Probabilistic bound to the number of edge disjoint triangles in a random graph Let $G$ be a random graph with $n$ vertices, and let $\delta(G)$ be the maximum number of triangles in $G$.

Question. How to prove the bound  $$P(\delta(G)) \leq m - t \sqrt{f(m)}) \leq 2e^{-t^2 / 4}$$ using Talagrand's concentration
Inequality? Here $m$ denotes the median of $\delta(G)$.

I am trying to use the convex distance inequality to get this bound.
 A: There is a (lovely) corollary of Talagrand's inequality contained in Alon and Spencer's Probabilistic Method which gives it immediately.  Here's a version of it:

Definition: Let $f: N \to N$ be a function.  A function $h$ is $f$-certifiable if whenever $h(x) \geq s$ there exists $I \subset \{1,\ldots,n\}$ with $|I| \leq f(s)$ so that for all $y$ that agree with $x$ on the coordinates $I$ we have $h(y) \geq s$.

In particular the number of triangles in a graph is $3s$ certifiable, since if there are $\geq s$ triangles, we can produce $3s$ edges which "certify" the existence of $s$ triangles.  Now we have a nice corollary of Talagrand's inequality:

Theorem (Talagrand): Let $X = h(\cdot)$ where $h$ is $f$ certifiable and $1$-Lipschitz.  Then for all $b$ and $t$ we have $$P(X \leq b - t\sqrt{f(b)}) P(X \geq b) \leq e^{-t^2 / 4}\,.$$

The number of edge disjoint triangles in a random graph is certainly $1$-Lipschitz, and is $3s$ certifiable.  Applying the above for $b = m$ completes the proof.
