First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part)

To begin with, note that all the numbers are greater or equal to one, so the denominators are positive and the question is well-posed.

As a first step in our solution, let's observe that since $a$ and $b$ lie in a closed interval of length $1$, we must have $(a - b)^2 \le 1$, which can be rearranged to
$$\frac{1}{a^2 + b^2 - 1} \ge \frac{1}{2ab}.$$
Applying similar bounds to all four terms, we see that it suffices to prove the inequality
$$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{cd} + \frac{1}{da} \ge 2.$$

But since the condition $(a + c)(b + d) = 8$ is equivalent to $ab + bc + cd + da = 8$, this follows from the inequality between the arithmetic and harmonic means (or, if one prefers, from the Cauchy-Schwarz inequality).

In order for equality to be attained, we must have equality in both our estimates. From the first one, we can infer that
$$|a - b| = |b - c| = |c - d| = |d - a| = 1.$$
Since all the numbers lie in an interval of length $1$, this leaves us with only two options where equality can be attained, $(1, 2, 1, 2)$ and $(2, 1, 2, 1)$, with an easy check confirming that we indeed get equality.

Conclusion. The inequality holds by the argument above and we get an equality precisely for quadruples $(1, 2, 1, 2)$ and $(2, 1, 2, 1)$.