3 sequences which are convergent 
Let $a_0, b_0, c_0\in \mathbb{R}$ and $k>0$. If
  \begin{align}
a_{n+1}=\frac{1}{k+1}b_n+\frac{k}{k+1}c_n\\
b_{n+1}=\frac{1}{k+1}c_n+\frac{k}{k+1}a_n\\
c_{n+1}=\frac{1}{k+1}a_n+\frac{k}{k+1}b_n
\end{align}
  Prove that $(a_n), (b_n),(c_n)$ are convergent.

I know that they are all convergent to $\frac{a_0+b_0+c_0}{3}$ and I have a geometrical solution. 
However, I'm interested in a purely analytical one. I got that $(a_n+b_n+c_n)$ is constant and then I tried to get a relation only in terms of say $(a_n)$, but this worked only when $k=1$.
 A: This proof is inspired by the geometrical intuition that each iteration of the sequence involves a weighted average of the points of the previous iteration.
Let $M_n = \max(a_n, b_n, c_n)$ and $m_n = \min(a_n, b_n, c_n)$. Let $I_n$ be the point intermediate to these two (if two or more points coincide, simply choose one to be $I_n$).  Let $\alpha = \max(1/(k+1), k/(k+1))$. Then we have
$$ \begin{align} M_{n+1} &\leq I_n + \alpha(M_n - I_n) \leq M_n\\
m_{n+1} &\geq m_n + (1-\alpha)(I_n - m_n) \geq m_n \end{align} $$
Since $M_n$ is a decreasing sequence bounded below (by $m_0$, say), it converges to $M$. Similarly, $m_n$ converges to $m$. $I_n$ is a sequence bounded by $m_0$ and $M_0$ and so by Bolzano-Weierstrass has a convergent subsequence. Let $n_j$ correspond to this subsequence, and $I$ its limit. We know that $n_{j+1} \geq n_j+1$, so we can write the above inequalities as
$$ \begin{align} M_{n_{j+1}} &\leq I_{n_j} + \alpha(M_{n_j} - I_{n_j})\\
m_{n_{j+1}} &\geq m_{n_j} + (1-\alpha)(I_{n_j} - m_{n_j})\end{align} $$
We can now freely take the limit $j \to \infty$ to yield
$$ \begin{align} M &\leq I + \alpha(M-I) \\
m &\geq m + (1-\alpha)(I - m)\end{align}$$
Rearranging these gives $M \leq I$ and $m \geq I$. This is only possible if $M = I = m$. Hence $a_n, b_n, c_n$ converge.
