SELECTION TESTS FOR THE 2019 JBMO

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 $(1-x)(1-y)(z-1)\ge 0\iff xyz+2\ge xy+yz+xz+3-(x+y+z)\ge x+y+z$, the last inequality being equivalent to $2(xy+yz+zx)+6=(x+y+z{)}^2+4\ge 4(x+y+z)$.

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Alternative solution.

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