Cauchy's Inequality/Proof 1

Theorem

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

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

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


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

Now:
 * $f \left({\lambda}\right) \ge 0$

because it is the sum of squares of real numbers.

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

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


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

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


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

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

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

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

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:
 * $\forall \lambda \in \R: f \left({\lambda}\right) \ge 0$

so the discriminant can not be positive.

Thus:
 * $\displaystyle 4 \left({\sum {r_i s_i}}\right)^2 - 4 \sum {r_i^2} \sum {s_i^2} \le 0$

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