the max value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$ What is the max value of the term $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$
If $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6$,
$a,b,c,d \in \mathbb{R}$
I tried to get from the equation to the term but it's too complicated.
Any idea?
Thanks.

Let $w=\frac{a}{b}, x=\frac{b}{c}, y=\frac{c}{d}, z=\frac{d}{a}$, so that $wxyz=1, w+x+y+z=6$,  then $$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=wx+xy+yz+zw=(w+y)(x+z) \leq \frac{((w+y)+(x+z))^2}{4}=9$$
(Note $st \leq \frac{(s+t)^2}{4} \Leftrightarrow 0 \leq (s-t)^2$)
Equality is achieved when $w+y=x+z$, so $w+y=x+z=3, wxyz=1$. We can achieve this for example by $y=x=\frac{3-\sqrt{5}}{2}, w=z=\frac{3+\sqrt{5}}{2}$, $(a, b, c, d)=(1, \frac{3-\sqrt{5}}{2}, 1, \frac{3+\sqrt{5}}{2})$.
