For which $a$ is $n\lfloor a\rfloor+1\le \lfloor na\rfloor$ true for all sufficiently large $n$? Inspired by this question I ask this. For which $a$ is $n\lfloor a\rfloor+1\le \lfloor na\rfloor$ true for all sufficiently large $n$?
The original question concerned $a=e$, the usual $2.71828\ldots$, where the inequality holds by a power series argument.  In the discussion Ross Millikan pointed out that if $a$ is natural, the inequality fails, and he also mentioned that if $a$ is just above a rational, the inequality fails.  My followup calculation is below.
Suppose $a$ is irrational.  Let $a=\frac{m}{n}+\delta$, where $0<\delta<\frac{1}{n}$. Write $m=nq+r$, where $0\le r<n$.  Then the LHS is $n\lfloor a \rfloor+1=n\lfloor \frac{m}{n}+\delta \rfloor+1 = n\lfloor \frac{nq+r}{n}\rfloor + 1 = nq+1$.
The RHS is $\lfloor n (\frac{m}{n}+\delta)\rfloor=m=nq+r$.  Hence the desired inequality is false so long as $r=0$.
However, this does not answer the question of which $a$, if any, manage to avoid $r=0$ consistently.  To avoid simple counterexamples, I ask for all sufficiently large $n$ rather than all $n$.  Perhaps this can be answered in one sentence with a continued fraction result.
Clarification: I intended to ask whether there was some $N_a$, depending on $a$, for which the inequality would hold for all $n>N_a$.  The calculation was for a specific $n$, because that was the case I had considered previously, so it was only tangentially relevant.  fgp kindly solved both the question I meant to ask and the question it looked like I was asking.
 A: The question seems to be a bit confused about whether $n\lfloor a\rfloor + 1 \leq \lfloor an \rfloor$ is supposed to hold for all $n$, for all $n \geq N$ where $N$ is chosen independent from $a$, or for all $n \geq N_a$ (i.e., what is sufficiently large may depend on $a$).
In the latter case, the following shows that the inequality is true for all $a \in \mathbb{R}^+\setminus\mathbb{N}$ if $n \geq N_a$. Assume $a \in \mathbb{R}^+\setminus\mathbb{N}$ and let $$
  \langle a \rangle := a - \lfloor a \rfloor
$$ then $\langle a\rangle \neq 0$ and $$
  na = n\lfloor a\rfloor  + n\langle a\rangle\text{.}
$$ Since $\lfloor m + x\rfloor = m + \lfloor x \rfloor$  if $m\in\mathbb{N}$ it follows that $$
  \lfloor na \rfloor = n\lfloor a\rfloor + \lfloor n\langle a\rangle\rfloor 
$$
and since $\lfloor n\langle a\rangle\rfloor \geq 1$ for $n \geq \frac{1}{\langle a\rangle} =: N_a$ you have for those $n$ that $$
  n\lfloor a\rfloor + 1 \leq \lfloor an \rfloor \text{.}
$$
The argument about $a$ being close to a natural only applies if you choose $n$ first (i.e. independent from $a$). Otherwise, for an $a \in \mathbb{R}\setminus\mathbb{N}$, you can always pick an $n$ such that $a$ is more than $\frac{1}{n}$ bigger than any smaller natural. If you do choose $N$ first, the above shows that it's only those $a$ which lie too close to a natural which cause problems, since the inequality $\lfloor n\langle a\rangle\rfloor \geq 1$ is the only one where the value of $n$ enters the picture. Thus, if you pick $N$ first, the inequality holds exactly if $$
  \langle a \rangle  = a - \lfloor a \rfloor \geq \frac{1}{N} \text{.}
$$
For completness' sake, it should be stated that the inequality quite obviously holds for no $a$ if $n \leq 1$, and for $n > 1$ it follows from the above that $n\lfloor a\rfloor + 1 \leq \lfloor an \rfloor$ for all $n > 1$ if and only if $$
  \langle a \rangle  = a - \lfloor a \rfloor \geq \frac{1}{2} \text{.}
$$
