This is a messy problem, but maybe could be interesting for someone who has the inclination. It came out analyzing a subroutine of an optimization algorithm. I'll post it here as a question, even though I'm currently working on it, so, maybe, I'll be the one posting the answer when the right idea pops up.
Let $c_1,c_2,c_3,c_4,c_5>0$ and $n_1,n_2,n_3,n_4,n_5 \in \{0,1,2,\dots\}$.
Define $n_1(0):=n_1, n_2(0):=n_2, n_3(0):=n_3, n_4(0):=n_4,n_5(0):=n_5.$
In the following we will adopt the convention that $c/0 = +\infty$ whenever $c>0$.
For each $i \in \{1,2,3\}$ and each $n\in \{0,1,2,\dots\}$ we define $f_i(n):=c_i/\sqrt{n}$.
For each $i \in \{4,5\}$ and each $n \in \{0,1,2,\dots\}$ we define $$f_i(n) :=\frac{c_4}{\sqrt{2^{\lfloor\log_2(n+1) \rfloor}-1}}+\frac{c_5}{2^{\lfloor\log_2(n+1) \rfloor}},$$ where $\lfloor\cdot\rfloor$ is the floor function, so that $2^{\lfloor\log_2(n+1) \rfloor}$ is equal to $2^m$ where $m$ is the largest integer such that $m \le \log_2(n+1)$, from which it follows that $\frac{n+1}{2}\le 2^{\lfloor\log_2(n+1)\rfloor} \le n+1$.
The algorithm works as follows.
At each time $t \in \mathbb{N}$ it selects $$i^\star\in\operatorname{argmax}_{i \in \{1,2,3,4,5\}}f_i\big(n_i(t-1)\big)$$
and update $n_i(t):=n_{i}(t-1)$ for $i \neq i^\star$, and $n_{i^\star}(t) = n_{i^\star}(t-1)+1$.
The goal is to prove that there exist two constants $c>0$ and $t_0\in \mathbb{N}$ such that, regardless of the initial choice of $n_1,n_2,n_3,n_4,n_5$, we have that, for all $t \ge t_0 +1$, $$\max_{i\in\{1,2,3,4,5\}}f_i\big(n_i(t)\big) \le \frac{c}{\sqrt{t-t_0}} \;.$$
Does anyone have any hint on how to attack it?