Prove the Cauchy Schwarz equality using vector dot product.

Question

Prove the Cauchy Schwarz inequality using vector dot product. I can't seem to make any progress on it. To prove this inequality; $(a_1b_1+....a_nb_n)^2<(a_1^2+....a_n^2)(b_1^2+....b_n^2)$

Answer 1

$$|v\cdot w|=|\|v\|\|w\|\cos(\theta)|\leq \|v\|\|w\|$$

Answer 2

HINT: observe that

$$0\le (x- \alpha y|x-\alpha y)=(x|x)+(\alpha y|\alpha y)- 2\Re (x|\alpha y)=(x|x)+|\alpha|^2(y|y)-2\Re(\bar\alpha(x|y))$$

by the properties of any dot product, for vectors $x,y$ and scalar $\alpha$. Now set $\alpha:=\frac{(x|y)}{(y|y)}$ and see what happen.

Answer 3

Consider $\alpha\in\mathbb{R}.$ Then we have

$$0\le (\alpha u+v)\cdot(\alpha u+v)=\alpha^2 |u|^2+2\alpha u\cdot v+|v|^2.$$ This is a polynomial of degree $2$ in $\alpha.$ It can't have two different roots (why?). So its discriminant must be nonpositive. That is $$4(u\cdot v)^2-4|u|^2|v|^2\le 0.$$ Take square roots and you are done.