4. The real numbers $a, b, c, d$ are positive and satisfy $(a+c)(b+d)=a c+b d$. Find the minimum of

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

Answer: 8.
Solution(Raphael):

After trying out some things one might suspect that the minimum is reached when $a=c$ and $b=d$. So we'll try to use an inequality with equality case $a=c$ and $b=d$. Going for the easiest such inequality one comes across:

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

Where we just used AM-GM on the terms $(a+c)$ and $(b+d)$ and then divided everything by $\sqrt{a b c d}$, to get the second inequality.

Now if $a=c$ and $b=d$ then $\frac{a}{b}=\frac{c}{d}$ and $\frac{b}{c}=\frac{d}{a}$. So another inequality to try is:

$$
\frac{a}{b}+\frac{c}{d}+\frac{b}{c}+\frac{d}{a} \geq 2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}}
$$

which is once again AM-GM. So combining these two inequalities one gets:

$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \geq 2 \cdot\left(\sqrt{\frac{a c}{b d}}+\sqrt{\frac{b d}{a c}}\right) \geq 2 \cdot 4=8
$$

with equality only when $\frac{a}{b}=\frac{c}{d}$ and $\frac{b}{c}=\frac{d}{a}$.

Now we still have to show that numbers $a=c, b=d$ exist which fulfil the condition $a c+b d=$ $(a+c)(b+d)$ Plugging in $c=a$ and $d=b$ we get:

$$
a^{2}+b^{2}=4 a b \Leftrightarrow\left(\frac{a}{b}\right)^{2}-4 \frac{a}{b}+1=0
$$

which has the solutions $\frac{a}{b}=2 \pm \sqrt{3}$. so using $a=c=(2+\sqrt{3}) \cdot b=(2+\sqrt{3}) \cdot d$ fulfils the condition.