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.
 A: 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}
A: Given that we are discussing the reals, $\mathbb{R}$, then the axioms of a field apply. Namely for :$x,y,z\in\mathbb{R}, \quad x+(-x)=0$; $x+(y+z)=(x+y)+z$; and $x+y=y+x$.
Start with $x=x+0=x+(-y+y)=(x-y)+y$. 
Then apply $|x| = |(x-y)+y|\leq |x-y|+|y|$.   By so-called "first triangle inequality."
Rewriting $|x|-|y| \leq |x-y|$ and $||x|-y|| \leq |x-y|$.
The item of Analysis that I find the most conceptually daunting at times is the notion of order $(\leq,\geq,<,>)$, and how certain sentences can be augmented into simpler forms.
Hope this helps and please give me feedback, so I can improve my skills.
Cheers. 
A: 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.
A: $$|x| + |y -x| \ge |x + y -x| = |y|$$
$$|y| + |x -y| \ge |y + x -y| = |x|$$
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:
$$|x-y| \ge \bigl||x|-|y|\bigr|.$$
A: Explicitly, we have
\begin{align}
\bigl||x|-|y|\bigr|
=&
\left\{
\begin{array}{ll}
|x-y|=x-y,&x\geq{}y\geq0\\
|x-y|=-x+y=-(x-y),&y\geq{}x\geq0\\
|-x-y|=x+y\leq-x+y=-(x-y),&y\geq-x\geq0\\
|-x-y|=-x-y\leq-x+y=-(x-y),&-x\geq{}y\geq0\\
|-x+y|=-x+y=-(x-y),&-x\geq-y\geq0\\
|-x+y|=x-y,&-y\geq-x\geq0\\
|-x+y|=-x-y\leq{}x-y,&-y\geq{}x\geq0\\
|-x+y|=x+y\leq{}x-y,&x\geq-y\geq0
\end{array}
\right\}\nonumber\\
=&|x-y|.\nonumber
\end{align}
A: WLOG, consider $|x|\ge |y|$. Hence:
$$||x|-|y||=||x-y+y|-|y||\le ||x-y|+|y|-|y||=||x-y||=|x-y|.$$
A: $\left| |x|-|y| \right|^2 - |x-y|^2 = \left( |x| - |y| \right)^2 - (x-y)^2 = |x|^2 - 2|x| \cdot |y| +y^2 - x^2 + 2xy-y^2 = 2 (xy-|xy|) \le 0 \Rightarrow \left| |x|-|y| \right| \le |x-y|.$
$"=" \iff xy\ge 0.$ Q.E.D.
The Triangle Inequality can be proved similarly.
