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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.