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Using the Cauchy-Schwarz inequality to prove all real values for a, b, and theta

  1. Dec 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Use the Cauchy-Schwarz inequality to prove that for all real values of a, b, and theta (which ill denote as θ),
    (a cosθ + b sinθ)2 ≤ a2 + b2



    2. Relevant equations
    so the Cauchy-Schwarz inequality is | < u,v>| ≤ ||u|| ||v||



    3. The attempt at a solution

    I'm having a difficult time figuring out what is my u and my v. Sorry for the stupid question.

    Thank you for any guidance.
     
    Last edited: Dec 15, 2011
  2. jcsd
  3. Dec 15, 2011 #2

    micromass

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    What is <u,v> and ||u|| in [itex]\mathbb{R}^2[/itex]??

    Can you rewrite the Cauchy-Schwarz inequality in [itex]\mathbb{R}^2[/itex]??
     
  4. Dec 15, 2011 #3
    <u,v> in R2 would be u1v1 + u2v2 I believe..

    and ||u|| in R2 is √u2

    I'm terribly sorry but I'm not sure what you're trying to get at. I know and I also don't want to be spoon fed the answer.
     
  5. Dec 15, 2011 #4

    micromass

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    Good

    No, that would be [itex]\sqrt{u_1^2+u_2^2}[/itex].

    Anybody, do you recognize something like [itex]u_1v_1+u_2v_2[/itex] in your OP??
     
  6. Dec 15, 2011 #5
    I see that a2 + b2 would be
    a12a12+b12b12
     
  7. Dec 15, 2011 #6

    micromass

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    Can you write out

    [tex]<u,v>^2\leq \|u\|^2\|v\|^2[/tex]

    ??

    Just write out what <u,v> and ||u|| mean...
     
  8. Dec 15, 2011 #7
    Sorry, I think you're losing patience with me but I did what you asked and here it is.
    Please correct me if I'm incorrect.

    <u,v>2≤∥u∥2∥v∥2
    = (u1v1+u2
    v2)(u1v1+u2
    v2) ≤ √(u12+ u22) √(v12 + v22)
     
  9. Dec 15, 2011 #8
    Well, thanks anyway. I got help from the wonderful people at yahoo answers. Thank you for trying to work with me.
     
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