lower bound for Ramsey number From Probabilistic extremal combinatorics:
"Assume that $e{2\choose \ell}{n-2 \choose \ell-2}2^{1-{\ell\choose 2}}<1\Rightarrow R(\ell,\ell)>n$. Prove $R(\ell,\ell)>(\sqrt{2}/e+o(1))\cdot \ell \cdot 2^{\ell/2}$ as $\ell\rightarrow \infty$."
So from large enough $\ell$ on, we must show that $R(\ell,\ell)>(\sqrt{2}/e+s_{\ell})\cdot\ell\cdot2^{\ell/2}$ for all $(s_{\ell})_{\ell}\rightarrow 0$.
In the theory I saw bounds like $R(\ell,\ell)\leq {2\ell-2\choose \ell-1}\sim 4^{\ell-1}/(\sqrt{\pi(\ell-1)})$ and $R(\ell,\ell)\geq \sqrt{2}^{\ell}$. I can't imagine the first inequality will help because that is an upper bound and not a lower bound, I have also tried the second without success. 
Maybe we can choose a reasonable $n$, but $n$ equal to $(\sqrt{2}/e+\varepsilon)\cdot\ell\cdot2^{\ell/2}$ with $\epsilon>0$ such that this becomes an integer, will this be s.t. the assumed implication holds? The result is an ugly expression.
Is there anybody who can help? I thank in advance because I can't thank in the comments (and of course I wanna be grateful).
 A: The point is not to use an existing bound on Ramsey numbers, but to use the implication
$$e{\ell \choose 2}{n-2 \choose \ell-2}2^{1-{\ell\choose 2}}<1 \implies R(\ell,\ell)>n$$
to deduce a lower bound on $R(\ell,\ell)$ of the asymptotic form $(\frac{\sqrt2}{e} + o(1)) \cdot \ell \cdot 2^{\ell/2}$.
That is, we want to plug an $n$ that has this asymptotic form into the left-hand side of the inequality, get something that's less than $1$, and conclude that $R(\ell,\ell) > n$.
(For context, the inequality itself arises from applying the Lovász local lemma to show that a uniformly random coloring of $K_n$ has no monochromatic $K_\ell$ with positive probability - as long as the inequality holds, which works out to be the condition for the local lemma to apply.)

The key idea is to use the upper bound $\binom nk \le \left(\frac{en}{k}\right)^k$ to make the expression $e{\ell \choose 2}{n-2 \choose \ell-2}2^{1-{\ell\choose 2}}$ simpler to analyze. All we have to do is to show that, for our chosen value of $n$, $$e{\ell \choose 2}\left(\frac{(n-2)e}{\ell-2}\right)^{\ell-2}2^{1-{\ell\choose 2}} < 1,$$ and then we can conclude that $e{\ell \choose 2}{n-2 \choose \ell-2}2^{1-{\ell\choose 2}} < 1$ as well.
With this in mind, let's set $n = \frac{\sqrt2}{e} \cdot t \cdot (\ell-2) \cdot 2^{\ell/2} + 2$, with a factor $t = t(\ell)$ we'll figure out later. Why? Because this gives us
$$\frac{(n-2)e}{\ell-2} = \sqrt2 \cdot t \cdot 2^{\ell/2} = t \cdot 2^{(\ell+1)/2}.$$
If you work through some character-building calculations, you get $$e{\ell \choose 2}\left(\frac{(n-2)e}{\ell-2}\right)^{\ell-2}2^{1-{\ell\choose 2}} = e \binom{\ell}{2} t^{\ell-2}$$ and it is possible to choose a value of $t$ which is $1 + o(1))$ (that is, $t \to 1$ as $\ell \to \infty$) and yet still guarantees that this expression is less than $1$ for all $\ell$.
Now the assumption we've started with guarantees that $R(\ell,\ell) > n$ for the value of $n$ we've chosen, and since $\frac{\sqrt2}{e} \cdot t \cdot (\ell-2) \cdot 2^{\ell/2} + 2 = (\frac{\sqrt2}{e} + o(1)) \cdot \ell \cdot 2^{\ell/2}$, we have the bound we wanted.
