A combinatorics problem associated with probability-method Let $\mathcal{F}\subseteq \binom{[n]}{3}$ be a family of $m$ subsets where $m\geq n/3$.
Prove that there exists a subset $A\subseteq [n]$ with
$$|A|\ge \frac{2n^{3/2}}{3\sqrt{3m}}$$ such that none of the sets in $\mathcal{F}$ is contained in $A$.
My idea is to delete $B\in\mathcal{F}$ that $B$ contains the best element $a_i\in[n]$ occur the most time step by step, then get $$|\mathcal{F}_i|\leq\lfloor\frac{n-i-3}{n-i}|\mathcal{F}_{i-1}|\rfloor$$ with $|\mathcal{F}_0|=|\mathcal{F}|=m$. Suppose at $k$-th step, $|\mathcal{F}_k|=0$ then $[n]-\{a_1,\cdots,a_k\}$ is the wanted set.
But I don't know how to prove the upper bound of $k$ (should be $n-\frac{2n^{3/2}}{3\sqrt{3m}}$).

I got an answer.
Let $S$ be a random set, for $i\in[n]$, $P[i\in S]=\sqrt\frac{n}{3m}\leq1$, then $E|S|=\frac{n\sqrt n}{\sqrt{3m}}$.
For $A$ be every element of $\mathcal{F}$, if $A\subseteq S$, delete one element from $A$ from $S$. $P[A\subseteq S]=(\frac{n}{3m})^{3/2}$ so the elements we delete $Y$ satisfy $E|Y|\leq\frac{n^{3/2}}{3\sqrt{3m}}$. After delete, $E|S|\geq\frac{2n\sqrt n}{3\sqrt{3m}}$, and none of $\mathcal{F}$ is contained in $S$. So there exists such $S$ with $|S|\geq\frac{2n^{3/2}}{3\sqrt{3m}}$.
