Intuition for the Cauchy-Schwarz inequality

In the Cauchy–Schwarz (CS) inequality $|u\cdot v|\le \|u\|\|v\|$, let's assume $v$ is a normalised vector, i.e., $\|v\|=1$. Then the CS inequality becomes $|u\cdot v|\le \|u\|$. Now, it's a trivial matter to show that these two forms of the CS inequality are in fact equivalent, in the sense that if $|u\cdot v|\le \|u\|$ for all normalised vectors $v$, then the usual CS inequality holds for all vectors. So, let us restate the CS inequality as stating that $|u\cdot v|\le \|u\|$ for all normalised vectors $v$. Now, the physical/geometric interpretation of $u\cdot v$ in this case is that it is the component of the vector $u$ in the direction $v$ (since $v$ is assumed normalised, that's all it is, a direction), while $\|u\|$ is the magnitude of $u$. So the CS inequality is merely stating the intuitively obvious fact that the component of a vector $u$ in a single direction is bounded by the magnitude of $u$.

Incidentally, this line of thought carries on to produce a very short and elegant proof of the full CS inequality. But, as you are not looking for a proof, I'll leave that out as an exercise.

By definition, the "dot" product of two vectors, say $\vec A$ and $\vec B$ is

$$\vec A\cdot \vec B=|\vec A||\vec B|\cos \theta$$

where $\theta$ is the angle between $\vec A$ and $\vec B$. That is to say, that the inner product is the projection of one vector onto the other. Visually, the projection is like a "shadow" that one vector casts along the direction of the other.

One can show that in Euclidean space, the angle $\theta$ between two vectors $v,w$ (in the sense of Euclidean geometry) satisfies

$$\cos(\theta)=\frac{v \cdot w}{\| v \| \| w \|}.$$

This is basically the law of cosines applied to an appropriate triangle. This equation only makes sense for every $v,w$ if the Cauchy-Schwarz inequality holds.