Why is one proof for Cauchy-Schwarz inequality easy, but directly it is hard? Let's say you are in $\mathbb{R}^n$ and you define the norm as $||x||=\sqrt{x_1^2+x_2^2...+x_n^2}$. This we recognize as the usual norm from the inner product: $||x|| = \sqrt{\langle x, x \rangle}$, where $\langle x, y \rangle = x_1 y_1 + x_2 y_2+ \cdots + x_n y_n$. It is easy to check that this satisfies all the axioms for an inner product. Then we may define orthogonality as a zero inner product, and we get the Pythagorean theorem, we define projection, and then the proof for Cauchy–Schwarz is pretty straight forward. 
But now comes my problem. Lets say you do not want to go through the inner product, but you still want to prove Cauchy–Schwarz. When you do not have an inner product, Cauchy–Schwarz do not make much sense, but I want the part where we have replaced the inner-product part.
I mean, Cauchy–Schwarz says:
$|\langle x, y \rangle| \le ||x|| \cdot ||y||$. This equation makes sense even without inner-products for our case:
$\left| x_1 y_1 + x_2 y_2 + \cdots + x_n y_n \right| \le \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} \cdot \sqrt{y_1^2 + y_2^2 + \cdots + y_n^2}$
However I am not able to prove this inequality. For me, it is easier going through the inner product for proving this, but I want to be able to prove this inequality directly, how am I supposed to do that?
That is, my problem is proving that
$\left| x_1 y_1 + x_2 y_2 + \cdots + x_n y_n \right| \le \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} \cdot \sqrt{y_1^2 + y_2^2 + \cdots + y_n^2}$
without going through the inner product. Is this hard? Would you say it is easier defining the inner-product and proving it in that way? It seems weird that it should be easier to define a lot of new terms just to prove an inequality.

Essentially Dan's prove doesn't avoid inner products, since he proved that $\|x\|^2\|y\|^2-\langle x,y\rangle^2=$ Gram-determinant of $(x,y)$.  So what qualifies my statement as an answer?  Even if you don't use the notation of an inner product, it's inherently present at all.
