Belarus — Final Round

It is evident that
$$|(a-b)(b-c)(c-d)(d-a)| \le \frac{abcd}{4} \iff$$
$$\frac{(a-b)^2}{ab} \cdot \frac{(b-c)^2}{bc} \cdot \frac{(c-d)^2}{cd} \cdot \frac{(d-a)^2}{da} \le \frac{1}{16}. \qquad (*)$$

Note that
$$\frac{(a-b)^2}{ab} \le \frac{1}{2}. \qquad (1)$$
Indeed, we have
$$\frac{(a-b)^2}{ab} \le \frac{1}{2} \Leftrightarrow 2(a-b)^2 \le ab \Leftrightarrow 2\left(\frac{a}{b}\right)^2 - 5\left(\frac{a}{b}\right) + 2 \le 0 \Leftrightarrow \\ 2\left(\frac{a}{b} - 2\right)\left(\frac{a}{b} - \frac{1}{2}\right) \le 0.$$
The last inequality holds since $a, b \in [1, 2]$.
Similarly we have
$$\frac{(b-c)^2}{bc} \le \frac{1}{2}, \qquad (2)$$
$$\frac{(c-d)^2}{cd} \le \frac{1}{2}, \qquad (3)$$
$$\frac{(d-a)^2}{da} \le \frac{1}{2}. \qquad (4)$$
Multiplying (1), (2), (3), (4), we obtain the required inequality (*).

Note that the equality occurs when
$$\{(a, b, c, d) = (2, 1, 2, 1), (1, 2, 1, 2)\}.$$