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Help with Cauchy-Schwartz Inequality proof.

  1. Jan 26, 2004 #1
    Help with Cauchy-Schwarz Inequality proof.

    Argh!!!

    I've been playing around with this and I can't get it...

    Here's what I have thus far:

    Given [tex]\mid u \cdot v \mid \leq \parallel u \parallel \parallel v \parallel [/tex]

    [tex] (u \cdot v)^2 \leq (u_1^2 + u_2^2)(v_1^2 + v_2^2)[/tex]

    [tex] (u_1v_1+u_2v_2)^2 \leq (u_1^2 + u_2^2)(v_1^2 + v_2^2)[/tex]

    [tex] (u_1v_1+u_2v_2)^2 \leq (u_1^2 + u_2^2)(v_1^2 + v_2^2)[/tex]

    [tex]u_1^2v_1^2+u_2^2v_2^2+2u_1v_1u_2v_2 \leq u_1^2v_1^2 + u_2^2v_1^2+u_1^2v_2^2 + u_2^2v_2^2[/tex]

    [tex]2u_1v_1u_2v_2 \leq u_2^2v_1^2+u_1^2v_2^2[/tex]

    This is where I get stumped which means I messed up somewhere earlier in my proof. Any help here would be greatly appreciated.

    Thanks a lot.
     
    Last edited: Jan 26, 2004
  2. jcsd
  3. Jan 26, 2004 #2
    Ok, Try number 2. Does this look right?

    given: [tex]\mid u \cdot v \mid \leq \parallel u \parallel \parallel v \parallel [/tex]

    [tex]\frac{\mid u \cdot v \mid}{\parallel u \parallel \parallel v \parallel} \leq 1[/tex]

    since [tex] \cos \theta = \frac{\mid u \cdot v \mid}{\parallel u \parallel \parallel v \parallel} \leq 1[/tex]

    [tex] \Rightarrow \cos \theta \leq 1 [/tex]

    thus the proof is true because by definition, [tex]\cos \theta \leq 1[/tex] for all values [tex] 0 \leq \theta \leq 2\pi[/tex]
     
  4. Jan 26, 2004 #3
    This is a perfect square and would be >=0

    [tex]-2u_1v_1u_2v_2 + u_2^2v_1^2+u_1^2v_2^2 \geq 0[/tex]

    Now u can proceed
     
    Last edited: Jan 26, 2004
  5. Jan 26, 2004 #4
    Yes both the prove are right
     
  6. Jan 26, 2004 #5
    I didn't notice the perfect square... Thanks.
     
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