Slovenia — National Math Olympiad

For all real $a$ we have $|a| \ge a$ and $|a| \ge -a$. So,
$$|x + y| \ge -(x + y), \quad |x + 1| \ge x + 1, \quad |y + 1| \ge y + 1, \quad (1)$$

$|x + y| + |x + 1| + |y + 1| \geq -(x + y) + (x + 1) + (y + 1) = 2.$
Assume that the equality holds. Then the equality case occurs in all three inequalities (1) and so $x + y \leq 0$, $x + 1 \geq 0$ and $y + 1 \geq 0$. We get $-1 \leq x \leq -y \leq 1$ or $-1 \leq x \leq 1$. If $-1 \leq x \leq 1$ and $y = -x$, then
|x + y| + |x + 1| + |y + 1| = 2.
$$$$
We can conclude that only for all real $x \in [-1, 1]$ there exists $y$, such that the equality holds.