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Proof of the Cauchy-Schwarz inequality

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If \vec{X} and \vec{Y} are vectors in \Re^n, then \lVert \vec{X} \cdot \vec{Y} \rVert \leq \lVert \vec{X} \rVert \lVert \vec{Y} \rVert


ProofEdit

If either \vec{X} or \vec{Y} are the zero vector, the statement holds trivially, so assume that both \vec{X} and \vec{Y} are nonzero. Let r be a scalar and \vec{Z} be defined as r\vec{X} + \vec{Y}. Since, for any nonzero vector \vec{V}, \vec{V} \cdot \vec{V} = \lVert \vec{X} \rVert^2 \geq 0 (NOTE: merits own proof)


\displaystyle 0 \leq (r\vec{X} + \vec{Y}) \cdot (r\vec{X} + \vec{Y})


\displaystyle {} \leq r^2(\vec{X} \cdot \vec{X}) + 2r(\vec{X} \cdot \vec{Y}) + (\vec{Y} \cdot \vec{Y})

\displaystyle {} \leq ar^2 + 2br + c


where a = \vec{X} \cdot \vec{X}, b = \vec{X} \cdot \vec{Y} and c = \vec{Y} \cdot \vec{Y}. It can be seen clearly that p(r) = ar^2 + 2br + c is a quadratic polynomial that is non-negative for any r. Consequently, the polynomial has two complex roots, or has a single distinct real root.[1]


Remember that the roots of p(r) are given by the quadratic formula


\displaystyle \frac{-2b \pm \sqrt{4b^2 - 4ac}}{2a}


In particular, the term 4b^2 - 4ac must either be negative, yielding two complex roots, or zero, yielding a single real root. Thus


4b^2 - 4ac \leq 0


4b^2 \leq 4ac

b^2 \leq ac

b \leq \sqrt{ac}


Substituting the values of a, b and c into the last of these inequalities, it can be seen that


\sqrt{(\vec{X} \cdot \vec{Y})^2} \leq \sqrt{\vec{X} \cdot \vec{X}}\ \sqrt{\vec{Y} \cdot \vec{Y}}


Failed to parse (unknown function\lvert): \lvert \vec{X} \cdot \vec{Y} \rvert \leq \lVert \vec{X} \rVert \lVert \vec{Y} \rVert



which is equal to the original statement.


QED


NotesEdit

  1. Intuitively, the graph of p(r) is either 'floating above' the horizontal axis, if it has two complex roots, or tangent if it has one real root. Since p(r) is non-negative for every r, it can't have two real roots because the graph of the function would have to 'pass under' the horizontal axis.

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