Answer
$ |a \cdot b|\leq |a| |b|$
Work Step by Step
Theorem $3$ states that: $a \cdot b=|a||b| cos \theta$ .... (1)
As $|a| \gt 0$ and $ |b| \gt 0$ , then $|a \cdot b|=||a||b| cos \theta|=|a||b| |cos \theta|$ .... (2)
(i) If $ 0 \leq \theta \leq \pi/2$, then If $ 0 \leq |a||b| |cos \theta| \leq |a||b| |$
(ii) If $ \pi/2 \leq \theta \leq \pi$, then If $ |a||b| \geq |a \cdot b| | \gt 0$
From the above two conditions, we have found that
$ |a \cdot b|\leq |a| |b|$