Cauchy's Inequality/Proof 2

Proof
From the Complex Number form of the Cauchy-Schwarz Inequality, we have:

As elements of $\R$ are also elements of $\C$, it follows that:
 * $\ds \sum \size {r_i}^2 \sum \size {s_i}^2 \ge \size {\sum r_i s_i}^2$

where all of $r_i, s_i \in \R$.

But from the definition of modulus, it follows that:
 * $\ds \forall r_i \in \R: \size {r_i}^2 = {r_i}^2$

Thus:
 * $\ds \sum {r_i}^2 \sum {s_i}^2 \ge \paren {\sum r_i s_i}^2$

where all of $r_i, s_i \in \R$.