Can we add absolute value to inequality that already have absolute value? I am trying to prove the following :
∀,  ∈ : ||| − ||| ≤ || + || 
given that we know :(1)->|| − || ≤ |+| and (2) ->| + | ≤ || + || 
so the way i tried to solve the problem is :
starting with what we know from (1):
|| − || ≤ |+| => apply absolute value =>
||| − ||| ≤ ||+|| =>  ||| − ||| ≤ |+| 
then by (2) => ||| − ||| ≤ |+| ≤ || + ||
i.e ||| − ||| ≤ || + || 
I am really not sure if we can add absolute value to an inequality. I have tried several numbers in order to test it out and it seems that it is working out.
 A: As Bungo and CY Aries have pointed out, we cannot add an absolute value to an inequality since $-2 < 1$ does not imply $|-2| < |1|$.
Method 1: We can use the piecewise definition of the absolute value
$$|t| = 
\begin{cases}
t & \text{if $t \geq 0$}\\
-t & \text{if $t < 0$}
\end{cases}
$$
to write 
$$||x| - |y|| = 
\begin{cases}
|x| - |y| & \text{if $|x| \geq |y|$}\\
-|x| + |y| & \text{if $|x| < |y|$}
\end{cases}
$$
We will establish that the inequality holds in each case.
Suppose $|x| \geq |y|$.  Then we must show that $|x| - |y| \leq |x| + |y|$.  Since $|y| \geq 0$, $-|y| \leq 0$.  Hence,
$$|x| - |y| \leq |x| \leq |x| + |y|$$ 
Suppose $|x| < |y|$.  Then we must show that $-|x| + |y| \leq |x| + |y|$.  Since $|x| \geq 0$, $-|x| \leq 0$.  Hence,
$$-|x| + |y| \leq |y| \leq |x| + |y|$$
Thus, $||x| - |y|| \leq |x| + |y|$. 
Method 2:  We use the definition of absolute value 
$$|t| = \sqrt{t^2}$$
\begin{align*}
||x| - |y|| & = \sqrt{(|x| - |y|)^2}\\
            & = \sqrt{|x|^2 - 2|x||y| + |y|^2}\\
            & \leq \sqrt{|x|^2 + |y|^2} && \text{since $|x|, |y| \geq 0 \implies -2|x||y| \leq 0$}\\
            & \leq \sqrt{|x|^2 + 2|x||y| + |y|^2} && \text{since $|x|, |y| \geq 0 \implies 2|x||y| \geq 0$}\\
            & = \sqrt{(|x| + |y|)^2}\\
            & = ||x| + |y||\\
            & = |x| + |y| 
\end{align*}
since $|x|, |y| \geq 0 \implies |x| + |y| \geq 0 \implies ||x| + |y|| = |x| + |y|$.
A: We cannot add absolute value to an inequality: $-2<-1$ but $|-2|\not<|-1|$.
We have $-(|x|+|y|)\le-|y|\le|x|-|y|\le |x|\le|x|+|y|$.
This implies that $||x|-|y||\le |x|+|y|$.
