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ParthKohli

Can anyone help me with this one?

  • 10 months ago
  • 10 months ago

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  1. ParthKohli
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    https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-ash4/389738_558445504198872_734761968_n.jpg Since I am too lazy to write everything down :-)

    • 10 months ago
  2. ParthKohli
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    @amistre64 @.Sam. ?

    • 10 months ago
  3. KingGeorge
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    Don't we also need the requirement that \(x,y,z\ge1\)?

    • 10 months ago
  4. ParthKohli
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    Yes.

    • 10 months ago
  5. ParthKohli
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    Oh, and it's not a homework or a test... just found it and wondered how to solve it. If you know the solution, then please solve it and show complete steps.

    • 10 months ago
  6. Bad2zBone
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    Are they integers?

    • 10 months ago
  7. ParthKohli
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    They are anything satisfying \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2\)

    • 10 months ago
  8. wmckinely
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    Well, @ParthKohli, it seems like you know more about how to solve this than anyone else here...-_-

    • 10 months ago
  9. ParthKohli
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    @wmckinely I just know the problem: nothing else ._.

    • 10 months ago
  10. Alchemista
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    What if you treat it as a minimization/maximization problem.

    • 10 months ago
  11. ParthKohli
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    that's a very good way to look at it.

    • 10 months ago
  12. Alchemista
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    http://en.wikipedia.org/wiki/Lagrange_multiplier

    • 10 months ago
  13. ParthKohli
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    But the biggest problem is how we proceed.

    • 10 months ago
  14. ParthKohli
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    Never mind, lol. I don't understand that mathematics. :-D

    • 10 months ago
  15. KingGeorge
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    Well, that might be a problem to understanding....

    • 10 months ago
  16. Alchemista
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    I'd have to actually do the math. But it looks like a simple application of the processed I linked you to. You have an equality constraint and function both over three variables.

    • 10 months ago
  17. Alchemista
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    Show that the minimum of the right hand of the inequality is greater than the maximum of left hand.

    • 10 months ago
  18. Alchemista
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    *greater or equal

    • 10 months ago
  19. ParthKohli
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    That is not what I am able to understand... how?

    • 10 months ago
  20. KingGeorge
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    Using some wolfram, this method does not seem to be a simple application of lagrange multipliers.

    • 10 months ago
  21. Alchemista
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    Ah ok

    • 10 months ago
  22. Alchemista
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    Did you use both constraints when plugging it into wolfram?

    • 10 months ago
  23. KingGeorge
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    I may have done it a bit wrong though. Let me double check.

    • 10 months ago
  24. Alchemista
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    Use the Minimize and Maximize routines.

    • 10 months ago
  25. asnaseer
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    I /think/ you can do this if you use:\[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}\]which I believe can be proved.

    • 10 months ago
  26. Alchemista
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    Lagrange, worked: Minimize[{Sqrt[x^2 + y^2 + z^2], 1/x + 1/y + 1/z == 2, x > 0, y > 0, z > 0}, {x, y, z}] = Maximize[{Sqrt[x - 1] + Sqrt[y - 1] + Sqrt[z - 1], 1/x + 1/y + 1/z == 2, x > 0, y > 0, z > 0 }, {x, y, z}]

    • 10 months ago
  27. Alchemista
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    I'll give you the link on wolfram alpha in a moment.

    • 10 months ago
  28. asnaseer
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    if you multiply both sides of your equation by xyz you get:\[yz+xz+xy=2xyz\]

    • 10 months ago
  29. asnaseer
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    then note that:\[(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=x^2+y^2+z^2+4xyz\] from above

    • 10 months ago
  30. Alchemista
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    Anyways, the solution using multivariable calculus is perfectly valid and works.

    • 10 months ago
  31. KingGeorge
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    The \(\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}\) was the fact I was looking for. I kept thinking the inequality was pointing the other direction :/

    • 10 months ago
  32. asnaseer
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    rearranging slightly we get:\[(x+y+z)^2=(x-1)^2+2x-1+(y-1)^2+2y-1+(z-1)^2+2z-1+4xyz\]\[=(x-1)^2+(y-1)^2+(z-1)^2+2x+2y+2z+4xyz-3\]

    • 10 months ago
  33. asnaseer
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    then take square roots of both sides to get:\[x+y+z=\sqrt{(x-1)^2+(y-1)^2+(z-1)^2+\text{other terms from above}}\]\[\ge\sqrt{(x-1)^2}+\sqrt{(y-1)^2}+\sqrt{z-1)^2}\]\[\ge(x-1)+(y-1)+(z-1)\]

    • 10 months ago
  34. asnaseer
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    then square root again to get:\[\sqrt{x+y+z}\ge\sqrt{(x-1)+(y-1)+(z-1)}\]\[\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\]

    • 10 months ago
  35. asnaseer
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    I /think/ this works

    • 10 months ago
  36. KingGeorge
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    Looks good to me.

    • 10 months ago
  37. help123please.
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    I dont understand that problem,Parth. Obviously.

    • 10 months ago
  38. asnaseer
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    thx @KingGeorge :)

    • 10 months ago
  39. help123please.
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    Hi,asnasser. Greetings,George.

    • 10 months ago
  40. Alchemista
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    In any case you will notice that Lagrange finds the local min/max at (1.5, 1.5, 1.5) for those functions subject to the given constraints.

    • 10 months ago
  41. experimentX
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    how is \[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b} \]

    • 10 months ago
  42. asnaseer
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    \[\begin{align} (\sqrt{a}+\sqrt{b})^2&=a+b+2\sqrt{ab}\\ \therefore a+b&=(\sqrt{a}+\sqrt{b})^2-2\sqrt{ab}\\ &\ge(\sqrt{a}+\sqrt{b})^2\\ \therefore \sqrt{a+b}&\ge\sqrt{a}+\sqrt{b} \end{align}\]

    • 10 months ago
  43. asnaseer
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    assuming a and b are both positive

    • 10 months ago
  44. asnaseer
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    hope that helps

    • 10 months ago
  45. asnaseer
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    it may not be a /strict/ proof but it is one that I used to convince myself that this relation might be valid.

    • 10 months ago
  46. KingGeorge
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    Wait...I'm no longer convinced. Let \(a=9\), \(b=16\). Then \(\sqrt{a+b}=5\) and \(\sqrt{a}+\sqrt{b}=3+4=7>5\).

    • 10 months ago
  47. asnaseer
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    :(

    • 10 months ago
  48. asnaseer
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    where is the flaw in my proof for this?

    • 10 months ago
  49. experimentX
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    there is a slight error \[ (\sqrt a + \sqrt b)^2 - 2 \sqrt a \sqrt b \leq (\sqrt a + \sqrt b)^2 \] proves the opposite

    • 10 months ago
  50. asnaseer
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    D'oh! thank for clarifying

    • 10 months ago
  51. asnaseer
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    but does that mean that I have proved the opposite relation to that being asked for?

    • 10 months ago
  52. experimentX
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    no ... \[ \sqrt {x-1}+... \ge \sqrt{x-1 + ... } \le \sqrt{x+y+z}\] condition is not strong enough

    • 10 months ago
  53. asnaseer
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    Ah! I see now - thanks again @experimentX (and @KingGeorge) Guess I'll have to think again on this problem - maybe calculas is the way to go as you guys showed above.

    • 10 months ago
  54. Alchemista
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    Why is no one satisfied with treating this as an optimization problem?

    • 10 months ago
  55. asnaseer
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    It's not that I wasn't satisfied - I just wondered if there was a simpler way of arriving at the solution. :)

    • 10 months ago
  56. experimentX
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    Lagrange definitely gives solution but I am still looking for more elementary proof.

    • 10 months ago
  57. KingGeorge
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    There's just something more appealing about the elementary proofs.

    • 10 months ago
  58. Alchemista
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    I suppose so. At the very least it would be more accessible.

    • 10 months ago
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