I don't understand why it's $|a-b|<c<a+b$ with the absolute value on the left.

So, we have this system of inequalities:

  • $a+b>c$
  • $a+c>b$
  • $b+c>a$

If I enter this into wolfram alpha, I get 2 solutions:

1) $a>0,b>a,b-a<c<a+b$

2) $a>0,0<b \leqslant a, a-b<c<a+b$

In the first one I don't understand why $b>a$, why can't it be less than $a$? Same for the second condition, I don't understand why b has to be in between $a$ and $0$.

  • $\begingroup$ If $b<a$, then $b-a$ is always less than $c$. Actually, the triangle inequality implies both $b-a<c$ and $a-b<c$. However only one of those is nontrivial depending on whether $b<a$ or $a<b$, which is why it is useful to distinguish the cases. (Or to write $|a-b|<c$ which covers both.) $\endgroup$
    – user856
    Sep 13, 2018 at 8:31
  • 1
    $\begingroup$ Recall that $|x|<c$ if and only if $x<c$ and $-x<c$. $\endgroup$ Sep 13, 2018 at 8:49

1 Answer 1


From $$a-b<c$$ and $$b-a<c$$ we get $$|a-b|<c$$ and since $$c<a+b$$ we get


  • $\begingroup$ Thanks, I have seen what you wrote, can you please explain why $a-b<c$ and $b-a<c$ implies that $|a-b|<c$? Sorry, I'm not very good at math. Trying to learn. $\endgroup$
    – Coder-Man
    Sep 13, 2018 at 8:44
  • $\begingroup$ Since we have $$a-b<c$$ and $$-(a-b)<c$$ and this is $$|a-b|<c$$ $\endgroup$ Sep 13, 2018 at 8:45
  • $\begingroup$ So, basically, if we have $-x<c$ and $x<c$, we can throw away $-x<c$, because $x<c$ is stricter? $\endgroup$
    – Coder-Man
    Sep 13, 2018 at 8:51
  • $\begingroup$ No, you can write $$|x|<c$$ $\endgroup$ Sep 13, 2018 at 8:54
  • $\begingroup$ No, I meant: A) if we consider that $x$ is positive and we have 1) $x<c$ 2) $-x<c$, we can throw away 2), because 1) is stricter. B) And in the case where $-x$ is positive, we can throw away 1), because 2) is stricter, am I right? $\endgroup$
    – Coder-Man
    Sep 13, 2018 at 8:55

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