What is the intuition behind the Cauchy-Schwarz inequality in the real numbers?

Here is the geometric intuition behind this for $n = 3$. You need not know much linear algebra, but you do need to know about vectors.

Given two vectors $x = (x_1,x_2,x_3)$ and $y = (y_1,y_2,y_3)$, we define their dot product $x \cdot y$ to be $|x||y|\cos \theta$, where $|x|$ and $|y|$ are the lengths of the vectors $x$ and $y$, and $\theta$ is the angle between them. (In theory, this definition may be slightly circular, since at an advanced level dot products are usually used to define angles. But if you accept the idea of an angle as intuitively meaningful, we needn't worry about this technicality.)

The dot product $x \cdot y$ is also given by the formula $x_1y_1 + x_2 y_2 + x_3 y_3$.

Then the Cauchy-Schwarz inequality is exactly equivalent to the statement that $|\cos \theta| \leq 1$.

An alternative interpretation without angles in general, but using perpendicularity, is the one given in A.S.'s comment.

If you'd like to see the details of this, have a look either at Chapter 12 of Apostol's Calculus or at Chapter 1 of Lang's Introduction to Linear Algebra.

Edit I can try to give a very imperfect algebraic "interpretation" of the inequality. I'm not convinced this is the best one, so I'll keep thinking about it.

If you look at the inequality $$\left(\sum_{i=1}^n x_i y_i\right)\left(\sum_{i=1}^n x_i y_i\right)\leq \left(\sum_{i=1}^n x_i^2\right) \left(\sum_{i=1}^n y_i^2\right),$$ note first that the general inequality follows from the special case where all the $x_i$'s and $y_i$'s are nonnegative, since $|\sum x_i y_i| \leq \sum |x_i||y_i|$. Next think about how the $x_i$'s and $y_i$'s match up. The inequality says that to make a product like the LHS and RHS as large as possible, it's better to match up the numbers $x_i$ and $y_i$ with themselves than with each other. This sort of makes sense, because if you have $x_1 < y_1$, and you go from $x_1 y_1$ (twice) to $x_1^2$ and $y_1^2$, you're better off with the latter. This is because $y_1^2$ is relatively large, and this is usually more than enough to compensate for the smaller $x_1^2$. Obviously, this is not a proof in any way. But it does make the inequality plausible.