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Homework Help: Schwarz Inequality Proof

  1. Jun 17, 2010 #1
    1. The problem statement, all variables and given/known data
    From Spivak's Calculus Chapter 1:

    "Suppose that [tex]y_1[/tex] and [tex]y_2[/tex] are not both [tex]0[/tex], and that there is no number λ such that [tex]x_1 =[/tex] λ[tex]y_1[/tex] and [tex]x_2 =[/tex] λ[tex]y_2[/tex]."

    Then [tex]0[/tex]<(λ[tex]y_1 - x_1)^2 + ([/tex]λ[tex]y_2 - x_2)^2[/tex].

    Using problem 18 (which involved proofs related to inequalities like [tex]x^2 + xy + y^2[/tex]), complete the proof of the Schwarz Inequality.

    2. Relevant equations

    None strike me.

    3. The attempt at a solution

    The thing that's really bothering me about this is that the problem I've given is just part a) of the problem. In part d) I am asked to "Deduce...that equality holds only when [tex]y_1 = y_2 = 0[/tex] or when there is a number λ [tex]\geq 0[/tex] such that [tex]x_1 =[/tex] λ[tex]y_1[/tex] and [tex]x_2 =[/tex] λ[tex]y_2[/tex]. Well, in a) he asked me to assume that both of those things were not true to start my proof. Doesn't this mean that, starting with those conditions, one cannot prove that equality is possible, and thus one can't prove the entirety of the Schwarz inequality (as in, the less than or equal to part)?
     
    Last edited: Jun 17, 2010
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  3. Jun 17, 2010 #2

    HallsofIvy

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    Yes assuming what is given, you could prove that [math]|<x, y>|< <x, x><y, y>[/math] while the "Cauchy-Schwarts" inequality only asserts "[itex]\le [/itex]".

    However, if you could prove "less than" you would have proved "less than or equal two". The latter is a subset of the former.
     
  4. Jun 17, 2010 #3
    I guess my issue is that when I think of proving that something is "less than or equal to" something else, I feel like I have to prove that it could be either less than or equal to that something else. It feels...sloppy not to. I guess that's my hang-up, though.
     
  5. Jun 18, 2010 #4

    HallsofIvy

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    It is certainly correct to say that "3 is less than or equal to 4". Of course, that is not a very "sharp" inequality!
     
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