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.
 A: Let $x=u+v$ and $y=u-v$, then 
$$|\|x\|^2-\|y\|^2|=|\|u+v\|^2-\|u-v\|^2|=4|u^\top v|$$
Then replace $u=\frac{x+y}{2}$ and $v=\frac{x-y}{2}$ in the above equation and you obtain
$$4|u^\top v|=|(x+y)^\top (x-y)|\leq \|x+y\|\|x-y\|$$
and the proof is complete.
A: 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 $|\|u\| - \|v\||^2 \leq \|u - v\|^2$.
A: 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
$$\|x-y\|\|x+y\| \geq \bigg|\|x\|^2 - \|y\|^2\bigg|$$
A: Note that
$$\|x-y\|\geq \bigg|\|x\|-\|y\|\bigg| \hspace{2em} \textrm{and} \hspace{2em} \|x+y\|\geq \bigg|\|x\|-\|y\|\bigg|  $$
by multiplying both inequalities we have
$$\|x-y\|\|x+y\| \geq \bigg| \|x\|^2 + \|y\|^2 - 2\|x\|\|y\|\bigg| = \|x\|^2 + \|y\|^2 - 2\|x\|\|y\| $$
