In one of the proofs of Cauchys Inequality that reduces vector inner product to a quadratic form:
$$0 \leq \langle u + \alpha, u + \alpha \rangle = \langle u , u \rangle + 2 \alpha\langle u ,v \rangle + \alpha^2\langle v,v \rangle$$
Since any quadratic $$ a \alpha^2+2b\alpha + c $$ takes its minimal value @ $ c - \frac{b^2}{a} $ when $ \alpha = -\frac{b}{2a}$ ---- (1)
and the inequality should hold for even this minimum value of the polynomial
$$ 0 \leq \langle u,u \rangle - \frac {\langle u,v\rangle^2}{\langle v,v \rangle} \iff \frac {|\langle u,v \rangle|}{||u|| ||v||} \leq 1 $$ --- (2)
Questions:
Main Q) Am not confident how the equivalence is same. Please confirm the following the is correct thinking:
$$ 0 \leq \langle u,u \rangle - \frac {\langle u,v\rangle^2}{\langle v,v \rangle} $$
Take the - to the left, we get:
$$ \frac {\langle u,v\rangle^2}{\langle v,v \rangle} \leq \langle u,u \rangle $$
Since in vector inner product of the same vector $\langle v,v \rangle$ = $|v|^2$ , thus:
$$ \frac {\langle u,v\rangle^2}{|v|^2} \leq |u|^2 $$
$$ \frac {\langle u,v\rangle^2}{|v|^2|u|^2} \leq 1 $$
Taking Square root on both sides:
$$ \frac {\langle u,v\rangle}{|v||u|} \leq 1 $$
Would the above be a correct mathematical deduction of how the equivalence is achieved?
Side Q) in (1) Is the minima of a Quadratic equation when its discriminant = 0? Any link that helps show this please suggest.