Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$

Question

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

Answer 1

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.

Answer 2

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