$3$ convergent sequences with the same limit Again, a problem from a 2013 Romanian GM.

Let $(x_n)_{n\geq0}, (y_n)_{n\geq0}, (z_n)_{n\geq0}$ three positive real numbers sequences. For every natural $n$ we have $$x_{n+1}\leq\frac{x_n+y_n+z_n}{3}$$
  $$y_{n+1}\leq\frac{x_ny_n+y_nz_n+z_nx_n}{x_n+y_n+z_n}$$
  $$z_{n+1}\leq\frac{3x_ny_nz_n}{x_ny_n+y_nz_n+z_nx_n}$$
  Prove that all three sequences are convergent and have the same limit.

I din't managed to show much in this problem. It is easy to see that $$\frac{x_n+y_n+z_n}{3}\geq\frac{x_ny_n+y_nz_n+z_nx_n}{x_n+y_n+z_n}\geq\frac{3x_ny_nz_n}{x_ny_n+y_nz_n+z_nx_n}$$  and using this we obtain that $$(x_n+y_n+z_n)_{n\geq0}, (x_ny_n+y_nz_n+z_nx_n)_{n\geq0}, (x_ny_nz_n)_{n\geq0}$$ are all decreasing, so convergent.
How should I solve this problem?
 A: As already noted in the OP, each of the sequences $(u_n=x_n+y_n+z_n),(v_n=x_ny_n+x_nz_n+y_nz_n)$ and $(w_n=x_ny_nz_n)$ is decreasing and positive, and so convergent ; denote the limits by $u,v$ and $w$ respectively. 
Let $a_n$ be the largest value in $\lbrace x_n,y_n,z_n \rbrace$, $b_n$ the second largest and $c_n$ the lowest. Then
$$
\begin{array}{l}
u_n=a_n+b_n+c_n=x_n+y_n+z_n, \\
v_n=a_nb_n+a_nc_n+b_nc_n=x_ny_n+x_nz_n+y_nz_n, \\
w_n=a_nb_nc_n=x_ny_nz_n \\
\end{array}\tag{1}
$$
and we have for $n\geq 2$,
$$
\begin{array}{l}
x_{n} \leq \frac{u_{n-1}}{3}=\frac{a_{n-1}+b_{n-1}+c_{n-1}}{3} \leq a_{n-1} \\
y_{n} \leq \frac{v_{n-1}}{u_{n-1}}=\frac{a_{n-1}b_{n-1}+a_{n-1}c_{n-1}+b_{n-1}c_{n-1}}{a_{n-1}+b_{n-1}+c_{n-1}} \leq a_{n-1} \\
z_{n} \leq \frac{3w_{n-1}}{v_{n-1}}=\frac{3a_{n-1}b_{n-1}c_{n-1}}{a_{n-1}b_{n-1}+a_{n-1}c_{n-1}+b_{n-1}c_{n-1}} \leq a_{n-1} \\
\end{array}\tag{2}
$$
It follows that $a_n \leq a_{n-1}$, so $(a_n)$ is nonincreasing and nonnegative, so it converges. If we denote by $a\geq b \geq c$ the roots of $P=T^3-uT^2+vT-w$ in decreasing order, the limit of $a_n$ can only be $a$. Next, notice that $b_n+c_n=u_n-a_n$ and
$b_nc_n=\frac{w_n}{a_n}$, so that $b_n=\frac{u_n-a_n+\sqrt{(u_n-a_n)^2-4\frac{w_n}{a_n}}}{2}$ converges to
$\frac{u-a+\sqrt{(u-a)^2-4\frac{w}{a}}}{2}=b$, and similarly $(c_n)$ converges to $c$.
From (2) we deduce that $x_n$, $y_n$, and $z_n$ are all $\leq \frac{a_{n-1}+b_{n-1}+c_{n-1}}{3}$, so $a_n \leq \frac{a_{n-1}+b_{n-1}+c_{n-1}}{3} \leq a_{n-1}$. Passing to the limit, we see that $a\leq \frac{a+b+c}{3} \leq a$, whence $a=b=c$.
Finally, since $c_n \leq x_n \leq a_n$, we have that $(x_n)$ converges to the same limit. Same thing for $y_n,z_n$.
