Prove that $|a-b|\ge \left|a\right|-\left|b\right|$ .

expeditiupc
2021-12-10
Answered

Prove that $|a-b|\ge \left|a\right|-\left|b\right|$ .

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Terry Ray

Answered 2021-12-11
Author has **50** answers

It is called the reverse triangle inequality.

$|a-b|\ge |\left|a\right|-|b\mid \mid$

$\left|a\right|=|a-b+b|\le |a-b|+\left|b\right|$

$\left|b\right|=|b-a+a|\le |a-b|+\left|a\right|$

Thus, we have:

$-|a-b|\le \left|a\right|-\left|b\right|\le |a-b|$

Thus, we have:

Tiefdruckot

Answered 2021-12-12
Author has **46** answers

The length of any side of a triangle is greater than the absolute difference of the lengths of the other two sides:

$|\left|a\right|-|b|\le |a-b\mid$

So,

$|a-b|\ge \left|a\right|-\left|b\right|$

$|a-b|\ge \left|b\right|-\left|a\right|$

That's it.

So,

That's it.

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