For #1 The inequality rewrite in the form \[(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a})^2(a+b+c+d)^2\ge 3(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d})(ab+ca+ad+bc+bd+cd))\] This inequality is true by computer! But I will not prove it.

rightways wrote: Given positive real numbers $a,b,c,d$ such that $ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6 \quad \text{and} \quad \frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}=36.$Prove the inequality ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}>ab+ac+ad+bc+bd+cd.$ $\frac{ab+cd}{ac}+\frac{bc+da}{bd}=6 , \frac{bc+da}{ac}+\frac{ab+cd}{bd}=36$ $\implies$ $ab+bc+cd+da=\frac{42abcd}{ac+bd} ,|a-c||b-d||ac-bd|=30abcd .$ By AM-GM, $\left((a-c)^2+(b-d)^2\right)(ac+bd)+(ac+bd)^2>2|a-c||b-d|(ac+bd)$ $>2|a-c||b-d||ac-bd|=60abcd>42abcd.$ $\implies \left((a-c)^2+(b-d)^2\right)(ac+bd)+(ac+bd)^2>42abcd$ $\iff({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}})(ac+bd)>(ac+bd)^2+42abcd$ $\iff{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}>ab+ac+ad+bc+bd+cd.$

Given real numbers $a,b,c,d$ such that $ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6 \quad \text{and} \quad \frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=8.$ Find all possible values of $ \frac{a}{b}+\frac{c}{d}.$

Also, some participants managed to prove that there are no such$a,b,c,d$ satisfying given equations :-P