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.

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, $a_n^2 = 2n + \frac{1}{2}\log n + O(1)$.
