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 $(2,1,2,1)$.