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Cauchy-Schwartz inequality

  1. Nov 4, 2015 #1
    The Cauchy-Schwartz inequality [itex](\sum_{i=1}^n x_i^2)(\sum_{i=1}^n y_i^2) - (\sum_{i=1}^n x_iy_i)^2 \geq 0 [/itex] holds with equality (or is as "small" as possible) if there exists an [itex]a \gt 0 [/itex] such that [itex]x_i=ay_i[/itex] for all [itex]i=1,...,n [/itex].

    But when is the inequality as "large" as possible? That is, can we say anything under what conditions [itex](\sum_{i=1}^n x_i^2)(\sum_{i=1}^n y_i^2) - (\sum_{i=1}^n x_iy_i)^2[/itex] is as large as possible?
     
  2. jcsd
  3. Nov 4, 2015 #2

    BvU

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    That's easy ! If all xy are negative. Can you interpret that ?
    Oops. Major glitch trying to reply too hasty.
    First of all welcome to PF :smile: ! And I'll be back soon to post something more useful -- if I don't get corrected sooner by someone more awake ...
     
  4. Nov 4, 2015 #3

    BvU

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    Take 2: That's easy ! If all xiyi are zero. Can you interpret that ?
     
  5. Nov 4, 2015 #4
    Sorry, I forgot to say that each [itex]x_i \gt 0[/itex] and [itex]y_i \gt 0[/itex]
     
  6. Nov 4, 2015 #5

    BvU

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    I remember Cauchy-Schwartz had something to do with vector products and you look at ##( \vec x + \vec y ) ^2 = \vec x^2 + \vec y ^2 + 2 \vec x \cdot \vec y = \vec x^2 + \vec y ^2 + 2 \,| \vec x | \, |\vec y| \cos \alpha##.

    The equals sign in CS is if ##|\cos\alpha| = 1 ## so I have a feeling that what you asked for might be found at ##\cos\alpha= 0 ##.

    In post #4 you exclude =0 but then your extreme can be approached as close as desired by letting ##\cos\alpha \rightarrow 0 ##.

    --
     
  7. Nov 4, 2015 #6
    If you write the summations as dot products, you can readily see the answer. Let [itex]\vec{x} = \langle x_1, x_2, \dotsm, x_n \rangle[/itex] and [itex]\vec{y} = \langle y_1, y_2, \dotsm, y_n \rangle[/itex]. Then the Cauchy-Schwartz inequality can be restated as:

    [itex]\underbrace{(\vec{x} \cdot \vec{x})}_{\text{dot product}} \cdot
    \underbrace{(\vec{y}\cdot \vec{y} )}_{\text{dot product}} \ge \underbrace{\vec{x} \cdot \vec{y}}_{\text{dot product}}[/itex] or equivalently [itex]\| \vec{x} \|^2 \cdot \| \vec{y} \|^2 \ge \| \vec{x} \| \cdot \| \vec{y} \| \cdot \cos(\theta)[/itex]

    where [itex]\theta[/itex] is the angle between [itex] \vec{x} [/itex] and [itex] \vec{y} [/itex]. Since [itex]\vec{x}\cdot \vec{y} = 0[/itex] if the two vectors are orthogonal, that's where you will get the largest value.
     
  8. Nov 5, 2015 #7

    Erland

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    It can be arbitrarily large. Just take any xi:s and yi:s for which the difference is > 0 and multiply all the xi:s with an arbitrary lange t > 0.
     
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