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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 $\pm(\|x\|-\|y\|) \leq \|x-y\|$.2011-11-05

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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\| $