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.

This is the case for inner product spaces:
The following is valid for a Vector space over $\mathbb{R}$, because I use the fact that $\langle a,b\rangle = \langle b,a\rangle.$ I asume that is the case you're interested in, if you need a more general case let me know.
Say $\langle a ,b \rangle$ is the scalar product in the space, such that $\|a\|=\sqrt{\langle a,a \rangle}$. Then notice
$$\|x-y\|^2 = \langle x-y,x-y \rangle = \|x\|^2 - 2\langle x,y\rangle +\|y\|^2  $$
$$\|x+y\|^2 = \langle x+y,x+y \rangle = \|x\|^2 + 2\langle x,y\rangle +\|y\|^2  $$
So we have
$$(\|x-y\|\|x+y\|)^2 = \|x-y\|^2\|x+y\|^2 = \|x\|^4 + \|y\|^4 + 2\|x\|^2\|y\|^2 - 4\langle x,y\rangle^2 $$
Now by Cauchy-Schartz inequality ($\langle  x,y \rangle^2 \leq \|x\|^2\|y\|^2$) we have:
$$\|x\|^4 + \|y\|^4 + 2\|x\|^2\|y\|^2 - 4\langle x,y\rangle^2 \geq \|x\|^4 + \|y\|^4 - 2\|x\|^2\|y\|^2 = (\|x\|^2 - \|y\|^2)^2  $$
So in the end we have
$$(\|x-y\|\|x+y\|)^2 \geq (\|x\|^2 - \|y\|^2)^2$$
Hence
$$\|x-y\|\|x+y\| \geq \bigg|\|x\|^2 - \|y\|^2\bigg|$$
