<s>[INST] Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ Let $a$,$b$,$c$ and $d$ be non-zero, pairwise different real numbers such that $ \frac{a}{b} +\frac{b}{c} +\frac{c}{d} + \frac{d}{a}=4$ and   $ac=bd$  .
Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ and that $-12$ is the maximum.
I simplified the inequality to prove: $a^2+b^2+c^2+d^2\le -12ac$
But I am not sure what to do next.
Hints and solutions would be appreciated. 
Taken from the 2018 Pan African Math Olympiad http://pamo-official.org/problemes/PAMO_2018_Problems_En.pdf
 [/INST] Let $u = \frac{a}{b},\,v=  \frac{b}{c},$ then $\frac{c}{d}=\frac{1}{u},\,\frac{d}{a}=\frac{1}{v}$ we get $u+v+\frac{1}{u}+\frac{1}{v}=4,$ and
$$P = \frac{a}{c} + \frac{b}{d} + \frac{c}{a} + \frac{d}{b} = \left(u+\frac{1}{u}\right)\left(v+\frac{1}{v}\right).$$
Now see $u,v$ can't satisfied $uv > 0.$ Indeed if $u > 0,\,v>0$ then
$$4 = u+\frac{1}{u}+v+\frac{1}{v} \geqslant 2 + 2 =4.$$
Equality occur when $u=v=1$ (contradiction the distinct numbers).
If $u<0,\,v<0$ replace $(u,v)$ by $(-x,-y)$ with $x>0,\,y>0$ then
$$4 = -\left(x+y+\frac 1 x + \frac 1 y\right)< 0 . $$
Thefore $uv < 0,$ now
$$P = -12 + \frac{(u+1)^2(v+1)^2}{uv}-2\left(u+v+\frac{1}{u}+\frac{1}{v}-4\right) \leqslant -12.$$
See here or here
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