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Proof exercise

  1. Jul 31, 2013 #1

    reenmachine

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    1. The problem statement, all variables and given/known data

    Prove that if ##x \in ℝ## and ##0<x<4## , then ##\frac{4}{x(4-x)}≥1##.

    (I'm really not sure about that one but managed to attempt something)

    2. Relevant equations

    We first want to prove that ##\forall y \in \mathbb{Z}## , the largest product of any ##a,b \in ℝ## with ##a+b=y## is when ##a=b=\frac{y}{2}##.

    Suppose ##a=b=\frac{y}{2}##.We have ##\frac{y}{2} \cdot \frac{y}{2} = \frac{y^2}{4}##.

    Now suppose ##a≠b≠\frac{y}{2}##.This implies that ##\exists x \in ℝ \ \ a=(\frac{y}{2}-x) \ \ b=(\frac{y}{2}+x)## with ##a## and ##b## being interchangeable.We have ##(\frac{y}{2}-x)(\frac{y}{2}+x) = \frac{y^2}{4}-x^2##.Since ##\frac{y^2}{4} > \frac{y^2}{4}-x^2## , this proves that ##\forall y \in \mathbb{Z}## , the largest product of any ##a,b \in ℝ## with ##a+b=y## is when ##a=b=\frac{y}{2}##.

    To get back at our original proof , we know that ##x+(4-x)=4##.From our proof above , we know that the largest product from any ##a,b \in ℝ## such that ##a+b=4## is if ##a=b=\frac{4}{2}##.We have ##2 \cdot 2 = 4##.This gives us ##\frac{4}{4}=1##.Since all other products of any other ##a,b \in ℝ## will result in a smaller number , we know that ##\frac{4}{4-n} \ \ n \in \mathbb{R} = 1+c ## with ##c \in R^+##.This proves that if ##x \in ℝ## and ##0<x<4## , then ##\frac{4}{x(4-x)}≥1##.


    any help greatly appreciated!!! thank you!
     
    Last edited: Jul 31, 2013
  2. jcsd
  3. Jul 31, 2013 #2

    lurflurf

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  4. Jul 31, 2013 #3

    reenmachine

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    Unfortunately I don't know about these concepts yet.I just tried to solve it on top of my head.Thank you for answering! Does this mean that the proof is incorrect , or is it simply a clearer way of proving it?
     
  5. Jul 31, 2013 #4

    lurflurf

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    Your proof is correct (except I am slightly confused by your y why is it important that it is an integer and is it defined by a+b=y or ab<=y^2/4 (of course both are true, but one is assumed and the other shown)) . As I mentioned you proof is the same as writing x(4-x)=4-(x-2)^2. In that form it is easy to see that x(4-x)=4-(x-2)^2<=4 with equality when x=2. We can also generalize slightly and write
    $$ab=\left( \frac{a+b}{2} \right) ^2 - \left( \frac{a-b}{2} \right) ^2 \le \left( \frac{a+b}{2} \right) ^2 $$
    we see that a=b is required for equality.
     
  6. Jul 31, 2013 #5

    verty

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    My method for questions like this one is to simplify the formula hoping to reverse the steps later. In this case it works brilliantly.
     
  7. Jul 31, 2013 #6
    Can you use calculus? This seems like an easy place to use derivatives to just show the local minimum on ##x\in(0,4)## is greater than or equal to 1.

    Your proof seems alright though. There are a few parts where I'm a little hesitant, though. Your notation toward the end implies that ##n## is a real number equal to ##1+c## for some positive real number ##c##. Clear these kinds of things up and you should be fine.
     
  8. Jul 31, 2013 #7

    reenmachine

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    No I didn't learn calculus yet unfortunately.Perhaps this is the wrong section for me to be in , but I figured writing proof didn't belong in the pre-calculus section.

    (I edited the n in N for n in R at the end but meant n in R+)

    Thanks for the advices!
     
    Last edited: Jul 31, 2013
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