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experimentX

  • 3 years ago

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.

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  1. Aryang
    • 3 years ago
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    *

  2. completeidiot
    • 3 years ago
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    \[a^2c+ab^2+bc^2\ge a+b+c\]

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

  4. experimentX
    • 3 years ago
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    most likely this problem won't take more than AM-GM

  5. completeidiot
    • 3 years ago
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    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
    • 3 years ago
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    here's extra nice problem http://math.stackexchange.com/questions/275208/the-least-value-for-fracab354-fracbc354-fracca354

  7. sauravshakya
    • 3 years ago
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  8. experimentX
    • 3 years ago
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    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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