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Thread: Cauchy-Schwartz Inequality

  1. #1
    Fel
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    Cauchy-Schwartz Inequality

    Prove the Cauchy-Schwartz inequality |<u, v>| <= |u| |v|. (<= means less than or equal to)

    I attempted it a couple times, but didn't get anywhere. Help would be great, thanks.
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    MHF Contributor arbolis's Avatar
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  3. #3
    Fel
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    I've seen that already, but I couldn't understand it all that well.
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    Please post an attempt at a solution or clarify what you don't understand in the proof provided (preferrably the former).
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  5. #5
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    If $\displaystyle v=0_v,$ then we have the equality, so there's nothing to prove there. Thus let's take care about the interesting case when it's $\displaystyle v\ne0_v.$

    Under this assumption, put $\displaystyle t=u-\frac{\left\langle u,v \right\rangle }{\left\| v \right\|^{2}}\cdot v$ and then

    $\displaystyle \begin{aligned}
    \left\langle t,v \right\rangle &=\left\langle u-\frac{\left\langle u,v \right\rangle }{\left\| v \right\|^{2}}\cdot v,v \right\rangle \\
    & =\left\langle u,v \right\rangle -\frac{\left\langle u,v \right\rangle }{\left\| v \right\|^{2}}\cdot \left\langle v,v \right\rangle \\
    & =0.
    \end{aligned}$

    From here we have $\displaystyle 0\le\|t\|^2=\left\langle t,t \right\rangle =\left\| u \right\|^{2}-\frac{\left\langle u,v \right\rangle ^{2}}{\left\| v \right\|^{2}}.$ Hence,

    $\displaystyle 0\le \left\| t \right\|^{2}\cdot \left\| v \right\|^{2}=\left\| u \right\|^{2}\cdot \left\| v \right\|^{2}-\left\langle u,v \right\rangle ^{2}\implies \left| \left\langle u,v \right\rangle \right|\le \left\| u \right\|\cdot \left\| v \right\|.\quad\blacksquare$
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