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Proof of the Cauchy-Shwarz Inequality

  1. Jan 3, 2013 #1


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    1. Another approach to the proof of the Cauchy-Shwarz Inequality is suggested by figure 16 (sorry, I don't have the image), which shows that, in ℝ2 or ℝ3, llproj[itex]_u{}[/itex]vll ≤ llvll. Show that this inequality is equivalent to the Cauchy-Schwartz Inequality.

    2. Cauchy-Schawrtz Inequality: luvl ≤ llull llvll

    3. I substituted proj[itex]_u{}[/itex]v with (uv/uu)u but I can't think of what to do next.
    Last edited: Jan 3, 2013
  2. jcsd
  3. Jan 3, 2013 #2


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    Actually, I think I got it (please tell me if I am correct):

    ll(u•v/u•u)ull ≤ llvll
    lu•v/u•ul llull ≤ llvll
    lu•v/u•ul^2 llull^2 ≤ llvll^2
    lu•v/u•ul lu•v/u•ul (u•u) ≤ v•v
    (lu•vl*lu•vl)/(u•u) ≤ v•v
    lu•vl^2 ≤ (v•v)(u•u)
    lu•vl^2 ≤ llvll^2 llull^2
    lu•vl ≤ llvll llull
  4. Jan 3, 2013 #3


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    That's pretty hard to read. But it look ok to me. There's probably some extra steps in there you don't need. u.u=||u||^2, right?
  5. Jan 3, 2013 #4


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    Yeah, I put few extra steps that weren't necessary. I know its difficult to read; I am not really familiar with how to properly input some of the symbols.
  6. Jan 3, 2013 #5


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    It isn't that difficult

    ##\| (\frac{u\cdot v}{u\cdot u}) u \| \le \| v \|##
    Right click on it to see the code. Put ## at the beginning and end of the code.
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