TITLE: Prove that AM-GM-HM recursive function converge to the same value.
QUESTION [6 upvotes]: I am trying to do this question Let $0<a_1 < b_1 < c_1$ $$ a_n = HM(a_{n-1}, b_{n-1}, c_{n-1})$$$$  b_n = GM(a_{n-1}, b_{n-1}, c_{n-1})$$$$  c_n = AM(a_{n-1}, b_{n-1}, c_{n-1})$$
Prove that ${a_n}, {b_n}, {c_n}$ converges and converge to the same value.

From HM-GM-AM inequality, I managed to show that $a_n< b_n< c_n$ and that $a_{n}< a_{n+1} < b_{n+1} <c_{n+1} < c_n$. This implies that $a_n$ and $c_n$ converges due to being bounded and monotonic. However, I am unable to show that they converge to the same limit or that $b_n$ converges. Any help would be appreciated.

REPLY [4 votes]: From $a_n<a_{n+1}<c_{n+1}<c_n$ and $b_n<c_n$ for all $n$ we can deduce
$$\begin{align*}c_{n+1}-a_{n+1} &< c_{n+1}-a_n\\
&= \frac{1}{3}a_n+\frac{1}{3}b_n+\frac{1}{3}c_n-a_n \\
&< \frac{1}{3}a_n+\frac{1}{3}c_n+\frac{1}{3}c_n-a_n \\
&= \frac{2}{3}c_n-\frac{2}{3}a_n \\
&= \frac{2}{3}(c_n-a_n)
\end{align*}$$
hence $c_n - a_n \to 0$ and the limits of $(a_n)$, $(c_n)$ are the same, which must then also be the limit of $(b_n)$.