$\|x -y\|+\|y-z\|=\|x-z\|$ implies $y= a x + b z$ where $a +b =1$
Hint: Take $m=x-y$ and $n= y-z$.
Does this follow from standard properties of inner product spaces (linearity, symmetry, and positive definiteness?) Or does Cauchy-Schwarz help? This is from online notes from KU.
Edit: I then have $\|m\|^2 + \|n\|^2 + 2 \|m\| \cdot \| n \| = \|m + n \|^2$. I have Cauchy Schwarz for products, which term does the inequality factor into? Do I show it both ways to show equality?
I will sove it geometrically. Suppose $y$ is the point $A$, $x$ is the point $B$ and $z$ is the point $C$. Then the given relation says that the triangle $ABC $ satisfies $|AB|+|AC|= |BC|$, but this is only possible if $A$ lies on the side $BC$. So $y$ is a convex combination of $x$ and $z$.
$\textbf{Another approach:}$ Using the hint $$\| m\|^2+\| n\|^2+2\| m\|^\cdot \| n\|=\| m+n\|^2=\| m\|^2+\| n\|^2+2\langle m,n\rangle$$ So you have $$\langle m,n\rangle =\| m\|\cdot \|n\|$$ Hence from Cauchy-Schwarz inequality $m$ and $n$ are linearly dependent. So, $$x-y=r(y-z)$$ which implies $$\frac{1}{1+r} x +\frac{r}{1+r}z=y$$