This is related to the John von Neumann proof that norm that satisfies parallelogram law is induced by a scalar product. I have seen other post regarding this, but I didn't find anything on this particular part of this particular proof.
The step I'm stuck at is showing that the Cauchy-Schwarz inequality holds.
Show that for $r\in\mathbb{Q}$ and $v,w\in V$ (our real normed vector space with norm $||\cdot||$): $$1)\space r^2||v||^2+2rq(v,w)+||w||^2\geq0.$$ Explain why it implies: $$2)\space4|q(v,w)|^2-4||v||^2||w||^2\leq0$$ and hence the Cauchy-Schwarz inequality for q. (which is obvious)
where $q(v,w)=\frac{1}{4}(||v+w||^2-||v-w||^2)$. I am struggling most with the first one because I think I've showed the second one, but I didn't use the first as an implication, so I am confused.
What I did: $$||v+w||\leq||v||+||w||, ||v-w||\geq||v||-|w||$$ imply $$||v+w||^2\leq||v||^2+2||v||\cdot||w||+||w||^2$$$$||v-w||^2\geq||v||^2-2||v||\cdot||w||+||w||^2$$ so $$q(v,w)\leq\frac{1}{4}(4||v||\cdot||w||)=||v||\cdot||w||.$$
Which certainly implies the second equation. Any sort of help is appreciated.
Something I will add that might be useful: $q(v,w)$ was already shown to be positive definite, symmetric and additive. The last step was showing that it is homogeneous and homogeneity for rational numbers was simple, but then it required the Cauchy-Schwartz inequality for homogeneity with real numbers. This is the last step of the proof where we need to show that $|q(v,w)|≤||v||⋅||w||$.
Your proof of the inequality isn't quite complete, because you need not only $\ q(u,v)\le\|u\|\|v\|\ $, but also $\ {-}\|u\|\|v\|\le$$\,q(u,v)\ $ as well. The latter inequality is easy enough to get, however, by replacing $\ v\ $ with $\ {-}v\ $ in the one you've already proved: \begin{align} -q(u,v)&=q(u,-v)\\ &\le \|u\|\|{-}v\|\\ &= \|u\|\|v\|\ , \end{align} from which the required second inequality, $\ {-}\|u\|\|v\|\le$$\,q(u,v)\ $, now follows.
With this additional step, your proof seems fine to me. To some extent, therefore, I share your confusion about why you would use the much more cumbersome method suggested by the text you've quoted. For what it's worth, however, here's a proof using that method.
For any real $\ r\ $ you have $$ 2q(u,rv)=\frac{1}{2}\big(\|u+rv\|^2-\|u-rv\|^2\big) $$ from the definition of $\ q\ $, and $$ \|u\|^2+\|rv\|^2=\frac{1}{2}\big(\|u+rv\|^2+\|u-rv\|^2\big) $$ from the parallelogram law. Adding these two identities gives \begin{align} 0&\le\|u+rv\|^2\\ &=\|rv\|^2+2q(u,rv)+\|u\|^2\\ &=r^2\|v\|^2+2q(u,rv)+\|u\|^2\ . \end{align} Since you've already established the bilinearity of $\ q\ $ for rational $\ r\ $, it follows that \begin{align} r^2\|v\|^2+2rq(u,v)+\|u\|^2&=r^2\|v\|^2+2q(u,rv)+\|u\|^2\\ &\ge0 \end{align} for all rational $\ r\ $. However, since $\ r^2\|v\|^2+$$\,2rq(u,v)+$$\,\|u\|^2\ $ is continuous in $\ r\ $, and $\ \mathbb{Q}\ $ is dense in $\ \mathbb{R}\ $, the above inequality must hold for all $\ r\in \mathbb{R}\ $. For $\ r={-}\frac{q(u,v)}{\|v\|^2}\ $ (when $\ v\ne0\ $), the inequality becomes $$ {-}\frac{q(u,v)^2}{\|v\|^2}+\|u\|^2\ge0\ , $$ or, on rearranging and taking square roots, $$ |q(u,v)|\le\|u\|\|v\|\ , $$ which is the required Cauchy-Schwartz inequality.