If $\displaystyle b_{n+1} = b_{n}+\frac{1}{b_{n}}$ and $b_{1} = 1,$ then find the value of $\lfloor b_{100}\rfloor$.
My attempt: $\displaystyle b_{n+1}b_{n} = b^2_{n}+1\Rightarrow b_{n}b_{n+1}-b^2_{n} = 1$
$$\frac{1}{b_{n}b_{n+1}-b^2_{n}} = 1\Rightarrow \frac{1}{b_{n}(b_{n+1}-b_{n})} = 1.$$
I am not able to go further, could some help me, thanks.
This answer uses that $$\sqrt{2n}\le b_n\le\sqrt{2n}+\frac{1}{\sqrt[4]{2n}}\tag1$$ for $n\ge 2$.
The proof for $(1)$ is written at the end of this answer.
From $(1)$, we get that $$14=\sqrt{196}\le\sqrt{200}\le b_{100}\le \sqrt{200}+\frac{1}{\sqrt[4]{200}}\lt\sqrt{201.64}+\frac{1}{\sqrt[4]{81}}=14.2+\frac{1}{3}\lt 15$$ from which we have $$\lfloor b_{100}\rfloor=\color{red}{14}$$
It is easy to see that $(1)$ holds for $n=2,3,4$.
Let us prove $(1)$ for $n\ge 4$ by induction.
Suppose that $(1)$ holds for some $n\ (\ge 4)$.
Let $f(x)=x+\frac 1x$ which is increasing for $x\gt 1$.
All we need to prove is that $$\sqrt{2(n+1)}\le f(\sqrt{2n})\tag2$$ and that $$f\left(\sqrt{2n}+\frac{1}{\sqrt[4]{2n}}\right)\le \sqrt{2(n+1)}+\frac{1}{\sqrt[4]{2(n+1)}}\tag3$$
For $(2)$ :
$$\begin{align}(2)&\iff \sqrt{2n}+\frac{1}{\sqrt{2n}}-\sqrt{2n+2}\ge 0\\\\&\iff 2n+1\ge 2\sqrt{n(n+1)}\\\\&\iff (2n+1)^2\ge 4n(n+1)\end{align}$$ which is true for $n\ge 1$.
For $(3)$ :
$$(3)\iff \sqrt{2(n+1)}+\frac{1}{\sqrt[4]{2(n+1)}}-\sqrt{2n}-\frac{1}{\sqrt[4]{2n}}-\frac{1}{\sqrt{2n}+\frac{1}{\sqrt[4]{2n}}}\ge 0\tag4$$ Using $$\sqrt{2n+2}-\sqrt{2n}=\frac{2}{\sqrt{2n+2}+\sqrt{2n}}$$ $$-\frac{1}{\sqrt{2n}+\frac{1}{\sqrt[4]{2n}}}=-\frac{1}{\sqrt{2n}}+\frac{1}{2n\sqrt[4]{2n}+\sqrt{2n}}$$ $$\frac{2}{\sqrt{2n}+\sqrt{2n+2}}-\frac{1}{\sqrt{2n}}=-\frac{2}{\sqrt{2n}\ (\sqrt{2n}+\sqrt{2n+2})^2}$$ $$\begin{align}\frac{1}{\sqrt[4]{2n}}-\frac{1}{\sqrt[4]{2n+2}}&=\frac{2}{\sqrt[4]{2n}\sqrt[4]{2n+2}\ (\sqrt[4]{2n}+\sqrt[4]{2n+2})(\sqrt{2n}+\sqrt{2n+2})}\\\\&\le \frac{1}{\sqrt{2n}\sqrt[4]{2n}\ (\sqrt{2n}+\sqrt{2n+2})}\end{align}$$ we see that, in order to prove $(4)$, it is sufficient to prove that $$\frac{1}{2n\sqrt[4]{2n}+\sqrt{2n}}-\frac{1}{\sqrt{2n}\sqrt[4]{2n}\ (\sqrt{2n}+\sqrt{2n+2})}\ge \frac{2}{\sqrt{2n}\ (\sqrt{2n}+\sqrt{2n+2})^2}\tag5$$
Here, $$\begin{align}&\frac{1}{2n\sqrt[4]{2n}+\sqrt{2n}}-\frac{1}{\sqrt{2n}\sqrt[4]{2n}\ (\sqrt{2n}+\sqrt{2n+2})}\\\\&=\frac{\sqrt{2n}\sqrt[4]{2n}\ (\sqrt{2n}+\sqrt{2n+2})-2n\sqrt[4]{2n}-\sqrt{2n}}{(2n\sqrt[4]{2n}+\sqrt{2n})\sqrt{2n}\sqrt[4]{2n}\ (\sqrt{2n}+\sqrt{2n+2})}\\\\&=\frac{\sqrt{2n}\sqrt[4]{2n}\sqrt{2n+2}-\sqrt{2n}}{(2n\sqrt[4]{2n}+\sqrt{2n})\sqrt{2n}\sqrt[4]{2n}\ (\sqrt{2n}+\sqrt{2n+2})}\\\\&\ge \frac{(\sqrt[4]{2n}\sqrt{2n}-1)\sqrt{2n}}{(2n\sqrt[4]{2n}+\sqrt{2n})\sqrt{2n}\sqrt[4]{2n}\ (\sqrt{2n}+\sqrt{2n+2})}\\\\&=\frac{\sqrt[4]{2n}\sqrt{2n}-1}{(2n+\sqrt[4]{2n})\sqrt{2n}\ (\sqrt{2n}+\sqrt{2n+2})}\\\\&\ge\frac{2(\sqrt[4]{2n}\sqrt{2n}-1)}{(2n+\sqrt[4]{2n})(\sqrt{2n}+\sqrt{2n+2})^2}\end{align}$$
So, in order to prove $(5)$, it is sufficient to prove that $$\frac{2(\sqrt[4]{2n}\sqrt{2n}-1)}{(2n+\sqrt[4]{2n})(\sqrt{2n}+\sqrt{2n+2})^2}-\frac{2}{\sqrt{2n}\ (\sqrt{2n}+\sqrt{2n+2})^2}\ge 0,$$ i.e. $$\frac{2y(y^2+1)(y^2-y-1)}{(2n+\sqrt[4]{2n})\sqrt{2n}\ (\sqrt{2n}+\sqrt{2n+2})^2}\ge 0$$ where $y=\sqrt[4]{2n}$.
This holds for $n\ge 4$. $\quad\blacksquare$
Note that $b_n$ is strictly increasing and for $n\geq 2$, $b_n^2\ge 2n$, because $b_2=2$ and $$b_{n+1}^2=\left(b_n+\frac{1}{b_n}\right)^2=b_n^2+2+\frac{1}{b_n^2}\geq 2n +2=2(n+1).$$ Moreover, for $n\geq 2$, $$\begin{align}b_n^2&=b_2^2+\sum_{k=2}^{n-1}(b_{k+1}^2-b_k^2)=4+\sum_{k=2}^{n-1}\left(2+\frac{1}{b_k^2}\right)\le 2n+\frac{1}{2}\sum_{k=2}^{n-1}\frac1{k}\\&\leq 2n+\frac{1}{2}\int_1^{n-1}\frac{dx}{x}< 2n+\frac{\ln(n-1)}{2}.\end{align}$$ Hence $$14^2=196<200\leq b_{100}^2< 200+\frac{\ln(99)}{2}< 203<225=15^2$$ which implies that $\lfloor b_{100}\rfloor={14}$.
