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Help with a basic proof

  1. Jul 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that if r is a real number such that 0<r<1, then

    [tex] \frac{1}{r(1-r)} \geq 4.[/tex]


    3. The attempt at a solution

    Assume that r is a real number such that 0<r<1. Then r= 1/n where nεZ such that n≥2. As such,

    [tex] \frac{1}{r(1-r)} = \frac{1}{\frac{1}{n}(1-\frac{1}{n})} = \frac{n^2}{n-1} \geq 4. [/tex]

    It can be shown by induction that this inequality is true for n≥2 where n is an integer.

    First the base case is proven where n=2.

    [tex] \frac{2^2}{2-1} = 4 \geq 4 [/tex]

    Now we let n=k such that [tex] \frac{k^2}{k-1} \geq 4 [/tex] and show that the inequality is true for k+1. Observe that

    [tex] \frac{(k+1)^2}{(K+1)-1} = \frac{k^2+2k+1}{k} \geq 4 [/tex] which we can rearrange as [tex] k^2 \geq 4k-2k-1 = 2k-1 = 2(k-\frac{1}{2}) .[/tex]

    From the induction hypothesis we see that [tex] k^2 \geq 4k-4 = 4(k-1) [/tex] which shows that the inequality is true for k+1. Thus [tex] \frac{n^2}{n-1} = \frac{1}{r(1-r)} \geq 4 [/tex] for nεZ such that n≥2.


    I apologize if this is a bit hard to read... this my first attempt at latex! I realize that this only takes into account rational numbers and disregards the rest of the reals. I made this attempt hoping I would see a way to complete the proof, but I am drawing a blank on how to include the rest of the real numbers. Any suggestions are greatly appreciated!
     
  2. jcsd
  3. Jul 27, 2012 #2
    It can be easier to manipulate the original inequality into a quadratic equation, and then find the minimum value on the range (0,1) to show that it is always larger than 4.
     
  4. Jul 28, 2012 #3
    As Villyer said, simply manipulating the inequality from what you are given will do the trick. Also, just because r is a real number DOES NOT mean that r = 1 / n for some integer n. That is not even true for rational numbers, so the proof you have right now is technically incorrect.

    [EDIT] Actually, now that I have read your "proof" more carefully, it is very incorrect since you use induction, which relies on the assumption that r = 1/n for some integer n, which is false.
     
  5. Jul 28, 2012 #4
    Ah. I again over-complicate my search! So now I have:

    Let rεR such that 0<r<1. Then we have
    [tex] \frac{1}{1(1-r)} \geq 4
    \longleftrightarrow 1 \geq 4r-4r^2
    \longleftrightarrow 0 ≥ -4r^2+4r-1
    \longleftrightarrow 0 ≥ -(2r-1)^2
    \longleftrightarrow 0 ≤ (2r-1)^2.
    [/tex]

    / end formal argument!

    I know that r= 1/2 here and that this is the minimum of the range between 0<r<1, which can be shown with first/second derivative tests. What is necessary for me state in order to show that 4 is the minimum value of the range on 0<r<1? Is there a simpler way for me to verify that this is the minimum value?

    @who_:
    I see my flaw! A number 0<r<1 could not only be 1/n but also 3/9, sqrt(6)/99..... I don't know where I got that idea from :redface:
     
  6. Jul 28, 2012 #5
    That's a much nicer and cleaner proof :)

    Haha, no worries - even the most talented mathematicians make mistakes like those.
     
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