Let $K_n$ denote a complete graph with $n$ vertices. Given any positive integers $k$ and $l$, the Ramsey number $R(k, l)$ is defined as the smallest integer $n$ such that in any two-coloring of the edges of $K_n$ by red and blue, either there is a red $K_k$ or a blue $K_l$.
(i) Prove if there is a real $0 \leq p \leq 1$ such that
$$\binom{n}{4}p^6 + \binom{n}{k}(1-p)^{\binom{k}{2}} < 1 $$ then we have $R(4,k) > n$.
(ii) Use the previous part to show that there exists a constant $c > 0$ such that $$ R(4,k) \geq c \Big(\dfrac{k}{log(k)}\Big)^{3/2} $$
I have no problems proving the first part. You let $p$ to be the probability of choosing a red edge, compute the probability of at least one red complete graph of size $k$, and blue graph of $l$, finally conclude that with positive probability there exists a graph such that it contains no red and blue complete graphs of the mentioned sizes, and therefore $R(4,k) > n$.
However, how to proceed for the second part? I am studying the probabilistic method and lots of exercises with finding constants. What should the sketch of the proof look like?
UPDATE: I have reduced the question to finding a constant satisfying $log(n) < n^{-2/3}k/4$. Also, the screenshot is from a claimed solution manual:
How is the right $n$ value concluded here?
To use part (i) to show part (ii), your goal is to find some $n$ of the required form, and some probability $p$, such that the inequality in part (i) holds.
If we're hoping for the best result possible, the two terms ought to contribute about equally: both $\binom n4 p^6$ and $\binom nk (1-p)^{\binom k2}$ ought to be constants. This motivates choosing $p = n^{-2/3}$ so that $\binom n4 p^6 < \frac{n^4 p^6}{24} = \frac{1}{24}$.
Next, we want to put some upper bounds on $\binom nk (1-p)^{\binom k2}$ to make it easier to work with. For the binomial coefficient, a good general bound to use is $\binom nk \le \left(\frac{ne}{k}\right)^k$. (A much weaker bound like $\binom nk \le n^k$ also works, but there's no way to know that ahead of time.) Meanwhile, $1-p \le e^{-p}$, so $(1-p)^{\binom k2} \le e^{-pk(k-1)/2}$.
Putting these together lets us collect the power of $k$: $$ \binom nk (1-p)^{\binom k2} \le \left(\frac{ne}{k}\right)^k e^{-pk(k-1)/2} = \left(\frac{ne}{k e^{p(k-1)/2}}\right)^k. $$ To make sure that this result is less than $\frac{23}{24}$, we need to make the base of the exponent less than $1$. The key to this is to make the $e^{pk/2}$ in the denominator cancel out the $n$ in the numerator. This is why $n^k$ is as good a bound on $\binom nk$ as $\left(\frac{ne}{k}\right)^k$ is: the $k$ isn't really helping. If we use $\binom nk \le n^k$, we get $$ \binom nk (1-p)^{\binom k2} \le n^k e^{-pk(k-1)/2} = \left(\frac{n}{e^{p(k-1)/2}}\right)^k. $$ Either way, we want to get $p \cdot \frac{k-1}{2} > \log n$, or $k > \frac{2 \log n}{p} + 1$. We don't care about the constant, though, so let's just ask for $k \ge \frac{3\log n}{p}$.
Since we set $p = n^{-2/3}$, this means that $k \ge 3 n^{2/3} \log n$. This is the answer, in a way, but it expresses the reverse relationship; we want a bound on $n$ in terms of $k$.
Assuming $n \le k^2$, we satisfy $k \ge 3 n^{2/3} \log n$ if we satisfy $k \ge 3n^{2/3} \log (k^2) = 6 n^{2/3} \log k$. Solving for $n$, this gives $n \le \left(\frac{k}{6\log k}\right)^{3/2}$.
(This, in addition to having the form we want, is stricter than $n \le k^2$ for all $k>1$, justifying our earlier assumption.)