Prove that $||x|-|y||\le |x-y|$ I've seen the full proof of the Triangle Inequality 
\begin{equation*}
|x+y|\le|x|+|y|. 
\end{equation*}
However, I haven't seen the proof of the reverse triangle inequality: 
\begin{equation*}
||x|-|y||\le|x-y|.
\end{equation*}
Would you please prove this using only the Triangle Inequality above?
Thank you very much.

We can write the proof in a way that reveals how we can think about this problem. 
The inequality $|a|\le M$ is equivalent to $-M\le a\le M$, which is one way to write the following two inequalities together:
$$
a\le M,\quad a\ge -M\;.
$$
Therefore, what we need to prove are (both of) the following:
$$
|x|-|y|\le |x-y|,\tag{1}
$$
$$
|x|-|y|\ge -|x-y|\;.\tag{2}
$$
On the other hand, the known triangle inequality tells us that "the sum of the absolute values is greater than or equal to the absolute value of the sum":
$$
|A|+|B|\ge |A+B|\;\tag{3}
$$
Observe that there are two (positive) quantities on the left of the $\ge$ sign and one of the right. Furthermore, (1) and (2) can be written in such a form easily:
$$
|y|+|x-y|\ge |x|\tag{1'}
$$
$$
|x|+|x-y|\ge |y|\tag{2'}
$$
Thus, we get (1') easily from (3), by setting $A=y$, $B=x-y$.
How about (2')? We don't, in general, have $x+(x-y)=y$. But wait, (2') is equivalent to 
$$
|x|+|y-x|\ge |y|\tag{2''}
$$
because $|x-y|=|y-x|$. Now we are done by using (3) again.
