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How to prove that $\|x-y\| \geq |\|x\|-\|y\||$?

I am only thinking of for the LHS, $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$ but not sure how to manipulate that and how to handle the RHS.

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Are you allowed to use the triangle inequality? If so $\|x\|=\|x-y+y\| \leq \|x-y\|+\|y\|$ so $\|x\|-\|y\| \leq \|x-y\|$ and $\|y\|=\|y-x+x\|\leq \|y-x\|+\|x\|=\|x-y\|+\|x\|$ so $\|y\|-\|x\| \leq \|x-y\|$. Thus $\pm(\|x\|-\|y\|) \leq \|x-y\|$. – Bill Cook Nov 5 '11 at 19:16

It is equivalent to the triangle inequality: $$ \|x\|=\|(x-y)+y\|\le\|x-y\|+\|y\| $$ Then subtract $\|y\|$ from both sides to get $$ \|x\|-\|y\|\le\|x-y\| $$ Similarly, we can show that $$ \|y\|-\|x\|\le\|x-y\| $$ to get $$ |\|x\|-\|y\||\le\|x-y\| $$

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