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For \( a, b, c > 0, abc =1\) show that \[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge a+b+c \] No Lagrange multiplier allowed.

Mathematics
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\[a^2c+ab^2+bc^2\ge a+b+c\]
another form would be it. or \[ \frac{ab^2+bc^2+ca^2}{a+b+c} \ge 1\]

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Other answers:

most likely this problem won't take more than AM-GM
i want to say the next step is reducing it to 2 variables but i dont want to think, good luck to the next person
here's extra nice problem http://math.stackexchange.com/questions/275208/the-least-value-for-fracab354-fracbc354-fracca354
it took nearly week for me to figure out it's solution. try using this technique http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Weighted_AM.E2.80.93GM_inequality it will be lot shorter.

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