Showing that $\left|x-1+\frac 1{1+x}-\left(y-1+\frac 1{1+y}\right)\right|\leq 2|x-y|$ I asked a question here the other day and one of the steps in the answer I encountered was: $$\left|x-1+\frac 1{1+x}-\left(y-1+\frac 1{1+y}\right)\right|\leq 2|x-y|$$
when $x,y\ge0$
I can't seem to figure out why this is true.. Cans someone help me out?
Thanks :) 

We have thanks to the triangular inequality
\begin{align*}\left|x-1+\frac 1{1+x}-\left(y-1+\frac 1{1+y}\right)\right|&= 
\left|x-1+\frac 1{1+x}-y+1-\frac 1{1+y}\right|\\
&=\left|x+\frac 1{1+x}-y-\frac 1{1+y}\right|\\
&\leq |x-y|+\left|\frac 1{1+x}-\frac 1{1+y}\right|\\
&= |x-y|+\left|\frac {1+y-(1+x)}{(1+x)(1+y)}\right|\\
&=|x-y|+\frac {|y-x|}{(1+x)(1+y)}\\
&\leq 2|x-y|.
\end{align*}
