From $x, y, z \in (0, 1)$ it follows that also $1-x, 1-y, 1-z \in (0, 1)$. We have
$$\sum (1-x)y = \sum x - \sum xy,$$
while
$$\prod x = \prod (1-x) \text{ translates into } (\sum x - \sum xy) + 2xyz = 1, \text{ hence } \sum (1-x)y + 2\sqrt{\prod (1-x)y} = 1 = 3 \cdot \frac{1}{4} + 2\sqrt{\left(\frac{1}{4}\right)^3}.$$
Now, either $\sum (1-x)y \ge 3 \cdot \frac{1}{4}$, or $\prod (1-x)y \ge \left(\frac{1}{4}\right)^3$, so at least one expression is at least $\frac{1}{4}$. In fact always $\prod (1-x)y = \prod (1-x)x \le \prod \left(\frac{(1-x)+x}{2}\right)^2 = \left(\frac{1}{4}\right)^3$.

Alternative Solution. The use of values $x$ and $1-x$, et al. does suggest looking around value $\frac{1}{2}$, the mid-value. There are four cases.

1) If none of $x, y, z$ is at least $1/2$, then $\prod x < \prod (1-x)$, absurd.

2) If exactly one of $x, y, z$ is at least $1/2$, say $x$, then $(1-z)x > 1/4$.

3) If exactly two of $x, y, z$ are at least $1/2$, say $y, z$, then $(1-x)y > 1/4$.

4) If all three of $x, y, z$ are at least $1/2$, then they all must be equal to $1/2$, otherwise $\prod x > \prod (1-x)$, absurd. Now all three expressions are equal to $1/4$.