Junior Balkan Mathematical Olympiad

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 $1/4$.