Cauchy's Inequality/Proof 1

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:

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:


 * $\displaystyle 4 \paren {\sum {r_i s_i} }^2 - 4 \sum {r_i^2} \sum {s_i^2}$

If this were (strictly) positive, then $\map f \lambda = 0$ would have two distinct real roots, $\lambda_1 < \lambda_2$, say.

If this were the case, then $f$ must be (strictly) negative somewhere.

To see this, note that $f$ must factor either as $\paren {\lambda - \lambda_1} \paren {\lambda - \lambda_2}$ or $-\paren {\lambda - \lambda_1} \paren {\lambda - \lambda_2}$.

When $\lambda^*$ is between $\lambda_1$ and $\lambda_2$, we have $\lambda^* - \lambda_1$ positive and $\lambda^* - \lambda_2$ negative, and so their product is negative.

When $\lambda^*$ is greater than both $\lambda_1$ and $\lambda_2$, both terms are positive and so their product is positive.

This means that the sign of $f$ must change at $\lambda_2$.

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

so the discriminant can not be positive.

Thus:
 * $\displaystyle 4 \paren {\sum {r_i s_i} }^2 - 4 \sum {r_i^2} \sum {s_i^2} \le 0$

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