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If $ \vec{x},\vec{y} $ are vectors in $ \R^n $ , then $ \|\vec{x}\cdot\vec{y}\|\le\|\vec{x}\|\|\vec{y}\| $

Proof

If either $ \vec{x} $ or $ \vec{y} $ are the zero vector, the statement holds trivially, so assume that both $ \vec{x},\vec{y} $ are non-zero. Let $ r $ be a scalar and $ \vec{z}=r\vec{x}+\vec{y} $ . Since, for any non-zero vector $ \vec{v} $ , $ \vec{v}\cdot\vec{v}=\|\vec{v}\|^2\ge0 $ (NOTE: merits own proof)

$ \begin{align} 0&\le(r\vec{x}+\vec{y})\cdot(r\vec{x}+\vec{y})\\ 0&\le r^2(\vec{x}\cdot\vec{x})+2r(\vec{x}\cdot\vec{y})+(\vec{y}\cdot\vec{y})\\ 0&\le ar^2+2br+c \end{align} $

where $ a=\vec{x}\cdot\vec{x},b=\vec{x}\cdot\vec{y},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

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

In particular, the term $ b^2-ac $ must either be negative, yielding two complex roots, or 0, yielding a single real root. Thus

$ \begin{align}&b^2-ac\le0\\&b^2\le ac\\&b\le\sqrt{ac}\end{align} $

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

$ \begin{align} \sqrt{(\vec{x}\cdot\vec{y})^2}&\le\sqrt{\vec{x}\cdot\vec{x}}\,\sqrt{\vec{y}\cdot\vec{y}}\\ \|\vec{x}\cdot\vec{y}\|&\le\|\vec{x}\|\|\vec{y}\| \end{align} $

which is equal to the original statement.

$ \blacksquare $

Notes

  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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