What is the importance of $ |A_{i}\cap A_{j}|\le 3 $ because number of elements in Ai is 3 What about m???? If nothing mentioned the worst case is when all the Ai's are distinct

Sorry, it should have been $\le 1$.

Yes, there is a typo. The condition ii) is $|A_i \cap A_j| \leq 1$ for any distinct $i$, $j$. This is Problem 16, Romanian Selection Test 5 of 1999, see http://www.artofproblemsolving.com/Forum/viewtopic.php?p=305430&sid=53e95b49d65b9f5beb8ca947c7ef3daf#p305430, with some solutions and discussions. Of course, the problem is a classic. I will present a short and sweet proof, making use of the trusted combinatorial extremal element method. Of course there exist subsets $Y \subset X$ which contain none of the $A_i$'s; just take any with $|Y| < 3$. Consider now such a set $Y$ of maximal cardinality $|Y| = c$. Then for any $x \in X \setminus Y$ there must exist (at least) a pair $\{y_1, y_2\} \subset Y$ such that $\{x,y_1,y_2\}$ is one of the $A_i$'s (otherwise we may augment $Y$ with $x$). Now, for distinct $x, x' \in X \setminus Y$ we cannot have a same pair $\{y_1, y_2\} \subset Y$ associated to both as above, since then the triplets $\{x,y_1,y_2\}$ and $\{x',y_1,y_2\}$ would intersect in $2$ elements, violating ii). Therefore the number of elements not in $Y$ is at most the number of pairs of elements from $Y$, i.e. $n - c \leq \binom {c} {2}$, or $2n \leq c^2 + c$. It means $c \geq \dfrac {-1 + \sqrt{1 + 8n}} {2} = \sqrt{2n + \dfrac {1} {4}} - \dfrac {1} {2}$, from which the conclusion follows. A very similar problem hase been used as Problem 4 at the 2009 Tuymaada Yakut competition, see http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1567161&sid=53e95b49d65b9f5beb8ca947c7ef3daf#p1567161, and my post there.

Can't we use induction here ?

Nice Problem! Take the subset $A$ of $X$ with $k=|A|$ as large as possible, so that none of $A_i$ ($i=1, 2, \dots, m$) is a subset of $A$. For any element $v \notin A$ there exists at least one pair $(a, b)$ such that $(a, b, v)=A_j$ for some index $j$. Since $|A_i \cap A_j| \le 1$ we see that this mapping from $A - X$ to pairs of elements of $X$ is an injection. It follows that $$\binom{k}{2} \ge n-k \Longrightarrow k(k+1) \ge 2n,$$so $k \ge \lfloor \sqrt{2n} \rfloor$ as desired.

gauss01 wrote: Can't we use induction here ? i mean, induction on 'n'?

sorry for the 3 year bumb, but would this mapping be bijective as well? Since for every $x$ in X but not is A, there must exist a pair in A, right? If not, wouldn't this contradict the minimality?