Problem: Let $A$ be a set with $|A|=225$, meaning that $A$ has $225$ elements.
Suppose further that there are eleven subsets $A_1, \dots, A_{11}$ of $A$
such that $|A_i|=45$ for $1\leq i\leq11$ and $|A_i\cap A_j|=9$
for $1\leq i<j\leq11$.
Prove that $|A_1\cup A_2\cup\dots\cup A_{11}|\geq 165$,
and give an example for which equality holds.

\end{enumerate}

Solution: Ignore the $225$ --- it is irrelevant.

Denote the elements of $A_1 \cup \dots \cup A_{11}$
by $a_1$, \dots, $a_n$,
and suppose that $a_i$ appears $x_i$ times among $A_i$
for each $1 \le i \le n$ (so $1 \le x_i \le 11$).
Then we have
\[ \sum_{i=1}^{11} x_i = \sum_{i=1}^{11} |A_i| = 45 \cdot 11 \]
and
\[ \sum_{i=1}^{11} \binom{x_i}{2} = \sum_{1 \le i < j \le 11}
  \left\lvert A_i \cap A_j \right\rvert
  = \binom{11}{2} \cdot 9. \]
Therefore, we deduce that
$\sum x_i = 495$ and $\sum_i x_i^2 = 1485$.
Now, by Cauchy Schwarz
\[ n \left( \sum_i x_i^2 \right) \ge \left( \sum x_i \right)^2 \]
which implies $n \ge \frac{495^2}{1485} = 165$.

Equality occurs if we let $A$ consist of the $165$
three-element subsets of $\left\{ 1, \dots, 11 \right\}$
(plus $60$ of your favorite reptiles
if you really insist $|A|=225$).
Then we let $A_i$ denote those subsets containing $i$,
of which there are $\binom{10}{2} = 45$,
and so that $|A_i \cap A_j| = \binom 91 = 9$.