Can we add absolute value to inequality that already have absolute value?

Question

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.

Answer 1

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 $||x|-|y||\le |x|+|y|$.

Answer 2

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 $|x|, |y| \geq 0 \implies |x| + |y| \geq 0 \implies ||x| + |y|| = |x| + |y|$.