smc

Suppose that the product $xyz$ is minimized at $(x,y,z)=(x_0,y_0,z_0).$ Without the loss of generality, let $x_0\le y_0\le z_0$ and fix $y=y_0.$ To minimize $x{y}_{0}z,$ we minimize $xz.$ Note that $x+z=1-y_0.$ By a corollary of the AM-GM Inequality (If two nonnegative numbers have a constant sum, then their product is minimized when they are as far as possible.), we get $z_0=2x_0.$ It follows that $y_0=1-3x_0.$ Recall that $x_0\le 1-3x_0\le 2x_0,$ so $\frac{1}{5}\le x_0\le \frac{1}{4}.$ This problem is equivalent to finding the minimum value of $$f(x)=xyz=x(1-3x)(2x)=2x^2(1-3x)$$ in the interval $I=\left[\frac{1}{5},\frac{1}{4}\right].$ The graph of $y=f(x)$ is shown below: Since $f$ has a relative minimum at $x=0,$ and cubic functions have at most one relative minimum, we conclude that the minimum value of $f$ in $I$ is at either $x=\frac{1}{5}$ or $x=\frac{1}{4}.$ As $f\left(\frac{1}{4}\right)=\frac{1}{32}\le f\left(\frac{1}{5}\right)=\frac{4}{125},$ the minimum value of $f$ in $I$ is $\boxed{\frac{1}{32}}.$