2015丝绸之路国际数学竞赛不等式题

longzaifei

$a,b,c,d>0$
$ \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}=6 \quad \text{and} \quad \dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}=36.$

证明：

${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}>ab+ac+ad+bc+bd+cd.$