For $g(n)$ a decreasing function, we have $\liminf Y_n/g(n) \leq 1$ if and only if $Y_n < g(n) (1+\epsilon)$ infinitely often for all $\epsilon$, which happens if and only if $X_n < g(n) (1+\epsilon)$ infinitely often for all $\epsilon$. This is because if $X_n$ satisfies that inequality then clearly $Y_n$ does, while if $Y_n$ satisfies the inequality then $X_m$  $X_m/ g(m) \leq X_m/g(n) = Y_n/g(n)$ for some $m \leq n$ but each $m$ can only satisfy $X_m=Y_n$ for finitely many $n$.

By Borel-Cantelli this happens if and only if $\sum_{n=1}^\infty g(n) $ is infinite.  

In particular, if this is true for $g(n)$ then it is true for $\epsilon g(n)$ for all $\epsilon>0$, i.e. if the lim inf is ever $\leq 1$ then it is $0$.

So there is no analogue of the law of the iterated logarithm here, at least not for the lim inf - there is no one best sequence to divide the lim inf by.

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For $g(n)$ a decreasing function, we have $\limsup Y_n/g(n)\geq 1$ if and only if $Y_n> g(n)(1-\epsilon)$ infinitely often. Based on [an approach suggested by Aleksei Kulikov][1], if $n \in [2^k, 2^{k+1}]$ then this implies $X_m > g(n) (1-\epsilon) \geq g(2^{k+1} ) (1-\epsilon)$ for all $m \leq 2^k $ for infinitely many $k$. This is an event with probability $\approx e^{ - 2^k g(2^{k+1} ) (1-\epsilon)}$ and it can only occur with positive probability for infinitely many $k$ if the sum of these probabilities $\sum_{k=1}^\infty e^{ - 2^k g(2^{k+1} ) (1-\epsilon)}$ is positive (this direction of Borel-Cantelli not requiring independence). 

So $\limsup Y_n/g(n)\geq 1$ with probability $0$ for $g(n) = C \log \log n/n$ as soon as $C>2$.

I suspect for some lesser value of $C$ one can prove that $\limsup Y_n/ (C \log \log n/n) \geq 1$ with probability $1$, for example by the second moment method, i.e. the Kochen-Stone lemma.

  [1]: https://mathoverflow.net/questions/484200/asymptotics-for-minimum-of-a-sequence-of-random-variables#comment1261226_484200