Prove the $x\leq \lvert x \rvert$ (absolute value)

Question

I want to prove $ x\leq \lvert x \rvert$.

We have two cases:

If $x\geq 0$: We have $x \leq \lvert x \rvert$ (or should it be $x \leq \lvert x \rvert=x$ ?)

If $x<0$: We have $-x<- \lvert x \rvert \iff x> \lvert x \rvert$ (Or maybe $x<- \lvert x \rvert \iff -x> \lvert x \rvert$ ?)

So I have $x\leq \lvert x \rvert$ and $x>\lvert x\rvert$, but now what?

When I check the two cases, should I only change the sign for the $x$ in absolute value-sign or also the normal $x$ (case 2)?

Thanks!

Answer 1

For $x\ge 0$ the equality $x=|x|$ holds. For $x<0$ we have $|x|=-x$, so $x<0$ and $|x|\ge 0$, hence trivially $x\le |x|$.

Answer 2

It's simple ...just remember the definition of absolute value function $$|x| = \begin{cases}{x} & x\geq0 \\ {-x} & x<0 \end{cases}$$

SO ...For $x \geq0$ The inequality is trivial.. $x\le x=|x|$ which is true

For $x<0$ which means $-x>0$

So,the inequality is as follows

$$x<0< -x=|x|$$