TITLE: Show that $(x_n)_n $ is bounded
QUESTION [5 upvotes]: Let $(x_n)_n, (y_n)_n $ two sequences s.t. $x_1,y_1>0$; $x_1 <y_1$
$x_{n+1}=x_n (1+\frac {1}{y_n}) $ 
$y_{n+1}=y_n (1+\frac {1}{x_n}) $ 
Show that $(x_n)_n $ is bounded.
I noticed that $x_n <y_n $ for any $n\in \mathbb {N} $.
Also, $(y_n)_n $ has limit $\infty $.

REPLY [2 votes]: Hints.
Step 1. Show, inductively, that $0<x_n<y_n$, for all $n\in\mathbb N$.
Step 2. Observe that,
$$
x_{n+1}=x_n\left(1+\frac{1}{y_n}\right)<x_n\left(1+\frac{1}{x_n}\right)=x_n+1,
$$
and hence $x_n<x_1+n-1$.
Step 3. Shoe that $y_n\to \infty$. This is done as follows:
$$
y_{n+1}=y_n\left(1+\frac{1}{x_n}\right)>y_n\left(1+\frac{1}{x_1+n-1}\right)=y_1\cdot\frac{x_1+n}{x_1+n-1}
$$
and hence
$$
y_n>y_1\prod_{k=1}^{n-1}\frac{x_1+k}{x_1+k-1}=y_1\frac{x_1+n-1}{x_1}=a+bn,
$$
where $b=y_1/x_1>1$.
Step 4. Observe that
$$
\log x_{n+1}=\log x_n+\log\left(1+\frac{1}{y_n}\right)<\log x_n+\frac{1}{y_n},
$$
and hence if $\sum\frac{1}{y_n}$, converges, then so does $\log x_n$ and also $x_n$.
Step 5. 
$$
x_{n+1}=x_n\left(1+\frac{1}{y_n}\right)<x_n\left(1+\frac{1}{a+bn}\right)<x_1\prod_{k=1}^n\left(1+\frac{1}{a+bk}\right)\le c(n+d)^{1/b}
$$
Step 6. Repetition of Step 3, with the estimate of Step 5 provides that 
$$
y_n>c'(n+d')^{1+1/b},
$$
and hence $\sum\frac{1}{y_n}<\infty.$