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Prove smallest Possible Value

  1. Aug 26, 2014 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    Here is one attempt


    But I'm stuck on this inequality. I can't go further.

    and here is another, but I don't know if I proved anything here.


    Really looking if anyone could help me on this or if I'm on the right track. Thanks.
  2. jcsd
  3. Aug 27, 2014 #2
    There are two components to the problem; (1) find a candidate for the solution and (2) prove that candidate is the right one.

    For (1) I recommend throwing away all this fancy schmancy algebra and calculus and just roll up your sleeves and try a few things. Also, recognizing that your problem is equivalent to maximizing ##\frac{1}{n}+\frac{1}{m}+\frac{1}{k}## subject to ##n,m,k## distinct and ##\frac{1}{n}+\frac{1}{m}+\frac{1}{k}<1## might make some of this work a little more manageable.

    Once you've found a triplet that works, try to prove that it's the best triplet. Don't get fancy, just think about it. If need be, find other triplets that work (in the sense that ##n,m,k## are distinct and ##\frac{1}{n}+\frac{1}{m}+\frac{1}{k}<1##) and try to see why your triplet is better.
  4. Aug 27, 2014 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    No, you are on the wrong track: you cannot take derivatives with respect to discrete (integer-valued) variables like n, m and k. Derivatives need continuous variables, and you don't have those in this problem.
  5. Aug 27, 2014 #4


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    Science Advisor

    And even if m, n, and k were continuous variables, you certainly cannot take the derivative with respect to three different variables as you did here.
  6. Aug 27, 2014 #5
    Thanks for pointing that out because I just remembered that it would be an implicit differentiation if I did take the derivative and I did not do that.

    Thanks for the hint, but I don't quite understand it. I thought I was trying to minimize it not maximize it.
  7. Aug 27, 2014 #6
    In this case if you maximize 1/n+1/m+1/k (while still being smaller than 1) you minimize your goal function, right?
  8. Aug 27, 2014 #7
    Thank you dirkmec1 I now understand!!
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