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More Schwarz inequality proofery

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the Schwarz inequality by first proving that

    [tex](x_{1}^{2} + x_{2}^{2})(y_{1}^{2} + y_{2}^{2}) = (x_{1} y_{1} + x_{2} y_{2})^{2} + (x_{1} y_{2} - x_{2} y_{1})^{2}.[/tex]
    2. Relevant equations

    [tex]x_{1} y_{1} + x_{2} y_{2} \leq \sqrt{x_{1}^{2} + x_{2}^{2}} \sqrt{y_{1}^{2} + y_{2}^{2}}[/tex]
    3. The attempt at a solution
    I'm not sure if my logic is right. I did the little proof above, and with that I can say

    [tex]-(x_{1} y_{2} - x_{2} y_{1})^{2} \leq (x_{1} y_{1} + x_{2} y_{2})^{2}.[/tex]
    Can I then sweep the LHS under the zero and say

    [tex](x_{1} y_{1} + x_{2} y_{2})^{2} \leq (x_{1}^{2} + x_{2}^{2})(y_{1}^{2} + y_{2}^{2}),[/tex]
    then take the square root to finish the proof?
     
  2. jcsd
  3. Mar 13, 2012 #2
    You don't need your relevant equation!

    You can prove the equality holds by simply starting with one side of the equation (remember not to assume they are equal, that would be assuming what you're trying to prove!) and end up with the other.

    You can start with either side to get the other but I'd suggest starting with the right hand side.

    See what you come up with!
     
  4. Mar 13, 2012 #3
    I did that part. I'm asking if I got from there to the Schwarz inequality, the "relevant equation", logically.
     
  5. Mar 14, 2012 #4
    Okay, sorry, I didn't see that line of text.

    So what you're doing (I think you left this step out) is saying that [itex](x_{1} y_{1} + x_{2} y_{2})^{2} + (x_{1} y_{2} - x_{2} y_{1})^{2} \geq 0[/itex] because the values are both positive and then you subtracted the one term. I'm still not sure how you went from the third equation in your post to the final equation. I think you can word that a little bit better? I get the sense you have to the right idea and understand why it's true, it just isn't obvious to me from reading it why you did it.
     
  6. Mar 14, 2012 #5
    Yeah, I started with

    [tex]0 \leq (x_{1} y_{1} + x_{2} y_{2})^{2} + (x_{1} y_{2} - x_{2} y_{1})^{2},[/tex]
    and since being squared mean both terms will always be positive, multiplying one of them by -1 makes that one always less than or equal to the other:

    [tex]-(x_{1} y_{2} - x_{2} y_{1})^{2} \leq (x_{1} y_{1} + x_{2} y_{2})^{2}.[/tex]
    Although looking at it now, I think I threw that in there just so I didn't disappear the LHS without comment. What I had in mind was this sorta thing:

    [tex]-(x_{1} y_{2} - x_{2} y_{1})^{2} \leq (x_{1} y_{1} + x_{2} y_{2})^{2} \leq (x_{1}^{2} + x_{2}^{2})(y_{1}^{2} + y_{2}^{2})[/tex]
    And then just cut out the [itex]-(x_{1} y_{2} - x_{2} y_{1})^{2}[/itex], because I'm not sure how else to get rid of it.
     
  7. Mar 14, 2012 #6
    Okay, I get what you are saying now. I think this way might be a little easier, you can be the judge.

    If [itex]a^{2} = b^{2} + c^{2}[/itex], then [itex]a^{2} \geq b^{2}[/itex]. I don't see any reason to "justify" getting rid of the [itex]c^{2}[/itex] term, it's just extra that you don't need.

    So, this is basically what you are doing except not trying to fiddle with the [itex]c^{2}[/itex] term. All you have to do is explain why an inequality appears when you get rid of the [itex]c^{2}[/itex] term.
     
  8. Mar 14, 2012 #7
    Okay, thanks.
     
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