If we define $(a_n)_{n\in N_+}$ recursively by $a_1=\frac{1}{4}$ and $a_{n+1}=\frac{2\root\of{a_n}}{a_n+1}$.
I want to show that $0<a_n<1$ $\forall n\in N_+$. I have tried to use induction and show that if we prove the base case; then assume that $0<a_n<1$ holds, then $a_{n+1}$ can only ever be something between 0 and 2 on top, and something between 1 and 2 on the bottom, meaning it too is within 0 and 1. But I don't feel like it is very rigorous as in the past I have been able to find more analytical methods.
Thanks!
Clearly if $a_n \gt 0, a_{n+1} \gt 0$, so we just need to show that if $0 \lt a_n \lt 1, a_{n+1} \lt 1$. Intuitively, this can only fail if $2 \sqrt {a_n} \gt 1+a_n$. Note that $1+a_n-2\sqrt {a_n}=(1-\sqrt {a_n})^2 \gt 0$ and you are done.