Reverse Triangle Inequality Proof 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.
$$|x| + |y -x| \ge |x + y -x| = |y|$$

$$|y| + |x -y| \ge |y + x -y| = |x|$$

Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. We get

$$|y -x| \ge |y| - |x|$$

$$|x -y| \ge |x| -|y|.$$

From absolute value properties, we know that $|y-x| = |x-y|,$ and if $t \ge a$ and $t \ge −a$ then $t \ge |a|$. 

Combining these two facts together, we get the reverse triangle inequality:

$$|x-y| \ge \bigl||x|-|y|\bigr|.$$
For all $x,y\in \mathbb{R}$, the triangle inequality gives
\begin{equation}
|x|=|x-y+y| \leq |x-y|+|y|,
\end{equation}

\begin{equation}
|x|-|y|\leq |x-y| \tag{1}.
\end{equation}
Interchaning $x\leftrightarrow y$ gives
\begin{equation}
|y|-|x| \leq |y-x|
\end{equation}
which when rearranged gives
\begin{equation}
-\left(|x|-|y|\right)\leq |x-y|. \tag{2}
\end{equation}
Now combining $(2)$ with $(1)$, gives
\begin{equation}
-|x-y| \leq |x|-|y| \leq |x-y|.
\end{equation}
This gives the desired result
\begin{equation}
\left||x|-|y|\right| \leq |x-y|. \blacksquare
\end{equation}