Prove that&nbsp;||x|−|y||≤|x−y|Proof of the Triangle Inequality&nbsp;|x+y|≤|x|+|y|.Proof of the reverse triangle inequality:||x|−|y||≤|x−y|.

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Prove that&amp;nbsp;||x|−|y||≤|x−y|Proof of the Triangle Inequality&amp;nbsp;|x+y|≤|x|+|y|.Proof of the reverse triangle inequality:||x|−|y||≤|x−y|.

esfloravaou · Accepted Answer

|x|+|y−x|≥|x+y−x|=|y|&amp;nbsp;|y|+|x−y|≥|y+x−y|=|x|&amp;nbsp;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|≥|y|−|x|&amp;nbsp;|x−y|≥|x|−|y|.&amp;nbsp;From absolute value properties, we have that&amp;nbsp;|y−x|=|x−y|, and if&amp;nbsp;t≥a&amp;nbsp;&amp;nbsp;and&amp;nbsp;&amp;nbsp;t≥−a&amp;nbsp;then&amp;nbsp;t≥|a|.&amp;nbsp;Putting these two facts together, we get the reverse triangle inequality:|x−y|≥∣∣|x|−|y|∣∣.

Mason Hall · Accepted Answer

Consider&amp;nbsp;|x|≥|y|. Therefore:&amp;nbsp;||x|−|y||=||x−y+y|−|y||≤||x−y|+|y|−|y||=||x−y||=|x−y|.

nick1337 · Accepted Answer

Since we are thinking about the reals, R, then a field&#039;s axioms apply. specially forx,y,z∈R,x+(−x)=0;x+(y+z)=(x+y)+z;&amp;nbsp;and&amp;nbsp;x+y=y+x.Start with&amp;nbsp;x=x+0=x+(−y+y)=(x−y)+y.Then apply&amp;nbsp;|x|=|(x−y)+y|≤|x−y|+|y|. By so-called &quot;first triangle inequality.&quot;Rewriting&amp;nbsp;|x|−|y|≤|x−y|&amp;nbsp;and&amp;nbsp;||x|−y||≤|x−y|.The item of Analysis that I find the most conceptually daunting at times is the notion of order&amp;nbsp;(≤,≥,&amp;lt;,&amp;gt;), note how some sentences might be improved into shorter versions.

Prove that ||x|-|y|| \le |x-y| I've seen the full proof of

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