<s>[INST] Prove a relationship involving floor functions I am trying to prove that a particular expression is a lower bound for a very unusually-behaved function.  The whole proof will be complete if I can just nail down the details of one technical lemma involving the floor function.  I am completely confident that the lemma is true, but have had a devil of a time proving it.
The Lemma is as follows:

Let $n$ and $b$ be natural numbers, with $1 \leq b \leq n$.  Suppose that there exists an integer $k$ such that
  $$\frac{n + \sqrt{n^2 + b + 2}}{b+2} < k < \frac{n+\sqrt{n^2+b+1}}{b+1}$$
  Then there exists an integer lying strictly between
  $$ \left( \left\lfloor \frac{n+\sqrt{n^2+b+2}}{b+2} \right\rfloor +1 \right)\sqrt{n^2+b+1}$$
  and
  $$ \left( \left\lfloor \frac{n+\sqrt{n^2+b+2}}{b+2} \right\rfloor +1 \right)\sqrt{n^2+b+2}$$
  (where $\lfloor x \rfloor$ denotes the floor function).

The hypothesis can easily be restated in the form:

Suppose $ \left\lfloor \frac{n+\sqrt{n^2+b+2}}{b+2} \right\rfloor < \left\lfloor \frac{n+\sqrt{n^2+b+1}}{b+1} \right\rfloor$

and it's clear that the two inequalities are closely related, but I have not been able to get from one to the other, and  I would appreciate any assistance. If there's interest, I can provide more background to the problem.
Edited to add: In fact, I am now almost certain that the Lemma above is actually if-and-only-if, and would also be happy to have help with the proof of the converse.
 [/INST] Fix $n,b$ with $1\le b\le n$.
Let $k\in\mathbb Z$. For $i\in\{1,2\}$ define
$$\tag1\delta_i:=k- \frac{n+\sqrt{n^2+b+i}}{b+i}.$$
Then $k$ is strictly between the bounds iff $\delta_1<0<\delta_2$. 
Then for $i\in\{1,2\}$
$$\begin{align}k\sqrt{n^2+b+i}&=\left( \frac{n+\sqrt{n^2+b+i}}{b+i}+\delta_i\right)\sqrt{n^2+b+i}\\
&=\frac{n^2+b+i+n\sqrt{n^2+b+i}}{b+i}+\delta_i\sqrt{n^2+b+i}\\
&=1+n\cdot \frac{n+\sqrt{n^2+b+i}}{b+i}+\delta_i\sqrt{n^2+b+i}\\
&\tag2=1+nk+\delta_i(\sqrt{n^2+b+i}-n).\end{align}$$
Note that
$$n<\sqrt{n^2+b+i}\le\sqrt{n^2+n+2}\le\sqrt{n^2+2n+1}=n+1  $$
so that $$\tag30<\sqrt{n^2+b+i}-n\le 1$$
with equality on the right if and only if $n=b=1$ and $i=2$. In other words, we have
$$\tag{4}0<\sqrt{n^2+b+i}-n< 1\rlap{\qquad\text{if $n>1$}.}$$
Lemma. Let $k\in\mathbb Z$. Then $k$ is strictly between $\frac{n+\sqrt{n^2+b+2}}{b+2}$ and $\frac{n+\sqrt{n^2+b+1}}{b+1}$ if and only if $1+nk$ is strictly between $k\sqrt{n^2+b+1}$ and $k\sqrt{n^2+b+2}$.
Proof.
Because of $(3)$ and $(2)$, the conclusion holds if and only if $\delta_1<0<\delta_2$. But that is equivalent to the premise. $_\square$
Proposition. Let $k=\left\lfloor\frac{n+\sqrt{n^2+b+2}}{b+2}+1\right\rfloor$. There exists an integer strictly between $\frac{n+\sqrt{n^2+b+2}}{b+2}$ and $\frac{n+\sqrt{n^2+b+1}}{b+1}$ 
if and only if there exists an integer strictly between $k\sqrt{n^2+b+1}$ and $k\sqrt{n^2+b+2}$.
Proof.
Assume  there exists an integer strictly between $\frac{n+\sqrt{n^2+b+2}}{b+2}$ and $\frac{n+\sqrt{n^2+b+1}}{b+1}$.
Then $k$ is certainly such an integer because it is the smallest integer strictly larger than $\frac{n+\sqrt{n^2+b+2}}{b+2}$.
Then by the lemma, $1+nk$ is  strictly between $k\sqrt{n^2+b+1}$ and $k\sqrt{n^2+b+2}$.
Conversely assume that $m$ is an integer strictly between $k\sqrt{n^2+b+1}$ and $k\sqrt{n^2+b+2}$. 
For $n=b=1$ this means that $2\sqrt 3<m<4 $, which is impossible. Therefore $n>1$. Define $\delta_{1,2}$ per $(1)$. 
Then by definition of $k$, we have $0<\delta_2\le 1$.
Then from $(2)$ and $(4)$
$$m<k\sqrt{n^2+b+2}=1+nk+\underbrace{\delta_2(\sqrt{n^2+b+2}-n)}_{\in(0,1)}$$
so that $m\le 1+nk$. Using $(2)$ again we get
$$ 1+nk\ge m>k\sqrt{n^2+b+1}=1+nk+\delta_1(\sqrt{n^2+b+1}- n)$$
so that $\delta_1(\sqrt{n^2+b+1}- n)<0$ and by $(3)$ ultimately $\delta_1<0$. But $\delta_1<0<\delta_2$ is equivalent to $k$ being strictly between 
 $\frac{n+\sqrt{n^2+b+2}}{b+2}$ and $\frac{n+\sqrt{n^2+b+1}}{b+1}$.$_\square$
</s>