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

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

Consider ${\displaystyle \left|x\right|\ge \left|y\right|}$. Therefore: 
${\displaystyle |\left|x\right|-|y\left||=|\right|x-y+y|-|y\left||\le |\right|x-y|+|y|-|y\left||=|\right|x-y||=|x-y|}$.

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