Three, let $a, b, c, d$ be four distinct real numbers such that
$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$, and $a c=b d$.
Find the maximum value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$.
Three, let $x=\frac{a}{b}, y=\frac{b}{c}$, then from $a c=b d$ we know $\frac{c}{d}=\frac{b}{a}=\frac{1}{x}, \frac{d}{a}=\frac{c}{b}=\frac{1}{y}$. Thus, the problem becomes finding the maximum value of $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}$ under the constraint $x \neq 1, y \neq 1, x+y+\frac{1}{x}+\frac{1}{y}=4$.
Let $x+\frac{1}{x}=e, y+\frac{1}{y}=f$, then $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}=e f$.
$\because$ When $t>0$, $t+\frac{1}{t} \geqslant 2$;
When $t<0, y<0$, then
$$
f \leqslant-2, e=4-f \geqslant 6, e f \leqslant-12 \text{. }
$$

When and only when $y=-1, x=3 \pm 2 \sqrt{2}$, the equality holds. Specifically, when $a=3+2 \sqrt{2}, b=1, c=-1, d=-(3+2 \sqrt{2})$, the equality holds.
$$
\therefore(e f)_{\max }=-12 \text{. }
$$