[3] - Algebraic Inequalities for Contests

Έστω πραγματικοί αριθμοί $a\ge b\ge c\ge 1\ge d\ge e>0$, τέτοιοι ώστε  

$ab+bc+cd+de+ea=54.$ 

Να αποδειχθεί ότι 

$${\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}+{\displaystyle \frac{1}{d}}+{\displaystyle \frac{1}{e}}+10\ge 3(a+b+c+d+e).$$

Vasile Cîrtoaje, Oil-Gas University of Ploies,ti, România