(IMO SL 2020 A3) Let $a, b, c, d$ be strictly positive real numbers satisfying $(a+c)(b+d)=a c+b d$. Determine the smallest value that

$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}
$$

can take.
The hypothesis suggests grouping the terms in $a$ and $c$ on one side, and the terms in $b$ and $d$ on the other. We can thus group as follows (and this preserves a certain symmetry in $a$ and $c$ and in $b$ and $d$):

$$
\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right)
$$

To make factors $a c$ and $b d$ appear, we use the AM-GM inequality (IAG):

$$
\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right) \geqslant 2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}}
$$

We would then like to use the AM-GM inequality again by recognizing a term of the form $x + 1/x$. However, this is very demanding: we have not yet used the hypothesis from the statement, and a small reasoning shows that there do not exist real numbers $a, b, c, d$ satisfying $a c = b d$ and $(a+c)(b+d) = a c + b d$. However, we can still set $x = \sqrt{a c / b d}$ and determine the smallest value that $x + 1/x$ can take. The function in $x$ being convex, it suffices to determine the set of values that $x$ can take to be able to bound $f$ from below.

According to the AM-GM inequality and the hypothesis,

$$
a c + b d = (a+c)(b+d) \geqslant 2 \sqrt{a c} \cdot 2 \sqrt{b d} \geqslant 4 \sqrt{a c b d}
$$

By dividing both sides by $\sqrt{a c b d}$, we obtain that $x + \frac{1}{x} \geqslant 4$. Thus, the sum is always greater than or equal to 8.

To verify that the value is indeed attainable, we use the equality cases established previously. We see that we need $a = c$ and $b = d$. Substituting into the hypothesis, this gives $4 a b = a^2 + b^2$. We derive a quadratic equation in $a / b$, which gives the solution $a / b = 2 \pm \sqrt{3}$. The value 8 is therefore attained, for example, for $b = d = 1$ and $a = c = 2 + \sqrt{3}$.