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.
 A: 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 $n\geq 0$).
A: tl;dr: You cannot prove the upper bound you want, as the true growth of the sequence is "slightly more" than $2Kn$:
$$
r_n^2 = 2Kn + \frac{K}{2}\log n + O(1)
$$

First, for the sake of "elegance", set $a_n := \frac{r_n}{\sqrt{K}}$ for all $n\geq 0$, so that the recurrence relation becomes
$$
a_n = a_{n-1} + \frac{1}{a_{n-1}}, \qquad n\geq 1 \tag{1}
$$
Squaring both sides, we get $a_n^2 = a_{n-1}^2 + 2 + \frac{1}{a_{n-1}^2}$, or, equivalently,
$$
a_n^2 - a_{n-1}^2 = 2 + \frac{1}{a_{n-1}^2}  \tag{2}
$$
Summing (2) from $1$ to $n$, we get
$$
a_n^2 - a_0^2 = 2n + \sum_{k=0}^{n-1}\frac{1}{a_k^2} \tag{3}
$$
and in particular $a_n^2 \geq 2n + a_0^2$ for all $n\geq 0$. We also have, analogously to (3), that
$$
a_n^2 - a_1^2 = 2(n-1) + \sum_{k=1}^{n-1}\frac{1}{a_k^2}
$$
and using our lower bound $a_k^2 \geq 2k$, we get
$$
a_n^2 \leq 2n-2+a_1^2 + \frac{1}{2}\sum_{k=1}^{n-1}\frac{1}{k}
\geq 2n + \frac{1}{2} \log n +a_1^2
$$
At this point, we have shown, for all $n\geq 1$,
$$
\boxed{2n + a_0^2 \leq a_n^2 \leq 2n + \frac{1}{2} \log n +a_1^2} \tag{4}
$$
This is great, but let us not stop there. Plugging (4) back into (3), we have
$$
a_n^2 - a_0^2 \geq 2n +\frac{1}{a_0^2} + \sum_{k=1}^{n-1}\frac{1}{2n + \frac{1}{2} \log n +a_1^2}
\geq 2n + \frac{1}{2}\log n + C \tag{5}
$$
for some constant $C\in\mathbb{R}$ which depends on $a_0,a_1$ only (and therefore on $a_0$ only, since $a_0$ determines $a_1$): where we used that
$$
\sum_{k=1}^{n-1}\frac{1}{2n + \frac{1}{2} \log n + a_1^2}
= \frac{1}{2}\sum_{k=1}^{n-1} \left(\frac{1}{n}\cdot \frac{1}{1+\frac{\log n + a_1^2}{n}}\right)
= \frac{1}{2}\sum_{k=1}^{n-1} \frac{1}{n} - \sum_{k=1}^{n-1}\left(\frac{\log n}{2n^2} + o\!\left(\frac{\log n}{n^2}\right)\right)
$$
and the second series is convergent.
Overall, we have therefore established that there exist a constant $C\in\mathbb{R}$ (depending on the initial term $a_0$ only) such that, for all $n\geq 1$,
$$
\boxed{2n + \frac{1}{2}\log n + C \leq a_n^2 \leq 2n + \frac{1}{2} \log n + a_1^2} \tag{6}
$$
that is, $a_n^2 = 2n + \frac{1}{2}\log n + O(1)$.
