How can we show that $ (a_n), (b_n), (c_n) $ are convergent and have the same limit? We have the following for $a \le b \le c >0$:
$A(a,b,c)=\frac{a+b+c}{3}, B(a,b,c)= (abc)^{1/3}, C(a,b,c)=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}  $.  
Then we define the sequences $(a_n),(b_n), (c_n)$ by
$a_1=a, b_1=b, c_1=c,$
$a_{n+1}=A(a_n,b_n,c_n), b_{n+1}=B(a_n,b_n,c_n), c_{n+1}=C(a_n,b_n,c_n)$.
How can we show that $(a_n),(b_n), (c_n)$ are convergent and have the same limit?
I understand that $A(a,b,c)\ge B(a,b,c)$ and $B(a,b,c)\ge C(a,b,c)$
 A: We have
$$
A(x,y,z) = \mbox{ arithmetic mean of } x,y,z;
$$
$$
B(x,y,z) = \mbox{ geometric mean of } x,y,z;
$$
$$
C(x,y,z) = \mbox{ harmonic mean of } x,y,z.
$$
It is well known that, for the same arguments,
$$
\mbox{ harmonic mean } \le
\mbox{ geometric mean } \le
\mbox{ arithmetic mean } 
\tag{1}
$$
(see e.g. this Wikipedia article).
Using these inequalities we see that the largest (arithmetic) means $a_n$ form a non-increasing sequence, while the smallest (harmonic) means $c_n$
form a non-decreasing sequence. Both sequences are bounded: all terms are within the interval $[a,c]$. If a sequence is monotonic and bounded, it has a limit.
Can you finish by proving that the limits of $a_n$ and $c_n$ are the same?
(Then $b_n$ necessarily has the same limit too because of the double inequality $(1)$ and the squeeze theorem.)
For $n>1$ we have
$$
a_{n+1}={a_n+b_n+c_n\over3}\le{a_n+a_n+c_n\over3}, \tag{2}
$$
$$
c_{n+1} \ge c_n. \tag{3}
$$
Subtracting $(3)$ from $(2)$ we find
$$
a_{n+1}-c_{n+1} \le {2\over3} (a_n-c_n).
$$
We observe that the intervals $[c_n,a_n]$ form a 
sequence of nested closed intervals, and the interval lengths tend to zero (no slower than a decreasing geometric progression).
Therefore these intervals have a (unique) common point, and this point must be the limit of all three sequences because $a_n,b_n,c_n \in [c_n,a_n]$ for all
$n>1$. (The common point is unique because the size of intervals $[c_n,a_n]$ tends to zero.) This completes the proof.
A: hint: try proving $c_n$ is increasing and $a_n$ is decreasing and use squeeze theorem.
A: Assuming $a>0.$  For any positive $a',b',c'$  we have $$\max (a',b',c')\geq A(a',b',c')\geq B(a',b','c)\geq C(a',b',c')\geq \min (a',b',c').$$  Let $U_n=\max (a_n,b_n,c_n)$ and $L_n=\min (a_n,b_n,c_n).$ $$\text {We have }\quad  U_n\geq U_{n+1}\geq L_{n+1}\geq L_n.$$  So $(U_n)_n$ is a deceasing sequence bounded below by $L_1$ and $(L_n)_n$ is an increasing sequence bounded above by $U_1 .$
Let $U=\lim_{n\to \infty}U_n$ and $L=\lim_{n\to \infty}L_n.$ Obviously $U\geq L\geq L_1>0.$ 
It suffices to show that $U=L.$ 
Let $r_n=U_n-L_n .$ Note that $U_n>r_n.$ $$\text {We have }\quad U_{n+1}\leq \frac {2U_n+L_n}{3}$$
 $$\text {and }\quad L_{n+1}\geq \frac {3}{\frac {1}{U_n}+\frac {2}{L_n}}=\frac {3U_nL_n}{2U_n+L_n}.$$ $$\text {Therefore }\quad  r_{n+1}\leq \frac  {2U_n+L_n}{3}-\frac {3U_nL_n}{2U_n+L_n}=$$ $$=\frac {(4U_n-L_n)(U_n-L_n)}{3(2U_n+L_n)}=$$ $$=\frac {(3U_n+r_n)r_n}{9U_n-3r_n}\leq$$ $$\leq  \frac {(4U_n)r_n}{6U_n}=\frac {2}{3}r_n.$$  
So $U-L=\lim_{n\to \infty}r_n=0$ because $0\leq r_{n+1}\leq \frac {2}{3}r_n.$   
