Cauchy's Inequality/Proof 1

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


 * $\ds \map f \lambda = \sum {\paren {r_i + \lambda s_i}^2}$

Now:
 * $\map f \lambda \ge 0$

because it is the sum of squares of real numbers.

Hence:

This is a quadratic equation in $\lambda$.

From Solution to Quadratic Equation:


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

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


 * $\ds D := 4 \paren {\sum {r_i s_i} }^2 - 4 \sum { {r_i}^2} \sum { {s_i}^2}$

$D$ is (strictly) positive.

Then $\map f \lambda = 0$ has two distinct real roots, $\lambda_1 < \lambda_2$, say.

From Sign of Quadratic Function Between Roots, it follows that $f$ is (strictly) negative somewhere between $\lambda_1$ and $\lambda_2$.

But we have:
 * $\forall \lambda \in \R: \map f \lambda \ge 0$

From this contradiction it follows that:


 * $D \le 0$

which is the same thing as saying:
 * $\ds \sum { {r_i}^2} \sum { {s_i}^2} \ge \paren {\sum {r_i s_i} }^2$