1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove smallest Possible Value

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

    problemsoftheweek1112_zpsd91e9ac0.png


    2. Relevant equations

    n/a



    3. The attempt at a solution

    Here is one attempt

    problemsoftheweek111_zps83c1ef7d.png

    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.

    problemsoftheweek_zpsc4547f53.png

    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

    User Avatar
    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Prove smallest Possible Value
  1. Eigen value 0 prove (Replies: 21)

Loading...