Finding the maximum of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$ If $a,b,c,d$ are distinct real numbers such that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}=4$ and $ac=bd$.
Then how would we calculate the maximum value of $$\dfrac{a}{c}+\dfrac{b}{d}+\dfrac{c}{a}+\dfrac{d}{b}.$$
I was unable to proceed due to the 'distinct'.

Let $w=\frac{a}{b}$, $x=\frac{b}{c}$, $y=\frac{c}{d}$, $z=\frac{d}{a}$. Then 
$$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=wx+xy+yz+zw=(x+z)(w+y)\leqslant\Bigl(\frac{w+x+y+z}{2}\Bigr)^2$$
