[Algebra]
Problem: Suppose that $a, b, c, d$ are positive real numbers satisfying $(a+c)(b+d)=a c+b d$. Find the smallest possible value of

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

Solution: To show that $S \geqslant 8$, apply the AM-GM inequality twice as follows:

$$
\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right) \geqslant 2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}}=\frac{2(a c+b d)}{\sqrt{a b c d}}=\frac{2(a+c)(b+d)}{\sqrt{a b c d}} \geqslant 2 \cdot \frac{2 \sqrt{a c} \cdot 2 \sqrt{b d}}{\sqrt{a b c d}}=8 .
$$

The above inequalities turn into equalities when $a=c$ and $b=d$. Then the condition $(a+c)(b+d)=a c+b d$ can be rewritten as $4 a b=a^{2}+b^{2}$. So it is satisfied when $a / b=2 \pm \sqrt{3}$. Hence, $S$ attains value 8 , e.g., when $a=c=1$ and $b=d=2+\sqrt{3}$.
By homogeneity we may suppose that $a b c d=1$. Let $a b=C, b c=A$ and $c a=B$. Then $a, b, c$ can be reconstructed from $A, B$ and $C$ as $a=\sqrt{B C / A}, b=\sqrt{A C / B}$ and $c=\sqrt{A B / C}$. Moreover, the condition $(a+c)(b+d)=a c+b d$ can be written in terms of $A, B, C$ as

$$
A+\frac{1}{A}+C+\frac{1}{C}=b c+a d+a b+c d=(a+c)(b+d)=a c+b d=B+\frac{1}{B} .
$$

We then need to minimize the expression

$$
\begin{aligned}
S & :=\frac{a d+b c}{b d}+\frac{a b+c d}{a c}=\left(A+\frac{1}{A}\right) B+\left(C+\frac{1}{C}\right) \frac{1}{B} \\
& =\left(A+\frac{1}{A}\right)\left(B-\frac{1}{B}\right)+\left(A+\frac{1}{A}+C+\frac{1}{C}\right) \frac{1}{B} \\
& =\left(A+\frac{1}{A}\right)\left(B-\frac{1}{B}\right)+\left(B+\frac{1}{B}\right) \frac{1}{B} .
\end{aligned}
$$

Without loss of generality assume that $B \geqslant 1$ (otherwise, we may replace $B$ by $1 / B$ and swap $A$ and $C$, this changes neither the relation nor the function to be maximized). Therefore, we can write

$$
S \geqslant 2\left(B-\frac{1}{B}\right)+\left(B+\frac{1}{B}\right) \frac{1}{B}=2 B+\left(1-\frac{1}{B}\right)^{2}=: f(B)
$$

Clearly, $f$ increases on $[1, \infty)$. Since

$$
B+\frac{1}{B}=A+\frac{1}{A}+C+\frac{1}{C} \geqslant 4
$$

we have $B \geqslant B^{\prime}$, where $B^{\prime}=2+\sqrt{3}$ is the unique root greater than 1 of the equation $B^{\prime}+1 / B^{\prime}=4$. Hence,

$$
S \geqslant f(B) \geqslant f\left(B^{\prime}\right)=2\left(B^{\prime}-\frac{1}{B^{\prime}}\right)+\left(B^{\prime}+\frac{1}{B^{\prime}}\right) \frac{1}{B^{\prime}}=2 B^{\prime}-\frac{2}{B^{\prime}}+\frac{4}{B^{\prime}}=8
$$

It remains to note that when $A=C=1$ and $B=B^{\prime}$ we have the equality $S=8$.
We present another proof of the inequality $S \geqslant 8$. We start with the estimate

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

Let $y=\sqrt{a c}$ and $z=\sqrt{b d}$, and assume, without loss of generality, that $a c \geqslant b d$. By the AM-GM inequality, we have

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

Substituting $x=y / z$, we get $4 x \leqslant x^{2}+1$. For $x \geqslant 1$, this holds if and only if $x \geqslant 2+\sqrt{3}$.

Now we have

$$
2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}}=2\left(x+\frac{1}{x}\right)
$$

Clearly, this is minimized by setting $x(\geqslant 1)$ as close to 1 as possible, i.e., by taking $x=2+\sqrt{3}$. Then $2(x+1 / x)=2((2+\sqrt{3})+(2-\sqrt{3}))=8$, as required.

Final answer: 8