Let $n>2$ be an integer and let $\ell \in\{1,2, \ldots, n\}$. A collection $A_{1}, \ldots, A_{k}$ of (not necessarily distinct) subsets of $\{1,2, \ldots, n\}$ is called $\ell$-large if $\left|A_{i}\right| \geq \ell$ for all $1 \leq i \leq k$. Find, in terms of $n$ and $\ell$, the largest real number $c$ such that the inequality  $$ \sum_{i=1}^{k} \sum_{j=1}^{k} x_{i} x_{j} \frac{\left|A_{i} \cap A_{j}\right|^{2}}{\left|A_{i}\right| \cdot\left|A_{j}\right|} \geq c\left(\sum_{i=1}^{k} x_{i}\right)^{2} $$  holds for all positive integer $k$, all nonnegative real numbers $x_{1}, x_{2}, \ldots, x_{k}$, and all $\ell$-large collections $A_{1}, A_{2}, \ldots, A_{k}$ of subsets of $\{1,2, \ldots, n\}$.
The answer turns out to be  $$ c=\frac{n+\ell^{2}-2 \ell}{n(n-1)} $$   【 Rewriting as a dot product. For $i=1, \ldots, n$ define $\mathbf{v}_{i}$ by  $$ \mathbf{v}_{i}[p, q]:=\left\{\begin{array}{ll} \frac{1}{\left|A_{i}\right|} & p \in A_{i} \text { and } q \in A_{i} \\ 0 & \text { otherwise; } \end{array} \quad \mathbf{v}:=\sum_{i} x_{i} \mathbf{v}_{i}\right. $$  Then  $$ \begin{aligned} \sum_{i} \sum_{j} x_{i} x_{j} \frac{\left|A_{i} \cap A_{j}\right|^{2}}{\left|A_{i}\right|\left|A_{j}\right|} & =\sum_{i} \sum_{j} x_{i} x_{j}\left\langle\mathbf{v}_{i}, \mathbf{v}_{j}\right\rangle \\ & =\left\langle\sum_{i} x_{i} \mathbf{v}_{i}, \sum_{j} x_{i} \mathbf{v}_{i}\right\rangle=\left\|\sum_{i} x_{i} \mathbf{v}_{i}\right\|^{2}=\|\mathbf{v}\|^{2} . \end{aligned} $$   $$ \begin{aligned} & \langle\mathbf{e}, \mathbf{v}\rangle=\sum_{i} x_{i}\left\langle\mathbf{e}, \mathbf{v}_{i}\right\rangle=\sum_{i} x_{i} \\ & \langle\mathbf{1}, \mathbf{v}\rangle=\sum_{i} x_{i}\left\langle\mathbf{1}, \mathbf{v}_{i}\right\rangle=\sum_{i} x_{i}\left|A_{i}\right| . \end{aligned} $$  That means for any positive real constants $\alpha$ and $\beta$, by Cauchy-Schwarz for vectors, we should have  $$ \begin{aligned} \|\alpha \mathbf{e}+\beta \mathbf{1}\|\|\mathbf{v}\| & \geq\langle\alpha \mathbf{e}+\beta \mathbf{1}, \mathbf{v}\rangle=\alpha\langle\mathbf{e}, \mathbf{v}\rangle+\beta\langle\mathbf{1}, \mathbf{v}\rangle \\ & =\alpha \cdot \sum x_{i}+\beta \cdot \sum x_{i}\left|A_{i}\right| \\ & \geq(\alpha+\ell \beta) \sum x_{i} . \end{aligned} $$  Set $\mathbf{w}:=\alpha \mathbf{e}+\beta \mathbf{1}$ for brevity. Then  $$ \mathbf{w}[p, q]= \begin{cases}\alpha+\beta & \text { if } p=q \\ \beta & \text { if } p \neq q\end{cases} $$  SO  $$ \|\mathbf{w}\|=\sqrt{n \cdot(\alpha+\beta)^{2}+\left(n^{2}-n\right) \cdot \beta^{2}} $$  Therefore, we get a lower bound  $$ \frac{\|\mathbf{v}\|}{\sum x_{i}} \geq \frac{\alpha+\ell \beta}{\sqrt{n \cdot(\alpha+\beta)^{2}+\left(n^{2}-n\right) \cdot \beta^{2}}} $$  Letting $\alpha=n-\ell$ and $\beta=\ell-1$ gives a proof that the constant  $$ c=\frac{((n-\ell)+\ell(\ell-1))^{2}}{n \cdot(n-1)^{2}+\left(n^{2}-n\right) \cdot(\ell-1)^{2}}=\frac{\left(n+\ell^{2}-2 \ell\right)^{2}}{n(n-1)\left(n+\ell^{2}-2 \ell\right)}=\frac{n+\ell^{2}-2 \ell}{n(n-1)} $$  makes the original inequality always true. (The choice of $\alpha: \beta$ is suggested by the example below.)  【 Example showing this $c$ is best possible. Let $k=\binom{n}{\ell}$, let $A_{i}$ run over all $\binom{n}{\ell}$ subsets of $\{1, \ldots, n\}$ of size $\ell$, and let $x_{i}=1$ for all $i$. We claim this construction works.  To verify this, it would be sufficient to show that $\mathbf{w}$ and $\mathbf{v}$ are scalar multiples, so that the above Cauchy-Schwarz is equality. However, we can compute  $$ \mathbf{w}[p, q]=\left\{\begin{array}{ll} n-1 & \text { if } p=q \\ \ell-1 & \text { if } p \neq q \end{array}, \quad \mathbf{v}[p, q]= \begin{cases}\binom{n-1}{\ell-1} \cdot \frac{1}{\ell} & \text { if } p=q \\ \binom{n-2}{\ell-2} \cdot \frac{1}{\ell} & \text { if } p \neq q\end{cases}\right. $$  which are indeed scalar multiples, finishing the proof.