Alfred Martin

Answered

2021-12-16

Prove that $|\left|x\right|-|y||\le |x-y|$

Proof of the Triangle Inequality $|x+y|\le \left|x\right|+\left|y\right|$.

Proof of the reverse triangle inequality:

$|\left|x\right|-|y||\le |x-y|$.

Answer & Explanation

esfloravaou

Expert

2021-12-17Added 43 answers

$\left|x\right|+|y-x|\ge |x+y-x|=\left|y\right|$

$\left|y\right|+|x-y|\ge |y+x-y|=\left|x\right|$

Move |x| to the right hand side in the first inequality and |y| to the right hand side in the second inequality. We receive

$|y-x|\ge \left|y\right|-\left|x\right|$

$|x-y|\ge \left|x\right|-\left|y\right|$.

From absolute value properties, we have that $|y-x|=|x-y|$, and if $t\ge a\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}t\ge -a\text{}then\text{}t\ge \left|a\right|$.

Putting these two facts together, we get the reverse triangle inequality:

$|x-y|\ge \mid \mid \left|x\right|-\left|y\right|\mid \mid$.

Mason Hall

Expert

2021-12-18Added 36 answers

Consider $\left|x\right|\ge \left|y\right|$. Therefore:

$|\left|x\right|-|y\left||=|\right|x-y+y|-|y\left||\le |\right|x-y|+|y|-|y\left||=|\right|x-y||=|x-y|$.

nick1337

Expert

2021-12-28Added 573 answers

Since we are thinking about the reals, R, then a field's axioms apply. specially for$x,y,z\in R,x+(-x)=0;x+(y+z)=(x+y)+z;\text{}and\text{}x+y=y+x$.

Start with $x=x+0=x+(-y+y)=(x-y)+y.$

Then apply $|x|=|(x-y)+y|\le |x-y|+|y|$. By so-called "first triangle inequality."

Rewriting $|x|-|y|\le |x-y|\text{}and\text{}||x|-y||\le |x-y|.$

The item of Analysis that I find the most conceptually daunting at times is the notion of order $(\le ,\ge ,<,>)$, note how some sentences might be improved into shorter versions.

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