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Help with the proof

  1. May 24, 2013 #1
    If a and b are real numbers, then a^2 + b^2 >= 2 a b
    Proof:
    if a>=0 and b>=0
    then a-b>=0
    (a-b)^2>=0
    a^2 + b^2 -2ab >=0
    a^2 + b^2 >= 2ab
    Q.E.D.

    at http://answers.yahoo.com/question/index?qid=20081104233240AApFu9W

    someone has proved more or less similar question, but at

    (a + b)^2 > 0
    a^2 + 2ab + b^2 > 0
    a^2 + b^2 ≥ 2ab

    when 2ab goes from L.H.S. to R.H.S then it should be -2ab , am I correct?
     
  2. jcsd
  3. May 24, 2013 #2

    Fredrik

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    Yes, and that's why you should use ##(a-b)^2## instead of ##(a+b)^2## to prove this result.

    You should remove the first two lines from the first proof. Do you see why?
     
  4. May 24, 2013 #3
    Because a and b could be equal, right?

    If I am correct then it should be:

    Proof:
    If a-b>=0
    then (a-b)^2 >= 0
    a^2 + b^2 - 2ab >= 0
    implies a^2 + b^2 >= 2ab
    Q.E.D.

    But if a and b could be equal to each other then ">=" justifies that so the proof in my original post remains correct, isn't?
     
  5. May 24, 2013 #4

    HallsofIvy

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    This is not true. Given that a and b are arbitrary real numbers, a- b can be any number, positive, negative, or 0.
    However, since a square of a real number is never negative, the next step is correct:

    If a and b are not allowed to be equal, yes. otherwise (a+ b)^2>= 0 as you had before.

    Yes, but since a and b can be any real numbers, either 2ab< 0, in which case, since a^2+ b^2 is not negative, it is a trivial statement, or 2ab> 0 so that the sign doesn't matter.
     
  6. May 24, 2013 #5
    Ahaan i see your point.

    Proof:
    If a-b >=0
    then (a-b)^2 >=0
    a^2 +b^2 -2ab >=0
    a^2 + b^2 >=2ab
    Q.E.D.

    What do you say?
     
  7. May 24, 2013 #6

    Office_Shredder

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    You seem to be stuck on this idea of a-b ≥ 0 that is completely unnecessary. Compare your proof to the following:

    If the sky is blue
    then (a-b)^2 >=0
    a^2 +b^2 -2ab >=0
    a^2 + b^2 >=2ab
     
  8. May 24, 2013 #7
    Ohh got it. It says "a and b are real numbers".

    Proof:
    If (a-b)^2 >=0
    then a^2 + b^2 >=0
    Q.E.D.

    Right?
     
  9. May 24, 2013 #8

    Fredrik

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    No, the problem is that if the proof relies on the assumption that ##a,b\geq 0##, then the result will only be valid for such values of ##a## and ##b##

    Don't forget that the statement you want to prove is that "For all real numbers ##a## and ##b##, we have ##a^2+b^2\geq 2ab##. It's for all real numbers, not just for the non-negative ones.


    What makes you think that ##(a-b)^2## may be less than 0, or that it needs to be greater than or equal to zero to ensure that ##a^2+b^2\geq 0##?
     
  10. May 25, 2013 #9
    Proof:
    If a,b>=0
    then (a-b)^2 >=0
    a^2 +b^2 -2ab >= 0
    a^2 + b^2 >= 2ab
    QED

    I don't understand that how come (a+b)^2 >=0 can lead us to prove that a^2 + b^2 >= 2ab because (a+b)^2 >= 0 will lead us to the resut a^2 +b^2 >= -2ab , please explain to me in a simple language becaueee I am a beginner.

    PS. I want to prove this thm by using direct proof's if & then technique.

    Thanks in advance.
     
  11. May 25, 2013 #10

    micromass

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    Can you give me some real numbers such that ##(a-b)^2<0##?
     
  12. May 25, 2013 #11

    Fredrik

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    I didn't express myself very clearly here. I meant that if you assume that ##a,b\geq 0## and then find that ##a^2+b^2\geq 2ab##, you will only have proved that ##a^2+b^2\geq 2ab## for all non-negative real numbers ##a## and ##b##.

    But your goal was to prove that ##a^2+b^2\geq 2ab## for all real numbers ##a## and ##b##.
     
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