Showing that the product of vector magnitudes is larger than their dot product QUESTION
Show that $$|\mathtt{u} \cdot \mathtt v| \le |\mathtt u||\mathtt v|$$
ATTEMPT
Let $ \mathtt {u,v} \in \mathbb R^n$ such that $ \mathtt u = u_1 x_1 + u_2 x_2 + ... + u_n x_n, \mathtt v = v_1 x_1 + v_2 x_2 + ... + v_n x_n$ 
Then  $$|\mathtt{u} \cdot \mathtt v| = | u_1 v_1 + u_2 v_2 + ... + u_n v_n | $$
$$ |\mathtt u||\mathtt v| = \sqrt{ {u_1}^2 + {u_2}^2 + ... + {u_n}^2}\sqrt{ {v_1}^2 + {v_2}^2 + ... + {v_n}^2}  $$
This is where I get a bit stuck. Instinctively I do know that (2) is in fact larger or equal to (1). I'm just not sure which lemma/theorem/rule to use to show this. 
Or am I going about this just the wrong way?
 A: This is fairly simple if you are allowed to assume that $a\cdot b = |a||b|\cos\theta$, because $|\cos \theta|$ is bounded between 0 and 1.
The proof of the $\cos \theta$ rule can be found in this answer.
A: The simplest approach is probably to consider the following polynomial in $\lambda$:
$$
|u+\lambda v|^2=|u|^2+2\lambda u\cdot v +\lambda^2|v|^2.  \quad \mbox{(By linearity of scalar product)}
$$
This polynomial in $\lambda$ cannot become strictly negative, because it is equal to a squared norm (left side).
As a consequence, its discriminant is negative or zero, i.e.:
$$
\Delta=(2 u\cdot v)^2 - 4 |u|^2 |v|^2 \leq 0.
$$
This inequality is equivalent to the desired result.
A: I think the standard method is by taking $\displaystyle a=u, b={u\cdot v\over |v|^2}v$ and then noting $$|a-b|^2=|a|^2-2a\cdot b+|b|^2=|u|^2-2{(u\cdot v)^2\over |v|^2}+{(u\cdot v)^2\over |v|^2}\ge 0$$
A: From a purely axiomatic/abstract approach, the question has already been answered very efficiently, for example in walcher's answer.
I want to provide here an answer that, although harder to verify, is "discoverable", meaning that every step in the derivation "feels" justified. Basically, how one could prove the Cauchy–Schwarz inequality, inspired by the geometric interpretation of the dot product, without knowing a proof already.
The Cauchy–Schwarz inequality:
$\def\u{{\bf u}}$
$\def\v{{\bf v}}$
$\def\up{{\bf u_{\parallel}}}$
$\def\ux{{\bf u_{\perp}}}$
$\def\vu{{\bf \hat{v}}}$
$$ |\u\cdot \v| \le |\u||\v|.$$
is one that you probably have a strong geometric intuition for why it is true, in the context of simple vectors in 2D for example.
The intuition that will be used here is that $\u\cdot \v$ is the product of the magnitude of the projection of $\u$ on the line of $\v$ - called $\up$ - and the magnitude of $\v$, with the detail that the result is negative if $\up$ points on the opposite direction to $\v$, and so
$$|\u \cdot \v| = |\up||\v|.\tag{1}\label{1}$$
The sign of $\u\cdot\v$ depends on if $\up$ and $\v$ point on the same direction or not. It is clear that multiplying $\v$ by $\u\cdot\v$ and dividing by $|\v|^2$ should give $\up$, as the magnitudes of $\v$ cancel out, remaining only the magnitude of $\up$ and the minus sign in case they point in opposite directions:
$$((\u\cdot\v)/|\v|^2)\v = (\u\cdot \vu)\vu = (\pm|\up|)\vu = \up, \tag{2}\label{2}$$
where $\vu = \v/|\v|$ is simply the unit vector - $|\vu| = 1$ - with the same orientation as $\v$.
You also probably can tell intuitively that the projection of $\u$ on any line cannot be bigger than
$\u$ itself, so $$|\up| \le |\u|. \tag{3}\label{3}$$
Together, \eqref{1} and \eqref{3} mean the inequality is intuitively true. If you already know they are true, then the proof is given. They will be proved next.
The goal is then, to prove these intuitive results with the properties of an arbitrary inner product (although I'll keep using the dot product in the notation), which are linearity on the first argument, conjugate symmetry and positive definiteness, as defined on wikipedia for example.
First we must prove that \eqref{2} actually gives the parallel part of $\u$ with regard to the line of $\v$. It is obvious that $\up =(\u\cdot \vu)\vu$ is parallel to $\v$ as the rest is a scalar and $\vu$ is parallel to $\v$. All is left to check is that $\ux = \u - \up$ actually is perpendicular to $\v$, which it is
$$\begin{equation}\begin{aligned}
\ux \cdot \v &= (\u-\up)\cdot\v = (\u\cdot\v) - (\up\cdot\v) \\
&= (\u\cdot\v) - ((\u\cdot \vu)\vu\cdot\v) = (\u\cdot\v) - (\u\cdot \vu)(\vu\cdot\v)  \\ 
&= (\u\cdot\v) - (\u\cdot \vu)|\v| = (\u\cdot\v) - (\u\cdot \v) = 0\end{aligned}\end{equation}$$
We have thusly proved that $\u = \up + \ux$ where $\up$ is parallel to $\v$ and $\ux$ is perpendicular to $\v$ (and $\up$) and $\up = (\u\cdot \vu)\vu$. We have left to prove \eqref{1} and \eqref{3}.
The intuitive result \eqref{1} can now be proved just by using the expression for $\up = (\u\cdot \vu)\vu$:
$$|\up||\v| = |\u\cdot \vu||\v| = |\u\cdot\v|.$$
To prove the intuitive result \eqref{3}, that a projection of a vector on a line cannot be bigger than the original vector, one can start with the expression for $|\u|$ and try to make $|\up|$ appear (you can also start with $|\up|$ or $|\ux|$ to get the same result. I chose this as it feels more obvious/natural). Since $|\u|= \sqrt{(\u\cdot\u)}$, you quickly discover it's easier to work with the squared magnitudes to avoid the square roots. Remembering Pythagoras' theorem might also be a good motivation to go down this path.
$$\begin{equation}\begin{aligned}
|\u|^2 &= \u\cdot\u  = (\up +\ux)\cdot(\up+\ux) \\
&= (\up\cdot\up) + (\up\cdot\ux)+(\ux\cdot\up) + (\ux\cdot\ux) \\
&= |\up|^2 + 0+0 + |\ux|^2 \\
\end{aligned}\end{equation}$$
As $|\ux|\ge0$, it is proved that $|\up|^2 \le |\u|^2$.
Note in the above derivation the similarity with walcher's proof. Their $a$, $b$ and $a-b$ are our $\u$, $\up$ and $\ux$. They start with $|\ux|^2$ instead to find it is equal to $|\u|^2 - |\up|^2$.
This method also makes it clear that $|\u\cdot\v| = |\up||\v| = |\u||\v|$ if and only if $|\up| = |\u|$, meaning that $\u$ and $\v$ are parallel.
