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I've got stuck on this problem

Let $a$, $b$, $c> 0$ and $x$ a real number such that $$|ax - b| \leq c,$$ $$|bx - c| \leq a$$ and $$ |cx - a| \leq b.$$

Prove that $0 \leq x \leq 2$.

What I've tried so far - to sum all three inequalities and to square the expresion. And the properties of modules of course ( such as $|x| + |y| >= |x+y| $) . It wasn't too helpful.

I would apreciate some hints.

Thanks!

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    $\begingroup$ What do you mean by "Module properties"? $\endgroup$ – Umberto P. Oct 2 '15 at 20:55
  • $\begingroup$ Properties such as $|x+y| <= |x| + |y|$. $\endgroup$ – scummy Oct 2 '15 at 21:08
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you need just add all the inequalities, and you would have: $$|ax-b|+|bx-c|+|cx-a|\leq a+b+c$$ Also, from the Module properties: $$|ax-b+bx-c+cx-a|\leq |ax-b|+|bx-c|+|cx-a|$$ $$|(a+b+c)x-(a+b+c)|\leq |ax-b|+|bx-c|+|cx-a|\leq a+b+c$$ Dividing by $(a+b+c)$, we have: $$|x-1|\leq 1$$ which proves that $0 \leq x \leq 2$.

Division would not affect the inequality because $a,b,c>0$

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Hint :

$$|a|\leq b\iff -b\leq a\leq b.$$

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