$|b-a|=|b-c|+|c-a| \implies c\in [a,b]$ We know that if $c\in [a,b]$ we have $|b-a|=|b-c|+|c-a|$. I'm trying to prove that if the norm is induced by an inner product, then the converse holds.
I need a hint or something.
Thanks in advance

Using one part of Kouba answer to simplify my answer, let $x=b-c$, $y=c-a$, then
$$ \langle x,y\rangle=\Vert x\Vert\cdot\Vert y\Vert$$
But every inner product is written in this form: $\langle x,y\rangle=\lambda xy$,where $\lambda\gt 0$ and the norm induced by it is $\Vert x \Vert=|\lambda x|$, then we have:
$$(b-c)(c-a)=|\lambda||b-c||c-a|\ge 0$$
which implies
$b-c\ge 0$ and $c-a \ge 0$ or $b-c\le 0$ and $c-a\le 0$
The second one is absurd, so $b-c\ge 0$ and $c-a \ge 0$, and $c\in [a,b]$
