# How to show the inequality?

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 .

• – Felix Marin Apr 28 '19 at 15:31

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.

• is there also a geometric meaning behind this ? – Gol D. Roger Apr 28 '19 at 15:25
• 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. – mxian Apr 28 '19 at 15:34

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 $$|\vec y|-|\vec x|\le|\vec x-\vec y|\,$$?

• is there also a geometric meaning behind this ? – Gol D. Roger Apr 28 '19 at 15:27
• 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. – Bernard Apr 28 '19 at 15:34