Let $x, y, z$ be real numbers such that $|x| \geq 2,|y| \geq 2,|z| \geq 2$. How small can $|x y z + 2(x + y + z)|$ be?
I. solution. We distinguish two cases.

1. case: All three numbers have the same sign. It is enough to deal with the subcase where each of them is positive. Then $x, y, z \geq 2$, so

$$
|x y z+2(x+y+z)|=x y z+2(x+y+z) \geq 8+2(2+2+2)=20
$$

2. case: Two numbers have the opposite sign of the third. Here, too, it is enough to deal with the subcase where two are negative and one is positive; for example, $x, y<0$, and $z>0$, so $x y z \geq 2 x y$ and $2(x+y+z) \geq 2(x+y)+4$, thus

$$
x y z+2(x+y+z)>2 x y+2(x+y)+4=2(x+1) \cdot(y+1)+2 .
$$

Here $x \leq-2$ and $y \leq-2$, so $x+1 \leq-1$ and $y+1 \leq-1$. This implies that $(x+1)(y+1) \geq 1$, and

$$
x y z+2(x+y+z) \geq 2(x+1)(y+1)+2 \geq 2+2=4
$$

Equality occurs when $x=-2, y=-2$ and $z=2$.

Thus, we have found that the minimum value of the expression $|x y z+2(x+y+z)|$ is 4, and this value is achieved precisely when the absolute value of each number is 2, and both positive and negative numbers occur among them.

II. solution. Here, too, we focus only on the two cases where $x, y, z>0$ or $x, y<0$. In the first case, the value of the expression is at least 20. In the second case, let $x=-2-a, y=-2-b, z=2+c, (a, b, c \geq 0)$. Then

$$
\begin{gathered}
x y z+2(x+y+z)=(-2-a)(-2-b)(2+c)+2((-2-a)+(-2-b)+(2+c))= \\
=8+4 a+4 b+4 c+2 a b+2 a c+2 b c+a b c+2(c-a-b-2)= \\
=4+2 a+2 b+6 c+2 a b+2 a c+2 b c+a b c \geq 4
\end{gathered}
$$

(since $a, b, c \geq 0$), and equality only occurs when $a=b=c=0$, that is, when two of $x, y$, and $z$ have the value 2 and one has the value -2, or vice versa.