Proving Cauchy-Schwartz inequality without the vanish assumption of inner products

Question

I've came accross the following question in a book that I'm studying from, about Hilbert spaces -

And the answer is -

What I don't understand is why

implies that - ?

Answer 1

If you have a quadratic of the form $$P(\lambda)=C+\lambda B + A\lambda^2$$ and you know that $P(\lambda)\geq 0$ for all $\lambda\in\mathbb{R}$, then you must have $B^2-4AC\leq 0$, or else the quadratic would have two real roots, and consequently, will assume negative values. Now put $C=\langle x,x\rangle$,$A=\langle y,y\rangle$ and $B=2|\langle x,y\rangle|$, and you get $$4|\langle x,y\rangle |^2\leq 4 \langle x,x\rangle \langle y,y\rangle$$ which gives the C-S inequality.