$ x_{n + 1} > x_n$. $ \rightleftarrows x_n^2 + n^2 > nx_n^2$. $ \rightleftarrows x_n^2 < \frac {n^2}{n - 1}$. Using induction proof: $ \sqrt {n} \leq x_n < {\sqrt {n + 1}}$.

Thjch Ph4 Trjnh wrote: $ x_{n + 1} > x_n$. $ \rightleftarrows x_n^2 + n^2 > nx_n^2$. $ \rightleftarrows x_n^2 < \frac {n^2}{n - 1}$. Using induction proof: $ \sqrt {n} \leq x_n < {\sqrt {n + 1}}$. Shorter !: Using simple proof, $ x_n$ is increasing and $ [x_n^2]=n$ More seriously, could you develop your proof for us, please ?

I'm sorry to insist, but could you kindly give us your solution, Thjch Ph4 Trjnh, please ? I dont succeed in showing your so simple induction. Thanks.

Assume holds for $ n$:$ \sqrt {n}\leq x_n\leq \sqrt {n + 1}$. Function $ f(x) = \frac {x}{n} + \frac {n}{x}$ increase on $ [\sqrt {n}, + )$. Thus $ f(\sqrt {n})\leq x_{n + 1} = f(x_n) < f(\sqrt {n + 1})$. $ f(\sqrt {n}) = \sqrt {n} + \frac {1}{\sqrt {n}}\geq \sqrt {n + 1}$, $ f(\sqrt {n + 1}) < \sqrt {n + 2}$.

Thjch Ph4 Trjnh wrote: Assume holds for $ n$:$ \sqrt {n}\leq x_n\leq \sqrt {n + 1}$. Function $ f(x) = \frac {x}{n} + \frac {n}{x}$ increase on $ [\sqrt {n}, + )$. Thus $ f(\sqrt {n})\leq x_{n + 1} = f(x_n) < f(\sqrt {n + 1})$. $ f(\sqrt {n}) = \sqrt {n} + \frac {1}{\sqrt {n}}\geq \sqrt {n + 1}$, $ f(\sqrt {n + 1}) < \sqrt {n + 2}$. Ok, thanks (I really did not see this ). Notice that $ f(\sqrt {n + 1}) < \sqrt {n + 2}$ is not true for the lowest values of $ n$ so the start of induction needs to be taken about $ 3$.