Property of absolute value How to prove for $x\in\Bbb R$, $-|x|\leq x\leq |x|$?I could prove it for the cases $x=0$ and for $x>0$.
$$|x| =
\left\{
 \begin{array}{ll}
         0 & \mbox{if } x =0 \\
   x  & \mbox{if } x > 0 \\
  -x & \mbox{if } x < 0
 \end{array}
\right.$$
Case 1:When $x=0$
$-|x|=x=|x|$ hence the inequality holds
Case 2:When $x>0$
We know that for all $x\in\mathbb R$:
$$\begin{align}
&|x|≥x && (i)\\
\Rightarrow&-|x|≤-x && (ii)
\end{align}$$
and when $x>0$
this implies $-x<0$
so $$–x<0<x\qquad (iii)$$
From (i),(ii) and (iii) We get
$$-|x|\leq-x<0<x\leq|x|\implies -|x|≤x≤|x|$$
hence the inequality holds
but I am not able to prove for the case $x<0$.

You are way overcomplicating the case when $x>0$. Remember, from the definition of absolute values, what is the absolute value of a positive number?
Hint: Take a look at some examples of positive numbers. What are $|1|, |2|, |1000|, |\pi|, |e|$?

For the negative numbers, the task should be even simpler. Remember that one of the fundamental properties of the absolute value is that it is always non-negative. So, from that, if $x<0$, can you see why $x<|x|$ (remember, again, $|x|$ is non-negative!)
