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experimentX

this is bugging me: for any \( a, b, c, d > 0\) prove that \[ {a^2 \over b} + {b^2 \over c} + {c^2 \over d} +{d^2 \over a} \geq a+ b+c+d \] Also if \( a+b+c=ab+bc+ca \) \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} \]

  • one year ago
  • one year ago

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  1. mukushla
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    man u most be on tensor and pde now...lol :)

    • one year ago
  2. experimentX
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    on ... just for fun. if possible quick ... man

    • one year ago
  3. experimentX
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    just this is bugging me so badly Exercise 1.1.8.

    • one year ago
  4. experimentX
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    Ctrl + F ... search 1.1.8. Don't have much time to waste

    • one year ago
  5. waterineyes
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    Actually, this question is very easy.. @experimentX

    • one year ago
  6. experimentX
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    how ??

    • one year ago
  7. waterineyes
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    Did not you get the question clearly ?? It is saying to prove that thing..

    • one year ago
  8. experimentX
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    how do you prove that?

    • one year ago
  9. waterineyes
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    Actually, I am talking about question.. Its solution is easy or not I don't know.. Ha ha ha..

    • one year ago
  10. experimentX
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    already wasted 1 hrs ... can't waste another ... lol

    • one year ago
  11. waterineyes
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    Actually, that pdf is quite useful for me.. Thanks for giving me that..

    • one year ago
  12. experimentX
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    some fool upvoted me :D ... that doesn't work at all .... lol

    • one year ago
  13. mukushla
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    man second one's RHS will simplify to\[a^2+b^2+c^2\]

    • one year ago
  14. experimentX
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    yeah ... i know that ... just need to prove it.

    • one year ago
  15. experimentX
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    olypiad problems ... i find it extremely difficult even at my level man!! i need to dig a hole hide.

    • one year ago
  16. experimentX
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    what kind of technique is this http://www.wolframalpha.com/input/?i=Minimize[{x+%2B+2+n%2Fx%2C+n+%3C%3D+3+%26%26+x+%3E%3D+3}%2C+{x}]

    • one year ago
  17. mukushla
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    for the first one use AM_GM 4 times\[\frac{a^2}{b}+b\ge 2a\]\[\frac{b^2}{c}+c\ge 2b\]...

    • one year ago
  18. experimentX
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    crazy crafty manipulation

    • one year ago
  19. experimentX
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    the second one ... i can prove half of the original problem without it.

    • one year ago
  20. experimentX
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    the last inequality i can get bu the use of quadratic formula ... but can't find any use of ... inequality n<=3

    • one year ago
  21. experimentX
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    particularly I mean this http://www.wolframalpha.com/input/?i=Minimize[{x+%2B+3+n%2Fx%2C++x+%3E%3D+3}%2C+{x}]

    • one year ago
  22. experimentX
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    what kind of problems is this x + 3n/x ... for x >= 3

    • one year ago
  23. mukushla
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    sorry man i was out...how u realte this to second one?

    • one year ago
  24. experimentX
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    can't ...have to do without it.

    • one year ago
  25. experimentX
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    don't know what property is used here ..

    • one year ago
  26. mukushla
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    i found it from one my books i will copy it here \[(a+b+c)(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}) \geq (a+b+c)(a^2+b^2+c^2)\]\[\frac{ab^3}{c}+\frac{bc^3}{a}+\frac{ca^3}{b} \geq b^2c+c^2a+a^2b\]use AM_GM 3 times\[\frac{ab^{3}}{c}+\frac{bc^{3}}{a}\ge 2b^{2}c\]...

    • one year ago
  27. experimentX
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    which book do you use man??

    • one year ago
  28. mukushla
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    oh sorry man original book is in persian and not translated to other languages :(

    • one year ago
  29. mukushla
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    this is 500 pages about geometry and inequalities

    • one year ago
  30. experimentX
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    you follow a very good writer!!

    • one year ago
  31. mukushla
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    yeah :)

    • one year ago
  32. mukushla
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    see u later santosh

    • one year ago
  33. experimentX
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    see ya :) thanks for help man

    • one year ago
  34. waterineyes
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    Are you Linkha Santosh @experimentX ?? Sorry if there is any spelling mistake in name..

    • one year ago
  35. experimentX
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    no mistake ... the first is my surname.

    • one year ago
  36. waterineyes
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    Santosh Linkha ..??

    • one year ago
  37. waterineyes
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    That day you said now a days, everyone is on Facebook and I was just wondering why experimentX who said those words is not on Facebook..?? Now, it is clear to me..

    • one year ago
  38. experimentX
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    lol ... what words?

    • one year ago
  39. waterineyes
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    I have written those.. "Everyone is on Facebook".. I think these you said to Yahoo that day..

    • one year ago
  40. experimentX
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    oh yeah ... kinda remembered on thing "if you are searched in Google instead of fB then you are successful person"

    • one year ago
  41. waterineyes
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    Ha ha ha... Yeah, this is true one..

    • one year ago
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