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Does anyone know how to start proving this inequality

$$ \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{4 \|x-y\|}{\|x\|+ \|y\|} $$

The norm is a random norm on a vector space $V$

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If you explain what you've tried so far you'll be more likely to get a helpful answer. – MasterOfBinary Nov 5 '13 at 14:48
in fact I tried to subtract the terms in the left hand side, but I couldnt see what to do next ! – Hiesenberg Nov 5 '13 at 14:57

Here is a somewhat complicated proof, I do not know if there is a simpler and more natural one.

Let $a=||x||,b=||y||,c=||x-y||$. The inequality is easy if $a=b$, so assume that it is not the case. Since the problem is symmetric in $x$ and $y$, we can assume $a < b$.

Then $2a < 2b$, so $a+b < 3b-a$, $ \frac{a+b}{3b-a}<1$, and hence $\frac{b^2-a^2}{3b-a} < b-a$.

The triangle inequality shows that $||y|| \leq ||y-x||+||x||$ ; we deduce $c \geq b-a > \frac{b^2-a^2}{3b-a}$. It follows that $(3b-a)c > b^2-a^2$, $ 4bc > b^2-a^2+(a+b)c$ and hence $\frac{4c}{a+b} > \frac{b-a+c}{b} $.

So it will suffice to show the following :

$$ \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{b-a+c}{b} \tag{1} $$

To show (1), we use the identity

$$ \bigg(\frac{x}{a}-\frac{y}{b}\bigg)= \frac{c}{b}\bigg(\frac{x-y}{c}\bigg)+ \frac{b-a}{b}\bigg(\frac{x}{a}\bigg) \tag{2} $$

It follows from (2) that

$$ \bigg(\frac{x}{||x||}-\frac{y}{||y||}\bigg)= \frac{c}{b}\bigg(\frac{x-y}{||x-y||}\bigg)+ \frac{b-a}{b}\bigg(\frac{x}{||x||}\bigg) \tag{3} $$

Applying the triangle inequality to (3), we see that

$$ \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{c}{b}+\frac{b-a}{b} \tag{4} $$

which concludes the proof.

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Thanks .. looks complicated .. especially identity (2) .. – Hiesenberg Nov 5 '13 at 16:13
Complicated, but at least it’s correct and shows your inequality is true. – Ewan Delanoy Nov 5 '13 at 16:16
exactly .. than you so much again – Hiesenberg Nov 5 '13 at 16:21
@Hiesenberg I rewrote (2) in a simpler form – Ewan Delanoy Nov 6 '13 at 3:09

To make life easier, first assume $\|x\| = \|y\| = 1$ which would reduce your inequality to a much more basic $$ \|x-y\| \leq 2 \|x-y\| $$ And now to prove the big result, you have it on the unit-scaled versions of the vectors. Can you take it from here?

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Thanks .. Could you explain the next step? – Hiesenberg Nov 5 '13 at 14:54
@Hiesenberg Think about what to do next -- take unit $x$ and non-unit $y$ and think of what would happen? – gt6989b Nov 5 '13 at 14:57
Can we then put the terms in the LHS on a common Denominator (norm(y)) and subtract !! – Hiesenberg Nov 5 '13 at 15:07
got stuck again !! – Hiesenberg Nov 5 '13 at 15:37
@gt6989b I'd be interested in some more details too. – gpo Nov 30 '13 at 19:51

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