ivybeibeidn

2022-10-07

Let $‖\cdot ‖:{\mathbb{R}}^{n}\to \mathbb{R}$ be the Euclidean norm. Prove that for any two vectors $x,y\in {\mathbb{R}}^{n}$ with $x\ne 0\ne y$ the following inequality holds:
$||\frac{x}{‖x‖}-\frac{y}{‖y‖}||\le \frac{2}{‖x‖+‖y‖}‖x-y‖.$
I'm very curious as to how I can attack this question. Any hints as well as a full answer is appreciated. I've tried using the triangle inequality (which is, basically, all I could think about) with no luck. I guess I could try using the definition of the euclidean norm letting $x=\left({x}_{1},\dots ,{x}_{n}\right)$ and $y=\left({y}_{1},\dots ,{y}_{n}\right)$ and try something, but I don't want to go down that path (and I don't think it would lead me anywhere, either).

Haylee Branch

Let x=ru and y=sv where u and v are unit vectors, $‖x‖=r$ and $‖y‖=s$. Then your inequality says

By homogeneity, we may assume r+s=1. Thus the inequality becomes

The right side is convex as a function of r. Now there is a linear isometry that interchanges u and v, so $‖ru-\left(1-r\right)v‖=‖\left(1-r\right)u-rv‖$
Thus the minimum must occur at $r=1/2$

Dana Russo

Let x=ur and y=sv where $r,s\in {\mathbb{R}}^{+}$ and $‖u‖=‖v‖=1.$
Let c=s/r. Then c>0.
The LHS of the inequality is $‖u-v‖.$. The RHS is $\frac{2}{1+c}‖u-cv‖.$
We have $‖u{‖}^{2}=‖v{‖}^{2}=1.$. And 1+c>0. So we have
$‖u-v‖\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\frac{2}{1+c}‖u-cv‖\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\left(1+c\right)\cdot ‖u-v‖\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}2‖u-cv‖\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\left(1+c{\right)}^{2}\cdot ‖u-v{‖}^{2}\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}4‖u-cv{‖}^{2}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\left(1+c{\right)}^{2}\cdot \left(‖u{‖}^{2}+‖v{‖}^{2}-2u\cdot v\right)\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}4\cdot \left(\phantom{\rule{thinmathspace}{0ex}}‖u{‖}^{2}+{c}^{2}‖v{‖}^{2}-2c\left(u\cdot v\right)\phantom{\rule{thinmathspace}{0ex}}\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\left(1+c{\right)}^{2}\cdot \left(2-2u\cdot v\right)\le 4\left(1+{c}^{2}-2c\left(u\cdot v\right)\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\left(-2u\cdot v\right)\cdot \left(\phantom{\rule{thinmathspace}{0ex}}\left(1+c{\right)}^{2}-4c\right)\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}4\left(1+{c}^{2}\right)-2\left(1+c{\right)}^{2}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\left(-2u\cdot v\right)\left(1-c{\right)}^{2}\le 2\left(1-c{\right)}^{2}.$
Now $|-2u\cdot v|\le 2‖u‖\cdot ‖v‖=2.$. So the absolute value of the LHS of the last line above is at most $2\left(1-c{\right)}^{2}.$

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