We consider all positive real numbers $a, b, c, d$ for which $(a+c)(b+d)=a c+b d$ holds.
Determine the smallest value that $S=\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$ can take.
The smallest possible value for $S$ is 8.
To show that $S \geq 8$ always holds, we apply the AM-GM inequality twice:
$\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right) \geq 2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}}=\frac{2(a c+b d)}{\sqrt{a b c d}}=\frac{2(a+c)(b+d)}{\sqrt{a b c d}} \geq 2 \cdot \frac{2 \sqrt{a c} \cdot 2 \sqrt{b d}}{\sqrt{a b c d}}=8$.
Equality holds at the points of the inequalities when $a=c$ and $b=d$. This allows us to rewrite $(a+c)(b+d)=a c+b d$ as $4 a b=a^{2}+b^{2}$. Solving for $a$ yields $a=b(2 \pm \sqrt{3})$, and thus a possible solution $S=8$ for $b=d=1$ and $a=c=2+\sqrt{3}$.