Cauchy's Inequality

Theorem
$$\sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$$ where all of $$r_i, s_i \in \mathbb{R}$$.

Proof
For any $$\lambda \in \mathbb{R}$$, we define $$f: \mathbb{R} \to \mathbb{R}$$ as the function:

$$f \left({\lambda}\right) = \sum {\left({r_i + \lambda s_i}\right)^2}$$.

Now $$f \left({\lambda}\right) \ge 0$$ because it is the sum of squares of real numbers.

Hence $$\forall \lambda \in \mathbb{R}: f \left(\lambda\right) \equiv \sum {r_i^2} + 2 \lambda \sum {r_i s_i} + \lambda^2 \sum {s_i^2} \ge 0$$.

This is a simple quadratic in $$\lambda$$, and we can solve it using Quadratic Equation, where:

$$a \lambda^2 + b \lambda + c = 0: a = \sum {s_i^2}, b = 2 \sum {r_i s_i}, c = \sum {r_i^2}$$

The discriminant of this equation (i.e. $$b^2 - 4 a c$$) is:

$$4 \left({\sum {r_i s_i}}\right)^2 - 4 \sum {r_i^2} \sum {s_i^2}$$

If this were positive, then $$f \left({\lambda}\right) = 0$$ would have two distinct real roots, $$\lambda_1$$ and $$\lambda_2$$, say.

If this were the case, then $$\exists \lambda_n: \lambda_1 < \lambda_n < \lambda_2: f \left({\lambda_n}\right) < 0$$.

But we have $$\forall \lambda \in \mathbb{R}: f \left({\lambda}\right) \ge 0$$.

Thus $$4 \left({\sum {r_i s_i}}\right)^2 - 4 \sum {r_i^2} \sum {s_i^2} < 0$$,

which is the same thing as saying $$\sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$$.