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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?

Keith
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1 Answers1

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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.

Fred
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  • Oh.............Thank you for your answer.... – Keith Aug 17 '18 at 06:52
  • Why the downvote ?????????????????????????????? – Fred Aug 17 '18 at 07:14
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    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. – Jyrki Lahtonen Aug 17 '18 at 10:16
  • Yes, my sheriff ! – Fred Aug 17 '18 at 10:20