Convex sum order If I have a strictly convex function $f(x)$ with $f''(x)>0$ 
and if I know that for some $a\le b \le c$ and $x \le y \le z$ I have
$$a+b+c = x+y+z$$
$$f(a)+f(b)+f(c)=f(x)+f(y)+f(z)$$
can I conclude that at least one of the following must be true about the order?
$$ a \le x \le y \le b \le c \le z$$
$$ x \le a \le b \le y \le z \le c$$
Perhaps an equivalent question is to consider coincident centroids of two triangles with vertices on the convex curve, looking something like this diagram with pairs of vertices on the inside of the order

though I realise that I cannot push this analogy too far as it would not be necessarily true if say the curve was a circle and the triangles equilateral

In geometry a triangle $(ABC)$ on the plane $xOy$ with coordinates $A=(a_1,a_2)$, $B=(b_1,b_2)$, $C=(c_1,c_2)$ has its gravity center (baricenter) the point 
$$
G=\left(\frac{a_1+b_1+c_1}{3},\frac{a_2+b_2+c_2}{3}\right).
$$
Hence what you ask is: 
Given a curve (function) $c_f:y=f(x)$ or $M=(x,f(x))$, with positive curvature,  find all triangles $(ABC)$ inscribed in $c_f$ such that they have the same gravity center.
In geometry, the gravity center of (ABC) is defined as the point of intersection of all the dimedians i.e. 
In $A$ edge the dimedian is the sigment $AM_1$ with $M_1$ being the median of $BC$ side. In $B$ edge the coresponding dimedian is $BM_2$, where $M_2$ is the median of $AC$ and in $C$ edge the sigment $CM_3$.
For a given baricenter there are finite number of triangles inscribed in $c_f$ with the same baricenter $G=\left(\frac{a+b+c}{3},\frac{f(a)+f(b)+f(c)}{3}\right)$.
A theorem of Leibniz says
Theorem.
If $M$ is an arbitrary point of plane of the triangle $(ABC)$ and $G$ is the baricenter, then
$$
MA^2+MB^2+MC^2=3MG^2+\frac{1}{3}(AB^2+BC^2+CD^2)
$$
Hence your broblem reduces to find all points $A=(a,f(a))$, $B=(b,f(b))$, $C=(c,f(c))$ of curve $y=f(x)$ such
$$
GA^2+GB^2+GC^2=\frac{1}{3}(AB^2+BC^2+CA^2)
$$
or equivalently for given $f$ find all $a,b,c$ such
$$
(g_1-a)^2+(g_2-f(a))^2+(g_1-b)^2+(g_2-f(b))^2+(g_1-c)^2+(g_2-f(c))^2=\frac{1}{3}\left((a-b)^2+(f(a)-f(b))^2+(b-c)^2+(f(b)-f(c))^2+(c-a)^2+(f(c)-f(a))^2\right),\tag 1
$$
where 
$$
g_1=\frac{a+b+c}{3}\textrm{ and }g_2=\frac{f(a)+f(b)+f(c)}{3}.\tag 2
$$
Hence we have three equations with three unknowns $a,b,c$. Hence the number of points will be finite.
For example assume the parabola $y=x^2$. Then equations $(2)$ become
$$
a=\frac{1}{2}\left(-c+3g_1-\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
$$
b=\frac{1}{2}\left(-c+3g_1+\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
or
$$
a=\frac{1}{2}\left(-c+3g_1+\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
$$
b=\frac{1}{2}\left(-c+3g_1-\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$ 
The value of $c$ is given from 
$$
\frac{4 a^3}{3}-\frac{2 a^2 b^2}{3}-\frac{2 a^2 c^2}{3}-\frac{2 a b}{3}-\frac{2 a c}{3}+\frac{4 b^3}{3}-\frac{2 b^2 c^2}{3}-\frac{2 b c}{3}+\frac{4 c^3}{3}=0
$$       
