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 
$$ S=\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} $$
(Israel) Answer: The smallest possible value is 8.
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 \( \frac{a}{b}=2 \pm \sqrt{3} \). Hence, \( S \) attains value 8, e.g., when \( a=c=1 \) and \( b=d=2+\sqrt{3} \).