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.
Best Answer
I thought I never did it directly, but now that I have found the solution below (quite quickly), I begin to suspect that I must have done something similar years ago.
Anyway, as a first step, let's square everything. Then we need to prove this: $$ \left(\sum_i x_i y_i\right)^2 \leq \left(\sum_i x_i^2\right)\left(\sum_j y_j^2\right). $$ Let's subtract the left from the right and open all the parentheses: $$ \begin{align*} \left(\sum_i x_i^2\right)\left(\sum_j y_j^2\right) - \left(\sum_i x_i y_i\right)^2 & = \sum_{i,j}x_i^2 y_j^2 - \sum_{i, j}x_i y_i x_j y_j \\ & = \sum_{i \neq j} x_i^2 y_j^2 - \sum_{i \neq j} x_i y_i x_j y_j \\ & = \sum_{i < j} (x_i^2 y_j^2 + x_j^2y_i^2 - 2 x_i y_i x_j y_j) \\ & = \sum_{i < j} (x_i y_j - x_j y_i)^2 \end{align*} $$ Here indices $i$ and $j$ always iterate from $1$ to $n$. We see that this is a sum of several squares, so it is nonnegative, proving the original inequality.
UPDATE: the very same thing can be done for complex numbers. We want to prove this: $$ \left|\sum_i x_i \overline{y_i}\right| \leq \sqrt{\sum_i x_i \overline{x_i}} \cdot \sqrt{\sum_j y_j \overline{y_j}}. $$ Let us square everything, keeping in mind that $|z|^2 = z\overline{z}$: $$ \left( \sum_i x_i \overline{y_i} \right) \left(\sum_j \overline{x_j} y_j\right) \leq \left( \sum_i x_i \overline{x_i} \right) \left(\sum_j y_j \overline{y_j}\right) $$ As before, we subtract the left from the right: $$ \begin{align*} & \left( \sum_i x_i \overline{x_i} \right) \left(\sum_j y_j \overline{y_j}\right) - \left( \sum_i x_i \overline{y_i} \right) \left(\sum_j \overline{x_j} y_j\right) \\ & = \sum_{i,j}x_i\overline{x_i}y_j\overline{y_j} - \sum_{i, j}x_i \overline{y_i} \overline{x_j} y_j \\ & = \sum_{i \neq j} x_i\overline{x_i}y_j\overline{y_j} - \sum_{i \neq j} x_i \overline{y_i} \overline{x_j} y_j \\ & = \sum_{i < j} (x_i \overline{x_i} y_j \overline{y_j} + x_j \overline{x_j} y_i \overline{y_i} - x_i \overline{y_i} \overline{x_j} y_j - x_j \overline{y_j} \overline{x_i} y_i) \\ & = \sum_{i < j} |x_i y_j - x_j y_i|^2 \end{align*} $$ As before, we have a sum of squares of real numbers, which is real and nonnegative. Done.