Finding value of $\lfloor b_{100} \rfloor $ in recursive relation 
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.

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 $\lfloor b_{100}\rfloor={14}$.
