Consider disks $s_1, \cdots, s_n$ in the plane and let $a_{ij}$ be the area of $s_i\cap s_j$. Is it true that for any real numbers $x_1,\cdots, x_n$ we have $$ \sum_{i,j=1}^n x_ix_j a_{ij} \geq 0$$
Equivalent formulation: one can put $a_{ij}$ into a matrix $A$ and ask whether it is positive semidefinite. For $n=2$ this is true since $$a_{12}^2\le \min(a_{11},a_{22})^2 \le a_{11}a_{22} $$
The question above has an affirmative answer. More generally, we have the following result:
Theorem: Suppose $s_1, \dots, s_n$ are measurable sets in the plane $\mathbf{R}^2$ with finite area. Let $a_{i,j}$ be the area of $s_i \cap s_j$. Then $$ \sum_{i,j=1}^n x_i x_j a_{i,j} \ge 0 $$ for all real numbers $x_1, \dots, x_n$.
Proof: In the inner product space $L^2(\mathbf{R}^2)$ (with the usual area measure), let $f_i$ be the characteristic function of $s_i$. Thus $$ a_{i,j} = \int_{s_i \cap s_j} 1 = \int_{\mathbf{R}^2} f_i f_j = \langle f_i, f_j \rangle. $$ Now suppose $x_1, \dots, x_n$ are real numbers. Let $f = x_1 f_1+ \dots + x_n f_n$. Then \begin{align*} \sum_{i,j=1}^n x_i x_j a_{i,j} &= \sum_{i,j=1}^n x_i x_j \langle f_i, f_j \rangle\\[6pt] &= \Bigl\langle \sum_{i=1}^n x_i f_i, \sum_{j=1}^n x_j f_j \Bigr\rangle\\[6pt] &=\langle f, f \rangle \\[6pt] &\ge 0, \end{align*} as desired.