Normed vector space inequality $|\|x\|^2 - \|y\|^2| \le \|x-y\|\|x+y\|$ I'm looking at an old qualifying exam, and one question is to prove the following inequality in any normed vector space:
$$ |\|x\|^2 - \|y\|^2| \le \|x-y\|\|x+y\| $$
My initial thought was that
$$ |\|x\|^2 - \|y\|^2| = |(\|x\|+\|y\|)(\|x\|-\|y\|)|=\left|(\|x\|+\|y\|)\right||(\|x\|-\|y\|)|,$$
and it's easy to show $|\|x\|-\|y\||$ is less than both $\|x-y\|$ and $\|x+y\|$, but it isn't true that $\|x\|+\|y\|$ is less than either in general (by the triangle inequality it's 'usually' larger than the latter), so I'm unsure what to do. Any guidance is appreciated.

We may assume w.l.o.g. that $\|x\|^2 \geq \|y\|^2$. Write $x = u + v$ and $y = u - v$. Now the inequality can be rewritten as
$$
\|u + v\|^2 \leq 4 \|u\| \|v\| + \|u - v\|^2.
$$
But this is the inequality one gets by combining $\|u + v\|^2 \leq (\|u\| + \|v\|)^2$ and $|\|u\| - \|v\||^2 \leq \|u - v\|^2$.
