Cauchy Schwarz Inequality

Theorem
In an inner product space, the absolute value of an inner product is less than or equal to the product of the lengths of its two vectors.

This is shown by:
 * $$|(u,v)| \le \|u\|*\|v\|$$

where $$u, v$$ are any two vectors in a Euclidean space.

Proof
Let $$t \in \R$$. Set $$w = u + t v$$. Then, $$(w,w) \ge 0$$ by Euclidean laws of inner product space.

$$(u+tv,u+tv)

= (u,u) + t(u,v) + t(v,u) + t^2(v,u)$$

$$=(u,u) + 2t(u,v) + t^2(v,v)$$

Let $$a=(v,v); b=(u,v); c=(u,v)$$

Therefore, $$at^2 + bt + c$$

We know that $$at^2 + bt + c \ge 0$$ and therefore can only have one distinct root.

The discriminant $$\Delta = 4b^2 - 4ac \le 0$$

Therefore, $$b^2 \le a*c$$

$$|(u,v)|^2 \le \|u\|^2*\|v\|^2$$

Taking the square root of both sides gives us:

$$|(u,v)| \le \|u\|*\|v\|$$