I am working on a random graphs problem, which is stated as follows:
Prove that there exists some positive constant $C$ such that with high probability (w.h.p.), every vertex belongs to a triangle in the random graph $G(n, p)$ when $p \geq C n^{- 2 / 3} \log^{1 / 3} n$.
I notice that Janson inequality would help. The following is my try. Denote the indicator $X_v = 1(\text{no triangle containing $v$})$, then we can calculate the expectation and pairwise probability as follows: $$ E X = \sum_{v} E X_v \geq n[(1 - p)^{n - 1} + (n - 1) (1 - p)^{n - 2}] $$ and $$ \Delta = \sum_{v \, \sim \, w} E [X_v X_w] = \sum_{v \, \sim \, w} P (\text{no triangle containing $v$ and $w$}) = \binom{n - 2}{2} (1 - p)^{2( n - 2)}. $$ Thus, Janson inequality will give $$ P (X = 0) \leq e^{- \mu + \Delta / 2} \leq \exp \left( - (1 - p)^{n - 2} \left[ n (n - p) - \frac{1}{2} \binom{n - 2}{2} (1 - p)^{n - 2} \right]\right) \sim \exp \left[ - n^2 e^{- C n^{1 / 3} \log^{1 / 3} n } \right] \rightarrow 0 $$ for any positive $C$.
I wonder whether I make mistakes in some steps or there fact is that the statement actually is true for any $C > 0$.