Alfred Martin

2021-12-16

Prove that $||x|-|y||\le |x-y|$
Proof of the Triangle Inequality $|x+y|\le |x|+|y|$.
Proof of the reverse triangle inequality:
$||x|-|y||\le |x-y|$.

esfloravaou

Expert

$|x|+|y-x|\ge |x+y-x|=|y|$
$|y|+|x-y|\ge |y+x-y|=|x|$
Move |x| to the right hand side in the first inequality and |y| to the right hand side in the second inequality. We receive
$|y-x|\ge |y|-|x|$
$|x-y|\ge |x|-|y|$
From absolute value properties, we have that $|y-x|=|x-y|$, and if
Putting these two facts together, we get the reverse triangle inequality:
$|x-y|\ge \mid \mid |x|-|y|\mid \mid$.

Mason Hall

Expert

Consider $|x|\ge |y|$. Therefore:
$||x|-|y||=||x-y+y|-|y||\le ||x-y|+|y|-|y||=||x-y||=|x-y|$.

nick1337

Expert

Since we are thinking about the reals, R, then a field's axioms apply. specially for.
Start with $x=x+0=x+\left(-y+y\right)=\left(x-y\right)+y.$
Then apply $|x|=|\left(x-y\right)+y|\le |x-y|+|y|$. By so-called "first triangle inequality."
Rewriting
The item of Analysis that I find the most conceptually daunting at times is the notion of order $\left(\le ,\ge ,<,>\right)$, note how some sentences might be improved into shorter versions.

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