<s>[INST] An upper bound for $\{r_n\}$ which satisfies $r_n = r_{n-1} + \frac{K}{r_{n-1}}$ I have a sequence $\{r_n\}$ of positive numbers defined recursively by
$$\tag{1} r_n = r_{n-1} + \frac{K}{r_{n-1}}.$$
for some positive $K$. The sequence is clearly strictly increasing and tends to $+\infty$ as $n\to \infty$. Also, by induction one can prove
$$\tag{2} r_n^2 \ge 2(n-1)K +r_1^2$$
for all $n$.

Question: Does there exists positive $K_2$ depending on $r_1, K$ such that
$$\tag{3} r_n^2 \le 2nK + K_2$$
for all large $n$?

Squaring (1) on both sides give
$$ r_n^2 = r_{n-1}^2 + 2K + \frac{K^2}{r_{n-1}^2},$$
If I try to prove (3) using induction, (2) gives
$$ \frac{K^2}{r_{n-1}^2} \le \frac{K^2}{ 2(n-2)K +r_1^2},$$
which is not very helpful, since $\sum \frac{1}{n-2}$ is not summable.
 [/INST] First, reproving the lower bound: we have, for all $n\geq 1$,
$$
r_{n-1}^2 +K = r_n r_{n-1} \leq \frac{r_n^2+ r_{n-1}^2}{2} \tag{1}
$$
using the AM-GM inequality; from which
$$
r_n^2 \geq r_{n-1}^2 + 2K \tag{2}
$$
leading to $r_{n}^2 \geq 2K n + r_0^2$ for all $n\geq 0$.
Then, the upper bound: We know, using the lower bound, that there exists $N\geq 1$ such that, for all $n\geq N$,
$$
r_n^2 \geq 2Kn
$$
so we get now use this to get
$$
r_n = r_{n-1}+\frac{K}{r_{n-1}} \leq r_{n-1}+\frac{\sqrt{K}}{\sqrt{2(n-1)}} \tag{3}
$$
so that
$$
r_n - r_1 = \sum_{k=2}^n (r_k - r_{k-1}) \leq \sum_{k=2}^n \sqrt{\frac{K}{2(k-1)}}
= \sqrt{\frac{K}{2}}\sum_{k=1}^{n-1} \frac{1}{\sqrt{k}}
\leq \sqrt{\frac{K}{2}}\int_{0}^{n-1} \frac{dx}{\sqrt{x}} = \sqrt{2K(n-1)}
$$
and so $(r_n-r_1)^2 \leq 2Kn$. Now, since $r_n \to \infty$, for every $\varepsilon>0$ there exists some $n_\varepsilon\geq 1$ such that $(r_n-r_1)^2 \geq (1+\varepsilon/2)^{-1}r_n^2$ for all $n\geq n_\varepsilon$.
Conclusion. For all $\varepsilon>0$, there exists $n_\varepsilon$ such that, for $n\geq n_\varepsilon$,
$$
\boxed{ 2Kn + r_0^2 \leq r_n^2 \leq (2+\varepsilon)Kn + r_1^2 }
$$
(the lower bound actually holds for all $n\geq 0$).
</s>