<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] The hint.
Prove that $$\frac{a}{b}+\frac{c}{d}=\frac{d}{c}+\frac{c}{d}\leq-2$$ or
$$\frac{b}{c}+\frac{d}{a}=\frac{a}{d}+\frac{d}{a}\leq-2$$ and use 
 $$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=\left(\frac{a}{b}+\frac{c}{d}\right)\left(\frac{b}{c}+\frac{d}{a}\right).$$
A full solution.
Let $\frac{a}{b}+\frac{c}{d}>0$ and $\frac{b}{c}+\frac{d}{a}>0$.
Thus, by AM-GM $$4=\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right)=\left(\frac{d}{c}+\frac{c}{d}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)\geq2+2=4,$$
which gives $c=d$ and $a=d$, which is impossible.
Thus, one of the expressions $\frac{a}{b}+\frac{c}{d}$ or $\frac{b}{c}+\frac{d}{a}$ is negative.
Let $u=\frac{a}{b}+\frac{c}{d}<0$.
Thus, $$u=\frac{a}{b}+\frac{c}{d}=\frac{d}{c}+\frac{c}{d}\leq-2$$ and we need to prove that
$$u(4-u)\leq-12$$ or
$$(u+2)(u-6)\geq0,$$ which is obvious.
The equality occurs for $\frac{a}{b}+\frac{c}{d}=-2$, $\frac{b}{c}+\frac{d}{a}=6$ and $ac=bd.$
Easy to see that it's possible, which says that $-12$ is a maximal value.
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