$|b-a|=|b-c|+|c-a| \implies c\in [a,b]$ 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
 A: I will prove that generally, in an inner product space $(H,\langle,\rangle)$ the equality
$$\Vert a-b\Vert=\Vert a-c\Vert+\Vert c-b\Vert$$
implies $c\in[a,b]$.
Indeed, let $x=b-c$, $y=c-a$, so that the hypothesis becomes $\Vert x+y\Vert=\Vert x\Vert+\Vert y\Vert$. Squaring and simplifying we get
$$
 \langle x,y\rangle=\Vert x\Vert\cdot\Vert y\Vert
$$
So
$$
\left\Vert \Vert y\Vert x-\Vert x\Vert y
\right\Vert^2= 2\Vert x\Vert^2\, \Vert y\Vert^2 -
2\Vert x\Vert\, \Vert y\Vert\langle x,y\rangle=0
$$ 
That is $\Vert y\Vert x-\Vert x\Vert y
 =0$. Going back to $a$, $b$ and $c$ this is equivalent to
$$
c=\lambda a+(1-\lambda) b \quad\hbox{with}\quad
\lambda=\frac{\Vert b-c\Vert}{\Vert a-b\Vert}.
$$
That is $c\in[a,b]$.$\qquad\square$
A: 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 $c\in [a,b]$
