<s>[INST] 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
 [/INST] For simplicity, let us assume that all the points are different, i.e., $a < b < c$, $x < y < z$ and the all the points $a,b,c$ are different from $x,y,z$.
(Maybe some arguments can be transferred to the general case).
Now, the order of the points can be described by a permutation of $abcxyz$ which preserves the order of $abc$ and $xyz$.
By symmetry, we can assume w.l.o.g. that the permutation starts with $a$.
Let us consider permutations starting with $ab$.


*

*$abc$ can only be completed to $abcxyz$ and this cannot satisfy $a + b + c = x + y + z$.

*$abx$ can be completed to

*
*

*$abxycz$: this cannot satisfy $a + b + c = x + y + z$.


*
*

*$abxcyz$: this cannot satisfy $a + b + c = x + y + z$.


*
*

*$abxyzc$: by your triangly analogy, the entire triangle $xyz$ lies below the segment $bc$, hence the centroids of $abc$ and $xyz$ cannot coincide.



These arguments show that the permutation has to start with $ax$. By similar arguments we can conclude that the permutation has to end with $zc$ or $cz$.
Hence, four possibilities remain:


*

*$ax by cz$

*$ax yb cz$

*$ax by zc$

*$ax yb zc$


The first case cannot satisfy $a + b + c = x + y + z$. The second case is what you would like to have. Therefore, it remains to rule out the last two cases. By symmetry, it is enough to rule out $axbyzc$.
The strict convexity gives
\begin{align}
 f(x) &< f(a) \frac{b-x}{b-a} + f(b) \frac{x-a}{b-a}, \\
 f(y) &< f(b) \frac{c-y}{c-b} + f(c) \frac{y-b}{c-b}, \\
 f(z) &< f(b) \frac{c-z}{c-b} + f(c) \frac{z-b}{c-b}.
\end{align}
Adding these inequalities and using $f(a) + f(b) + f(c) = f(x) + f(y) + f(z)$ gives
\begin{equation*}
 f(a) + f(b) + f(c)
 <
 f(a) \frac{b-x}{b-a} + f(b) \Big[ \frac{x-a}{b-a} + \frac{2c-z-y}{c-b}\Big] + f(c) \frac{z+y-2b}{c-b}.
\end{equation*}
Rearranging terms and using $a + b + c = x + y + z$
gives
\begin{equation*}
 0
 <
 (x-a)
 \Bigg[
  \frac{f(b) - f(a)}{b-a}
  -
  \frac{f(c) - f(b)}{c-b}
 \Bigg].
\end{equation*}
This, however, is a contradiction to the strict convexity which yields
\begin{equation*}
 \frac{f(b) - f(a)}{b-a}
 <
 \frac{f(c) - f(b)}{c-b}.
\end{equation*}
This rules out the case $axbyzc$.
Hence, if $a$ is the smallest point, the permutation has to be $axybcz$, i.e.,
$a < x < y < b < c < z$.
</s>