# Cauchy's Inequality/Proof 1

Jump to navigation
Jump to search

## Contents

## Theorem

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

## Proof

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

- $\displaystyle \map f \lambda = \sum {\paren {r_i + \lambda s_i}^2}$

Now:

- $\map f \lambda \ge 0$

because it is the sum of squares of real numbers.

Hence:

\(\displaystyle \forall \lambda \in \R: \map f \lambda\) | \(\equiv\) | \(\displaystyle \sum {\paren {r_i^2 + 2 \lambda r_i s_i + \lambda^2 s_i^2} } \ge 0\) | |||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle \sum {r_i^2} + 2 \lambda \sum {r_i s_i} + \lambda^2 \sum {s_i^2} \ge 0\) |

This is a quadratic equation in $\lambda$.

From Solution to Quadratic Equation:

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

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

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

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:

- $\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \paren {\sum {r_i s_i} }^2$

$\blacksquare$

## Source of Name

This entry was named for Augustin Louis Cauchy.

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $2.2$: Examples: Example $2.2.1$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 1$: Real Numbers: $\S 1.11$:*Example*