## experimentX Group Title 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. one year ago one year ago

1. Aryang Group Title

*

2. completeidiot Group Title

$a^2c+ab^2+bc^2\ge a+b+c$

3. experimentX Group Title

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

4. experimentX Group Title

most likely this problem won't take more than AM-GM

5. completeidiot Group Title

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

6. experimentX Group Title

here's extra nice problem http://math.stackexchange.com/questions/275208/the-least-value-for-fracab354-fracbc354-fracca354

7. sauravshakya Group Title

8. experimentX Group Title

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.