Small detail in proof of Lemma 1.2 in Frieze and Karoński's 'Introduction to Random Graphs' In the text Introduction to Random Graphs by Karoński and Frieze, the authors claim that
$$
(1+o(1))\sqrt{\frac{N}{2\pi m(N-m)}} \geq \frac{1}{10\sqrt{m}}
$$
in the proof of Lemma 1.2 where

*

*$m = m(n) \to \infty$

*$N = \binom{n}{2}$

*$N - m \to \infty$.

I have been trying to figure how they made this simplification for some time now and am not confident with my reasoning. Here is my attempt at a proof:
Since the $o(1)$ term comes from Stirling's formula, it must be positive. Hence
\begin{align*}
  (1+o(1))\sqrt{\frac{N}{2\pi m(N-m)}} &\geq \sqrt{\frac{N}{2\pi m(N-m)}} \\
&\geq  \sqrt{\frac{N-m}{2\pi m(N-m)}} \\
&= \frac{1}{\sqrt{2\pi m}} \\
&\geq \frac{1}{10\sqrt{m}}.
\end{align*}
For some reason this doesn't feel right to me. Any feedback or insight would be greatly appreciated.
Also, it's a bit confusing why they chose the constant $10$. Couldn't they have chosen a smaller constant so long as it was bigger than $\sqrt{2\pi}$ (for instance $3$)? Or was the reason they chose $10$ just because it's "nice".
EDIT: As was nicely pointed out by Calvin Khor below, I don’t think I can claim that the $o(1)$ term that comes from Stirling’s formula is positive. That is, it was not sound to say that the $o(1)$ term is positive just because it came from Stirling’s formula. Apologies for the error there.
 A: This is almost right, but (as pointed out by Calvin Khor in the comments) we should add the step that for large enough $n$, $1+o(1)$ is at least $\frac12$ (or any constant less than $1$) rather than say it is at least $1$.
More interesting (though not often useful) is the fact that the lemma actually holds for all $n \ge 2$ and $1 \le m \le \binom n2 - 1$, but this requires a slightly more careful approach than the proof in the textbook. (It is also not enough to say that $n! > \sqrt{2\pi n} (\frac ne)^n$, since we want lower bounds in the numerator but upper bounds in the denominator.)
According to Wikipedia (citing Robbins, A remark on Stirling's formula), we have $$\sqrt{2\pi n} \left(\frac ne\right)^n e^{\frac1{12n+1}} < n! < \sqrt{2\pi n} \left(\frac ne\right)^n e^{\frac1{12n}}$$ for all $n \ge 1$. Therefore in $\binom rs$ with $1 \le s \le r-1$, we get
$$
   \binom rs \ge e^{\frac1{12r+1} - \frac1{12s} - \frac1{12(r-s)}} \sqrt{\frac{r}{2\pi s(r-s)}} \frac{r^r}{s^s (r-s)^{r-s}}. 
$$
(We'd have the same expression with $e^{\frac1{12r} - \frac1{12s+1} - \frac1{12(r-s)+1}}$in the upper bound, which we don't need here.)
The probability that a $\text{Binomial}(r, \frac sr)$ random variable is exactly $s$ is $\binom rs (\frac sr)^s (\frac{r-s}{r})^{r-s}$, which is at least $e^{\frac1{12r+1} - \frac1{12s} - \frac1{12(r-s)}} \sqrt{\frac{r}{2\pi s(r-s)}}$. The square root is bigger than $\frac1{\sqrt{2\pi s}}$ and the worst case for the exponent is $r=2$, $s=1$, which gives $e^{-19/150}$ and a lingering sense that perhaps we didn't need to be quite this careful. For all practical purposes, it is  enough that $e^{-19/150} \frac1{\sqrt{2\pi}} > \frac13$, giving us a lower bound of $\frac1{3\sqrt s}$ for this probability.
In the specific application, we get
$$
\mathbb P(\mathbb G_{n,m} \in \mathscr P) \le \frac{\mathbb P(\mathbb G_{n,p} \in \mathscr P)}{\mathbb P(|E_{n,p}|=m)} \le 3 m^{1/2}\, \mathbb P(\mathbb G_{n,p} \in \mathscr P)
$$
whenever $1 \le m \le \binom n2 - 1$. (Actually, when $m = \binom n2$ as well, trivially.)
We can also show $\mathbb P(\mathbb G_{n,m} \in \mathscr P) \le n\, \mathbb P(\mathbb G_{n,p} \in \mathscr P)$ whenever $n \ge 2$. This requires bounding $\frac{N}{m(N-m)} \ge \frac{N}{(N/2)^2} \ge \frac 8{n^2}$, and arguing $n=2$ separately to get the right constant; also, it's only better for dense random graphs, and only slightly. But it is simpler.
