Selection Tests for the JBMO, 2019

Notice that $2xy \le x^2 + y^2 \le x^2 + y^2 + z^2 = 2$, therefore $xy \le 1$. Similarly, $xz \le 1$, and $yz \le 1$. We also have $(x+y)^2 = x^2 + y^2 + 2xy \le 4$, so $x+y \le |x+y| \le 2$. Also, $x+z \le 2$, and $y+z \le 2$. If one of the numbers is negative, let's say $z$, then the conclusion follows from the inequalities $x+y \le 2$ and $z \le xyz$.

In the case $x, y, z \in [0, 1]$, we have $z(1 - xy) \le 1 - xy \le 2 - x - y$ (the last inequality is equivalent to $(1-x)(1-y) \ge 0$).

We may have at most one number greater than 1, otherwise the product of two such numbers would be greater than 1. We are left with the case $x, y \in [0, 1]$ and $z > 1$. In this case $(1-x)(1-y)(z-1) \ge 0 \Leftrightarrow xyz + 2 \ge xy + yz + xz + 3 - (x+y+z) \ge x+y+z$, the last inequality being equivalent to $2(xy+yz+zx)+6 = (x+y+z)^2+4 \ge 4(x+y+z)$.

Using Cauchy-Buniakowsky-Schwarz we get $(x + y + z - xyz)^2 = ((x+y)+z(1-xy))^2 \le ((x+y)^2+z^2)(1+(1-xy)^2) = (2+2xy)(2-2xy+x^2y^2) = 4+2x^2y^2(xy-1) \le 4$ since $x^2 + y^2 \le 2$ and $xy \le 1$.