# 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 $$|a-b|^2=|a|^2-2a\cdot b+|b|^2=|u|^2-2{(u\cdot v)^2\over |v|^2}+{(u\cdot v)^2\over |v|^2}\ge 0$$