Inca o inegalitate

[DrAGos Calinescu]

Demonstrati ca \( \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{a+b+c}{\sqrt[3]{abc} \).

[Marius Mainea]

Din AM-GM

\( LHS=\frac{(2\frac{a}{b}+\frac{b}{c})+(2\frac{b}{c}+\frac{c}{a})+(2\frac{c}{a}+\frac{a}{b})}{3}\ge RHS \)