Consider three real numbers $a,b,c$ s.t:
$|a-b|\geq|c|$ and $|b-c|\geq|a|$ and $|c-a|\geq|b|$
Show either a , b or c is the sum of the remainign terms.
I have tried manipulating the inequalities using the triangle inequality , yet i can not get anywhere.
Any help would be greatly appreciated.
My try :
because of the symmetry , let's assume W.L.O.G : $a\geq b\geq c$
We then get : $$|b+c|\leq|2a| \Longrightarrow \frac{|b+c|}{2}\leq|a|$$
However :
$$|a|\leq |b-c|\leq |b|+|c|$$
So :$$\frac{(b+c)^2}{4} \leq (|b|+|c|)^2$$ $\Longleftrightarrow$ $$3(b^2+c^2+2bc)\leq 0$$ $\Longleftrightarrow$ $$b+c=0$$
But that is clearly wrong.
When any two of $a,b,c$ are equal then the remaining one must be zero and there is nothing more to prove.
Therefore, W.L.O.G we can assume $c<b<a\,.$ Then, since $c\le |c|\,,a\le |a|\,,b\le|b|$ we get by assumption $$ c\le a-b\,,\quad a\le b-c\,,\quad b\le a-c\,. $$ When $c\ge 0$ then we get the contradiction $$ b<a\le b-c\le b\,. $$ Therefore, $c<0\,.$ Then, since $|c|=-c$ we get by assumption $$ -c\le a-b\,,\quad a\le b-c\,,\quad b\le a-c\,. $$ In particular, $$ b\le a+c\,,\quad a+c\le b\, $$ which yields $$\boxed{\quad b=a+c\,\,.\quad}$$