First note that if \(\vec{v}=\vec{0}\) both sides of \(\eqref{cauchy}\) equal zero and so the inequality holds in this case. Therefore, it will be assumed in what follows that \(\vec{v}\neq \vec{0}\). listening on any, link-type LINUX_SLL (Linux cooked v1), capture size 262144 bytes
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The dot product satisfies the inequality \[\left\vert \vec{u}\bullet \vec{v}\right\vert \leq \| \vec{u}\| \| \vec{v}\| \label{cauchy}\] Furthermore equality is obtained if and only if one of \(\vec{u}\) or \(\vec{v}\) is a scalar multiple of the other. Proof Now this means the graph of \(y=f\left( t\right)\) is a parabola which opens up and either its vertex touches the \(t\) axis or else the entire graph is above the \(t\) axis. In the first case, there exists some \(t\) where \(f\left( t\right) =0\) and this requires \(\vec{u}+t\vec{v}=\vec{0}\) so one vector is a multiple of the other. Then clearly equality holds in \(\eqref{cauchy}\). In the case where \(\vec{v}\) is not a multiple of \(\vec{u}\), it follows \(f\left( t\right) >0\) for all \(t\) which says \(f\left( t\right)\) has no real zeros and so from the quadratic formula, \[\left( 2\left( \vec{u}\bullet \vec{v}\right) \right)