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Homework Help: AM-GM Inequality - Troubles with an example

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Let ##a## and ##b## real numbers such that ##a>b>0##.

    Determine the least possible value of ##a+ \frac{1}{b(a-b)}##

    I took this example from page 3 of this paper

    2. Relevant equations

    In the article previously linked, explaining the example, the author writes down:

    [itex]a+ \frac{1}{b(a-b)}=(a-b)+b+\frac{1}{b(a-b)}[/itex]

    Now, where does that come from?

    3. The attempt at a solution

    As the title says, at least I am aware of what the topic is... (!). So everything moves around the AM-GM inequality.

    [itex]\frac{a_1 + \dots + a_n}{n} \geq \sqrt[n]{a_1 \dots a_n}[/itex]

    I have to admit I have some troubles figuring out what's going on here, so it's not a matter of solving something, it's more about showing why I don't see the solution.

    I tried to manipulate a bit the first formula, but it's not about that I guess, cause I really cannot see how the equality in 2. stands.

    So, I am looking forward to any feedback. Thanks a lot.
  2. jcsd
  3. Jan 21, 2013 #2


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    Try moving the parens around on the RHS, you should be able to see the equality.
  4. Jan 21, 2013 #3
    Jeez. That's really bad... completely mathematically blind.
    I am quite ashamed of myself. :redface:

    Now that I see that 2+2=4, I have some problems that I am afraid will be challenging like 3+3=???.

    There are some questions I have related with other things I don't understand.

    1) Why do we need to use that trick I didn't understand? Why do we need it?
    2) Why in the LHS we have a 3 before the root? From that 3, it seems that the GM formula should be ##n \sqrt[n]{a_1 \dots a_n}##.
    3) In other words, why do we need 3 elements?
    4) Last dumb question, why are ##b##, ##(a-b)## and ##1/b(a-b)## our three elements? Shouldn't they be ##a## and ##b##?

    I guess that now it's clear that I have a real problem with inequalities.
  5. Jan 21, 2013 #4
    Ok, correct me if I am wrong.
    I think I see now what's going on here.

    1) we build up that trick I couldn't see to get rid of everything under the root in the GM side;
    2) we have a 3 in the LHS cause it comes from the AM in the RHS;
    3) yeah, we do need three elements to implement that trick on 1.;
    4) no, the elements are indeed three.

    Did I guess it right? :smile:
  6. Jan 25, 2013 #5
    Random question here : what ensure that the solution 3 is the LEAST possible solution?
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