Let u and v be two vectors in RR^2 . The Cauchy-Schwarz inequality states that |u * v| <= |u||v| We are able to transform the above inequality so that it also shows us that |u+v| <= |u|+|v| But I cannot find a way to show that |u−v| <= |u|+|v| even though I now this has to be true. Any ideas?

Lexi Mcneil 2022-07-22 Answered
Let u and v be two vectors in R 2
The Cauchy-Schwarz inequality states that | u · v | | u | | v |
We are able to transform the above inequality so that it also shows us that
| u + v | | u | + | v |
But I cannot find a way to show that | u v | | u | + | v |
even though I now this has to be true. Any ideas?
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Answers (1)

ri1men4dp
Answered 2022-07-23 Author has 14 answers
| u + v | | u | + | v |
is shown already, so let's use this.
| u v | = | u + ( v ) | | u | + | v | = | u | + | v |
and we're done.
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