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?

Solutions Collecting From Web of "Showing that the product of vector magnitudes is larger than their dot product"

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.

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.

I think the standard method is by taking $\displaystyle a=u, b={u\cdot v\over |v|^2}v$ and then noting $$|a-b|^2=|a|^2-2a\cdot b+|b|^2=|u|^2-2{(u\cdot v)^2\over |v|^2}+{(u\cdot v)^2\over |v|^2}\ge 0$$