Cauchy's Inequality

Theorem

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

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

Proof 1
For any $$\lambda \in \R$$, we define $$f: \R \to \R$$ as the function:


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

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

because it is the sum of squares of real numbers.

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

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


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

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


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

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

If this were the case, then $$\exists \lambda_n: \lambda_1 < \lambda_n < \lambda_2: f \left({\lambda_n}\right) < 0$$.

But we have $$\forall \lambda \in \R: f \left({\lambda}\right) \ge 0$$.

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

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

Proof 2
Let $$w_1, w_2, \ldots, w_n$$ and $$z_1, z_2, \ldots, z_n$$ be arbitrary complex numbers.

Take the Binet-Cauchy Identity:
 * $$\left({\sum_{i=1}^n a_i c_i}\right) \left({\sum_{j=1}^n b_j d_j}\right) = \left({\sum_{i=1}^n a_i d_i}\right) \left({\sum_{j=1}^n b_j c_j}\right) + \sum_{1 \le i < j \le n} \left({a_i b_j - a_j b_i}\right) \left({c_i d_j - c_j d_i}\right)$$

and set $$a_i = w_i, b_j = \overline {z_j}, c_i = \overline {w_i}, d_j = z_j $$.

This gives us:

$$ $$ $$ $$

Hence the result.

He first published this result in 1821.

It is a special case of the Cauchy–Schwarz Inequality.