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Math Help - 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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  2. #2
    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 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 v\ne0_v.

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

    \begin{aligned}<br />
   \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  \\ <br />
 & =\left\langle u,v \right\rangle -\frac{\left\langle u,v \right\rangle }{\left\| v \right\|^{2}}\cdot \left\langle v,v \right\rangle  \\ <br />
 & =0. <br />
\end{aligned}

    From here we have 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,

    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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