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B How to construct a proof?

  1. Jul 19, 2016 #1
    a and b are integers

    Prove that:
    2ab <= a2 + b2

    I have tested various values for a and b and determined that the statement seems to be generally true. I'm having a hard time though constructing a formal proof.

    It will not do to suppose the statement is wrong and then provide a counterexample. This would only prove that the statement is true for those particular values of a and b, and not for all values.

    I learned about mathematical induction but I think that method is used to prove statements about a well-ordered set. a and b can be any integer so I don't think that would work.

    I wondered what kind of statement may have simplified to produce a2 + b2 so I played with (a+b) * (a+b) which produced a2 + b2 + 2ab and also with (a-b) * (a-b) which produced a2 + b2 - 2ab. This looks similar to the original statement, and I feel like I may be onto some clues.

    Without giving me the answer, I wonder if someone could give me a push in the right direction?
  2. jcsd
  3. Jul 19, 2016 #2


    Staff: Mentor

    The last formula gives you the answer. What do you know about squares?
  4. Jul 19, 2016 #3
    Well I know that squares are always positive.
  5. Jul 19, 2016 #4


    Staff: Mentor

    Then gather all together. Start with "all square are positive". Then substitute a certain square of one of your formulas and rearrange the terms..
  6. Jul 19, 2016 #5
    I don't know what some of your comment means. I'm trying hard to figure it out. I'm looking at the term (a-b)2 and I think that may be the "certain square" you are talking about. It's equal to a2 + b2 - 2ab though so I don't know what I could substitute it for. Maybe I'm on the wrong track here altogether.
  7. Jul 19, 2016 #6


    Staff: Mentor

    No, you are done. The square is not negative (##≥ 0##) and all which remains, is to add ##2ab## on both sides.
  8. Jul 19, 2016 #7
    How is that a proof that 2ab is always less than or equal to a2 + b2 when a and b are integers?
  9. Jul 19, 2016 #8


    Staff: Mentor

    The only way to help you further is to type in the proof.

    Because you can only derive true statements from true statements, it is a proof:
    ##0 ≤ (a - b)^2## for all integers, and even for real numbers, too, is a true statement. The next step is to multiply the square which is of course also true. Shifting numbers by addition doesn't change the order, so you are allowed to add ##2ab## on both sides and get again a true statement, which is the one you want to have.
  10. Jul 19, 2016 #9
    Thanks. I didn't think about it like that. That is, starting with a separate true statement and transforming it into the one I want.
  11. Jul 20, 2016 #10


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    Science Advisor
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    Gold Member

    That's probably the simplest proof but you could have done it simply by contradiction as well:

    Suppose ## a^2 + b^2 < 2ab ## then

    ##a^2+b^2 -2ab <0##
    ##(a-b)^2 <0##

    Which is a contradiction.
  12. Jul 20, 2016 #11
    Nice one Perok. Contradictions are a lot more intuitive often times.
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