Irish Mathematical Olympiad

By hypothesis, $x$, $y$, $z \in [0, 1]$, and so $3xyz \le xy + yz + zx$. Also, $(x+y+z)^2 \le x+y+z$, i.e., $x^2+y^2+z^2+2(xy+yz+zx) \le x+y+z$, equivalently,
$$2(xy + yz + zx) \le x(1-x) + y(1-y) + z(1-z),$$
whence
$$6xyz \le 2(xy + yz + zx) \le x(1-x) + y(1-y) + z(1-z),$$
and there is equality iff $xyz = xy = xz = yz$ and $x+y+z = 0$ or $1$. This means that we have equality iff $(x, y, z)$ is one of $(0, 0, 0)$, $(0, 0, 1)$, $(0, 1, 0)$ and $(1, 0, 0)$.

Using the AM-GM inequality and the assumption $x+y+z \le 1$, we obtain
$$1 - x \ge y + z \ge 2\sqrt{yz}$$
$$1 - y \ge z + x \ge 2\sqrt{zx}$$
$$1 - z \ge x + y \ge 2\sqrt{xy}.$$
Multiplying these inequalities gives $(1-x)(1-y)(1-z) \ge 8xyz$ and so
$$xyz(1-x)(1-y)(1-z) \ge 8(xyz)^2 \ge (2xyz)^3$$
where we have used $1 \ge xyz$. Using AM-GM again, we obtain
$$x(1-x) + y(1-y) + z(1-z) \ge 3\sqrt[3]{x(1-x)y(1-y)z(1-z)} \ge 6xyz.$$
For the case of equality it is necessary that $x(1-x) = y(1-y) = z(1-z)$ and $xyz = 0$ or $1$. If $xyz = 1$, the first three inequalities above have no term equal to zero, hence all of them must be equalities, hence $x = y = z$ and $x+y+z = 1$. This implies $x = y = z = 1/3$ contradicting $xyz = 1$. Hence equality implies $xyz = 0$. Therefore, $x$, $y$, $z$ can only have the values $0$ or $1$. As $x+y+z \le 1$, at most one of them is $1$ and we have equality iff $(x, y, z)$ is one of $(0, 0, 0)$, $(0, 0, 1)$, $(0, 1, 0)$ and $(1, 0, 0)$.