Prove that $a ≤ x ≤ b ⇒ |x| ≤ |a|+|b|$ Can you please help me with this proof $$a ≤ x ≤ b ⇒ |x| ≤ |a|+|b|$$ ? I am literally stuck for hours.
This is what i thought, but i don't know if it counts as a proof.
First of all, if a ≤ x then  $x ≤ -a$ so $a ≤ x ≤ -a$ if $x ≤ -a$ then $|x| ≤ |-a| = |a|$
Now we look at b:
if x ≤ b then $-b ≤ x ≤ b$ so  $$|-b| = |b| ≤ |x|$$ OR  $$|x| ≤ |-b| = |b|$$
let's choose the "worst" option which is |b| ≤ |x|
so $$|b| ≤ |x| ≤ |a| ⇒  |x| ≤ |a| - |b|  ≤ |a| + |b| ⇒  |x| ≤ |a| +|b|$$
Is it correct? Please show me other ways to prove it.

Hint
$x$ can not be further away from $0$ than both $a$ and $b$.
Consider that $x$ must have the same sign as at least one of $a,b$.
