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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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