Establish the inequality $\frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} \geq (1-\frac{2\Omega}{N})^{\sqrt{N}}$ Establish the inequaity
$$ \frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} \geq (1-\frac{2\Omega}{N})^{\sqrt{N}}$$ where $N > \Omega > \sqrt{N}$ and $N$ is a perfect-square.
My attempt:
$$ \frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} = \frac{(N-\sqrt{N})(N-\sqrt{N}-1)\cdots (N-\sqrt{N}-\Omega+1)}{N(N-1)(N-2)\cdots (N-\Omega+1)}$$
Therefore, by manipulating the fraction we obtain:
$$ \frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} = (1-\frac{\sqrt{N}}{N})(1-\frac{\sqrt{N}}{N-1})\cdots (1-\frac{\sqrt{N}}{N-\Omega+1})$$
Since $\Omega > \sqrt{N}$, we will have:
$$(1-\frac{\sqrt{N}}{N})(1-\frac{\sqrt{N}}{N-1})\cdots (1-\frac{\sqrt{N}}{N-\Omega+1}) \geq (1-\frac{\Omega}{N})(1-\frac{\Omega}{N-1})\cdots (1-\frac{\Omega}{N-\Omega+1})$$
Which in turn gives:
$$  \frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} = (1-\frac{\Omega}{N})(1-\frac{\Omega}{N-1})\cdots (1-\frac{\Omega}{N-\Omega+1}) \geq (1-\frac{\Omega}{N-\Omega+1})^{\Omega}$$
If we also assume that $\Omega < \frac{N}{2}$, we get:
$$(1-\frac{\Omega}{N-\Omega+1})^{\Omega}=(\frac{N-2\Omega+1}{N-\Omega+1})^{\Omega}\geq (\frac{N-2\Omega}{N})^{\Omega} \geq (1-\frac{2\Omega}{N})^{\Omega}$$
But since $0<1-\frac{2\Omega}{N}<1$, exponentiation by $\sqrt{N} < \Omega$ will reverse the inequality and it won't finish the proof.
 A: According to the comments after the question, we assume $\sqrt{N}<Q<\frac{N}{2}$.
As you started,
$$
\dfrac{\binom{N-\sqrt{N}}{\Omega}}{\binom{N}{\Omega}}
= \prod_{k=0}^{\Omega-1} \frac{N-\sqrt{N}-k}{N-k}
= \prod_{k=0}^{\Omega-1} \left(1-\dfrac{\sqrt{N}}{N-k}\right) >
\\
> \left(1-\frac{\sqrt{N}}{N-Q}\right)^{\Omega}
> \left(1-\frac{\sqrt{N}}{N-N/2}\right)^{\Omega}
= \left(1-\frac2{\sqrt{N}}\right)^{\Omega}.
$$
By Bernoulli,
$$
\left(1-\frac2{\sqrt{N}}\right)^{\frac{\Omega}{\sqrt{N}}}
> 1 -\frac2{\sqrt{N}} \cdot \frac{\Omega}{\sqrt{N}} = 1 -\frac{2\Omega}{N},
$$
so
$$
\left(1-\frac2{\sqrt{N}}\right)^{\Omega}
> \left(1 -\frac{2\Omega}{N}\right)^{\sqrt{N}}.
$$

$\textbf{Update:}$ I realised that it can be done simpler, without using Bernoulli:
$$
\dfrac{\binom{N-\sqrt{N}}{\Omega}}{\binom{N}{\Omega}} 
= \prod_{k=0}^{\sqrt{N}-1} \frac{N-\Omega-k}{N-k} 
= \prod_{k=0}^{\sqrt{N}-1} \left(1-\dfrac{\Omega}{N-k}\right) >
\left(1-\frac{\Omega}{N-\sqrt{N}}\right)^{\sqrt{N}} >
\left(1-\frac{2\Omega}{N}\right)^{\sqrt{N}}.
$$
