $x,$ $y,$ and $z.$ Without loss of generality, assume that $x\le y\le z.$ Then $z\le 2x.$

Suppose $z<2x.$ Let $x_1=\frac{x+z}{3}$ and $z_1=\frac{2x+2z}{3}.$ Then $z_1=2x_1,$ and $x_1+z_1=x+z.$ (We do not change the value of $y.$) Note that $$\begin{array}{rl}xyz-x_{1}y{z}_1& =y\left(xz-\frac{x+z}{3}\cdot \frac{2x+2z}{3}\right)\\ & =y\cdot \frac{(2z-x)(2x-z)}{9}>0.\end{array}$$This means that if $z<2x,$ and we replace $x$ with $x_1$ and $z$ with $z_1,$ the value of the product $xyz$ decreases. (The condition $x+y+z=1$ still holds.) So, to find the minimum of $xyz,$ we can restrict our attention to triples $(x,y,z)$ where $z=2x.$

Our three numbers are then $x\le y\le 2x.$ Since the three numbers add up to 1, $3x+y=1,$ so $y=1-3x.$ Then $$x\le 1-3x\le 2x,$$so $\frac{1}{5}\le x\le \frac{1}{4}.$

We want to minimize $$xyz=x(1-3x)(2x)=2x^2(1-3x).$$This product is $\frac{4}{125}$ at $x=\frac{1}{5},$ and $\frac{1}{32}$ at $x=\frac{1}{4}.$ We can verify that the minimum value is $\frac{1}{32},$ as follows: $$\begin{array}{rl}2x^2(1-3x)-\frac{1}{32}& =-\frac{192x^3-64x^2+1}{32}\\ & =\frac{(1-4x)(48x^2-4x-1)}{32}.\end{array}$$Clearly $1-4x\ge 0,$ and both roots of $48x^2-4x-1$ are less than $\frac{1}{5}.$ Therefore, $$2x^2(1-3x)-\frac{1}{32}=\frac{(1-4x)(48x^2-4x-1)}{32}\ge 0$$for $\frac{1}{5}\le x\le \frac{1}{4},$ and equality occurs when $x=\frac{1}{4}.$ Thus, the minimum value is $\boxed{\frac{1}{32}}.$