Let $V$ be a normed vector space and I need to prove the follwing inequality $\mid \parallel x \parallel - \parallel y \parallel \mid$ $\leqq$ $\parallel x-y \parallel$ containing the norm and the absolute value of the real numbers. However things just get twisted and I cannot see how to prove it.... Could anyone please tell me how to prove it?
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$\begingroup$ The question that presents itself : what can you infer from this inequality ? $\endgroup$– Neil hawkingAug 17, 2018 at 6:54
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$\begingroup$ Search for triangle inequality. $\endgroup$– Jyrki LahtonenAug 17, 2018 at 7:00
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1 Answer
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$||x||=||x-y+y|| \le ||x-y||+||y||$, hence
$(1)$ $||x||-||y|| \le ||x-y||$.
In a similar way we get
$(2)$ $||y||-||x|| \le ||y-x||$.
Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.
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$\begingroup$ Why the downvote ?????????????????????????????? $\endgroup$– FredAug 17, 2018 at 7:14
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1$\begingroup$ The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation. $\endgroup$ Aug 17, 2018 at 10:16
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