# Cauchy's Inequality/Proof 1

## Theorem

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

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

## 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:

 $\ds \forall \lambda \in \R: \,$ $\ds \map f \lambda$ $\equiv$ $\, \ds \sum {\paren { {r_i}^2 + 2 \lambda r_i s_i + \lambda^2 {s_i}^2} } \,$ $\, \ds \ge \,$ $\ds 0$ $\ds$ $\equiv$ $\, \ds \sum { {r_i}^2} + 2 \lambda \sum {r_i s_i} + \lambda^2 \sum { {s_i}^2} \,$ $\, \ds \ge \,$ $\ds 0$

This is a quadratic equation in $\lambda$.

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

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

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

Aiming for a contradiction, suppose $D$ 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:

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

$\blacksquare$

## Source of Name

This entry was named for Augustin Louis Cauchy.