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.


  • 1
    $\begingroup$ 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

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 $a,b,c>0$


Hint :

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.