I would like hints as to how to solve this question:
Let $(V,\|\cdot\|)$ be a normed vector space over $\mathbb R$ or $\mathbb C$. Show that the norm $\|\cdot\|$ satisfies the parallelogram identity if, and only if, $$ \|u-v\|^2 + \|u+v\|^2\le 2\left(\|u\|^2+\|v\|^2\right) \qquad \forall u,v \in V. $$
I've tried
$ \displaystyle \|u-v\|^2 + \|u+v\|^2 \le (\|u\|+\|v\|)^2 + (\|u\|+\|v\|)^2 = 2\left(\|u\|^2 + 2\|u\|\|v\| + \|v\|^2\right) $
but I can't see where to go from there. I realise that I need to show that $$ 2\left(\|u\|^2+\|v\|^2\right) \le \|u-v\|^2 + \|u+v\|^2 . $$
Well if a vectorspace satisfies the parallelogram law then the inequality is trivial and we are reduced to deducing the parallelogram law from the inequality.
Consider $w=u+v$, $z=u-v$, then $w+z=2u$ and $w-z=2v$. Using the inequality:
$$\|2v\|^2+\|2u\|^2=\|w-z\|^2+\|w+z\|^2≤2(\|w\|^2+\|z\|^2)=2\|u+v\|^2+2\|u-v\|^2$$
Dividing by $2$ gives you what you want: $$2\|v\|^2+2\|u\|^2≤\|u-v\|^2+\|u-v\|^2$$