Lower bound for the Ramsey number $R(3,t)$ Define the Ramsey number $R(s,t)$ to be:
$\text{min}\{n \in \mathbb N \mid \text{colouring } E(K_n) \text{ blue and yellow yields a blue } K_s \text{ or a yellow } K_t\}$
Then I am asked to find that a lower bound for $R(3,t)$ as $t$ tends to infinity is $\Omega((\frac{t}{logt})^{\frac{3}{2}})$.
I am given the hint that I should modify a random graph $G \in \mathcal G(n,p)$ such that for all $n,p$ we have
$$
R(3,t) \geq n - {{n}\choose{3}}p^3 - {{n}\choose{t}}(1-p)^t.
$$
I recognise that ${{n}\choose{3}}p^3$ is the expected number of triangles in $G$ and that ${{n}\choose{t}}(1-p)^t$ is the expected number of $K_t$'s not in $G$. I then reasoned that if we remove a vertex for each of the expected triangles and $K_t$'s, then we would have a number of vertices on which it is not possible to have a triangle, or to not have a $K_t$.
Hence why the right hand side is a lower bound for $R(3,t)$. However, I am really unsure whether or not this reasoning is valid since it is taking something built on probability (expected numbers) and treating as something definite. 
Further I don't really know what to do now to achieve the lower bound that I want. 
I would imagine that I need a creative value of $p$ but I can't quite see what that might be. Any insight would be greatly appreciated, thank you! 
 A: Let $Y$ be the number of triangles in $G$, let $Z$ be the number of copies of $K_t$ with no edge in $G$, and let $X = Y + Z$.  As you have observed, by linearity of expectation,
$$
E(X) = E(Y) + E(Z) = {{n}\choose{3}}p^3 + {{n}\choose{t}}(1-p)^t.
$$
The key is that, with positive probability, $X \leq E(X)$.  Hence, there exists $G \in \mathcal G(n,p)$ such that $X \leq {{n}\choose{3}}p^3 + {{n}\choose{t}}(1-p)^t$.  Now you have a specific graph to which you can apply the vertex-deletion process that you described.
At this point, it's enough to show that there exists a positive constant $c$ such that for all $t \geq 3$, there exist $n$ and $p$ such that
$$
n - {{n}\choose{3}}p^3 - {{n}\choose{t}}(1-p)^t \geq c \biggl(\frac{t}{\log t}\biggr)^{\frac{3}{2}}. \tag{1}\label{eq:R3tBound}
$$
Here's a hint: Let $n = 2c\bigl(\frac{t}{\log t}\bigr)^{\frac{3}{2}}$.  If $p$ is a function of $n$ (and hence of $t$) such that
$$
{{n}\choose{3}}p^3 + {{n}\choose{t}}(1-p)^t \leq c \biggl(\frac{t}{\log t}\biggr)^{\frac{3}{2}},
$$
then the desired inequality \eqref{eq:R3tBound} follows immediately.
