Context
-------


I'm working on a problem involving Lovasz Local Lemma, for proving that there exists a graph with a certain property.


What I need to prove:
---------------------


There exists some constant $c$, and functions $p,a$ (which can depend on $n$) s.t.


$$ \frac{p^3}{a} \leq \left(1- \left[\frac{1-p}{(1-a)^n}\right]^{c^2k^2} \right)^{n^{ck}}$$


where $k = \sqrt{n}\log n$


Why I believe it is true:
-------------------------


Now, the RHS can be approximated as:


$$\exp \left( -n^{ck} \exp\left( (-p + an)c^2k^2\right) \right)$$


$$\exp
 \exp\left( ck( \log n + (an - p) ck)
 \right)
 $$


this then works if we take $p=O(n^{-1/2})$, and $a = O(n^{-3/2})$, and force the RHS to be close to 1, while the LHS to be < 1.


What I'm stuck on:
------------------


Now, I'm stuck on the following. I'm not sure how to make this rigorous. In particular, I'm making three approximations of the form 1-x \approx e^{-x}, but on the outer $X^{n^{ck}}$ "amplifies" the inner errors.


Question:
---------


How do I rigorize this?


Thanks!



How to Rigorize an inequalities argument
Take, say, $a=2p^3$, $p=1/2\sqrt{n}$. Then 
$$(1-p)/(1-a)^n=\exp(-p+na+O(p^2+na^2))=\exp(-1/4\sqrt{n}+O(1/n)).$$ 
You need lower estimate of the RHS, hence upper estimate of $(1-p)/(1-a)^n$, you mau use $\exp(-1/5\sqrt{n})$ for large enough $n$. Next, use an estimate like $1-p\geq e^{-2p}$ for $0 < p < 1/100$, and for large $n$ we estimate RHS from below as 
$$\exp(-2\exp(-c^2k^2/5\sqrt{n})n^{ck}).$$ So, it is enough to satisfy inequality
$$
10n^{ck} < \exp(c^2k^2/5\sqrt{n}),
$$
or, taking logarithm of both sides, 
$$
\log 10+ck\log n < c^2k^2/5\sqrt{n}.
$$
This holds for $c=10$ and large enough $n$.


