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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.

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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 $|\|u\| - \|v\||^2 \leq \|u - v\|^2$.

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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.

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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

$$\|x-y\|\|x+y\| \geq \bigg|\|x\|^2 - \|y\|^2\bigg|$$

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  • $\begingroup$ The question is about arbitrary normed spaces, not just inner product ones. $\endgroup$ – Andrés E. Caicedo May 30 '17 at 1:05
  • $\begingroup$ @AndrésE.Caicedo I see, I couldn't think of a general proof, I will add that coment to my answer, maybe it helps solve the general case... $\endgroup$ – The Giraffe Guy May 30 '17 at 1:18
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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

$$\|x-y\|\|x+y\| \geq \bigg| \|x\|^2 + \|y\|^2 - 2\|x\|\|y\|\bigg| = \|x\|^2 + \|y\|^2 - 2\|x\|\|y\| $$

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