Given the sequences $a_{n},b_{n},c_{n}$ all subsets of the interval $(0,1)$, and all converges to 0. Also we have $a_{n}\leq b_{n}^{\alpha}$, and $a_{n}\leq c_{n}$, for all $n=1,2,3,...$, for some $\alpha \geq 1$ (real number).
Now, for a given $n$, we could have $a_{n}\leq b_{n}^{\alpha} \leq c_{n}$ or we could have $a_{n} \leq c_{n} \leq b_{n}^{\alpha}$
Suppose that we have the choice to pick any $\alpha\geq 1$. Is it possible to pick an $\alpha$ large enough so that the first case is true for all $n$ (or at least for a subsequence of $n$)?.
Edit: "for some $\alpha$" instead of for all.
I'm trying the following:
If there is $n_{1}, n_{2}, n_{3},...$ such that $b_{n_{i}}^{\alpha} > c_{n_{i}} $, for $i=1,2,3,.. $ ,then we can choose $\alpha_{1}, \alpha_{2}, \alpha_{3},..$ so that $ b_{n_{i}}^{\alpha_{i}} \leq c_{n_{i}}$ for all $i=1,2,3,.. $.
Now, let $ \alpha=\sup \alpha_{i}$ then $ b_{n_{i}}^{\alpha} \leq c_{n_{i}}$ for all $i=1,2,3,.. $. But the problem in this method is that the "sup" could be $\infty$!!