Let $G$ a random graph with $n$ vertices.
Every vertex appears in the graph with a probability of $\frac{f(n)}{n}$ ($n$ is the number of vertices so the distribution is uniform),
while $f(n)$ is a function satisfying $\lim\limits_{n\to\infty}f(n) = 0$.
Prove the probability of the graph containing a triangle is $0$ as $n \to \infty$.
My try:
Using the union bound we can bound this probability with $$\binom{n}{3}\cdot \left(\frac{f(n)}{n}\right)^3$$ which tends to $0$ as $n \to \infty$.
I was wondering if there's a more elegeant or direct or any other way to solve this one.