I am having lots of trouble answering this question:
Define a sequence recursively as follows. $x_1 = 1$ and for $n ∈ N, x_{n+1} = \sqrt{(x_n)^2 + 1/(x_n)^2}$
Prove using mathematical induction that for all n ∈ N, $ 1 ≤ x_n ≤\sqrt{n}$
In order to solve this question, I feel you need to find a closed formula for this recursive sequence, but I cannot find that formula.
Base Case, clear.
Induction Hypothesis: Assume that $1 \leq x_n \leq \sqrt{n}$ for all $n \leq m$.
Induction step: Consider $x_m$. We note that $$x_m = \sqrt{(x_{m-1})^2 + \frac{1}{(x_{m-1})^2}}\leq \sqrt{(x_{m-1})^2 + 1} \leq \sqrt{(\sqrt{m-1})^2 +1} = \sqrt{m }$$
The "$ 1 \leq$" portion is easier and is left to you to figure out. Notice that we did not need to find a closed form formula.