$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b+c}{\sqrt[3]{abc}}$
Edited by halloffame, 31-12-2018 - 09:39.
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b+c}{\sqrt[3]{abc}}$
Started By nhatminhkh2602, 26-12-2018 - 15:35
#2
Posted 26-12-2018 - 18:11
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$$\sum\limits_{cyc}\left ( \frac{\frac{\it{a}}{\it{b}}+ \frac{\it{a}}{\it{b}}+ \frac{\it{b}}{\it{c}}}{\it{3}} \right ) \geqq \sum\limits_{cyc}\frac{a}{\sqrt[\it{3}\,]{\it{abc}}}$$
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