Putnam and Beyond Problem 328 I would like to ask about this Real Analysis question as intuitively I had tried AM-GM and assume wlog that $a_n>b_n>c_n$ but it doesn't get me anywhere. The question is as follows:

Let $a_0,b_0,c_0$ be real numbers. Define the sequences $(a_n)_n,
 ,(b_n)_n, (c_n)_n$ recursively by $a_\text{n+1}=\frac{a_n+b_n}{2}$,
  $b_\text{n+1}=\frac{b_n+c_n}{2}$, $c_\text{n+1}=\frac{c_n+a_n}{2}$,
  $n\ge0$. Prove that the sequences are convergent and find their
  limits.

Thanks!
 A: Firstly, notice that the sum $a_i+b_i+c_i$ is invariant, so using some meta-logic we can deduce that the limit must be $$L=\frac{a_0+b_0+c_0}{3}.$$ But we still need to prove this, obviously.
Anyways, the key idea is that the maximum distance between any two of $a_i,b_i,c_i$ keeps decreasing. Your WLOG is a good start -- assuming that, we perform pairwise subtractions, and notice that the term with the largest absolute value is $$|a_{n+1}-c_{n+1}|=\frac{a_n-c_n}{2}.$$
This means that the "largest gap" between any two of $a_i,b_i,c_i$ decreases by a factor of at least two each time! This is good news, because since the sum is fixed, all three sequences must converge towards $L$. I'll leave the details to you, but do comment if you need help.
Alternatively, here's a pretty cool solution, involving solving a generalisation of the problem.
Rather than having $a_0, b_0, c_0 \in \mathbb R$, let's instead prove it works $\mathbb R^2$ (obviously this implies the $\mathbb R$ case by simply considering the $x$-coordinate). Let's also not care about which one is which, and simply consider them as a $3$-tuple of points.
Then our process replaces a triangle with a the triangle formed by the midpoint of the sides; this is a homothety from the centroid $G$ with factor $- \frac 12$, and repeating this clearly means that the three points converge to the centroid.
A: You cannot assume that $a_n>b_n>c_n$ for all $n$ because that relation is not an invariant, consider the following example:
$$
 (8, 4, 0) \to (6, 2, 4) \to (4, 3, 5) \to \ldots
$$
But – as already observed in the other answer – the sum $s = a_n + b_n + c_n$ is an invariant for the iteration. Without loss of generality we can assume that $s=0$, otherwise replace $(a_n, b_n, c_n)$ by $(a_n-s/3, b_n-s/3, c_n-s/3)$.
Then
$$
 a_{n+3} = -\frac 12 c_{n+2} = \frac 14 b_{n+1} = -\frac 18 a_n
$$
which implies that the sequence $(a_n)$ converges to zero. The same argument works for the other two sequences.
