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Let $r \in \left(0,1\right),\ \phi \in \left(0,2\pi\right)$. $$ \mbox{For}\quad \vec{x} = \left(\vphantom{\large A} r_{1}\cos\left(\phi_{1}\right), r_{1}\sin\left(\phi_{1}\right)\right),\quad \vec{y} = \left(\vphantom{\large A} r_{2}\cos\left(\phi_{2}\right), r_{2}\sin\left(\phi_{2}\right)\right) $$ it holds $\displaystyle\left\vert\,{\vec{x} - \vec{y}}\,\right\vert \geq \left\vert\,{r_{1} - r_{2}}\,\right\vert$.

Has anyone got an idea ?. Thank you .

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You have to show $$\begin{split} |x-y|&\geq ||x|-|y|| \\ \Leftrightarrow (x-y)(x-y)&\geq ||x|-|y||^2 \\ &=(|x|-|y|)^2 \\ &=|x|^2-2|x||y|+|y|^2. \end{split}$$
But the left hand side is $$x\cdot x-2xy+y\cdot y=|x|^2-2xy+|y|^2$$ and it remains to show that $$-2xy\geq -2|x||y|\Leftrightarrow |x||y|\geq xy,$$ which is just the Cauchy-Schwarz inequality.

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  • $\begingroup$ is there also a geometric meaning behind this ? $\endgroup$ – Gol D. Roger Apr 28 '19 at 15:25
  • $\begingroup$ If $X$ and $Y$ are points in $\mathbb{R}^n$, then the length of the segment $\overline{XY}$ is greater than the difference of the distances of $X$ and $Y$ to the origin. $\endgroup$ – mxian Apr 28 '19 at 15:34
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Hint:

To show $|A|\le B$, you only have to show successively that $A\le B$ and $-A\le B$.

Write $\vec x=(\vec x-\vec y)+\vec y$ and use the triangle inequality to show $|\vec x|-|\vec y|\le |\vec x-\vec y|$.

Can you see why this also proves $|\vec y|-|\vec x|\le|\vec x-\vec y|\,$?

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  • $\begingroup$ is there also a geometric meaning behind this ? $\endgroup$ – Gol D. Roger Apr 28 '19 at 15:27
  • $\begingroup$ Oh! yes. It's from medschool: In a triangle, the length of a side is between the sum and the difference of the lengths of the two other sides. $\endgroup$ – Bernard Apr 28 '19 at 15:34

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