I encountered this small combinatorial problem and do not quite know how to solve it:

Consider a set $\mathbf N:=\left\{1,2,....,N \right\}.$ This set has $\binom{N}{2}$ many subsets of cardinality $2.$ Thus, we can introduce variables $x_1,...,x_{\binom{N}{2}}$ taking a value in $\mathbb R$ where each of the variables is associated to precisely one subset of $\mathbf N$ of cardinality $2.$

In other words there is a unique correspondence:

$x_i \leftrightarrow M_i$ where $M_i \subset \mathbf N$ and $\left\lvert M_i \right\rvert =2.$

I would like to know whether the following estimate is true:

$$ \frac{1}{N} \sum_{i,j=1}^{\binom{N}{2}} x_i x_j \le C \sum_{i,j=1}^{\binom{N}{2}} \left\lvert M_i \cap M_j \right\rvert x_i x_j ?$$ for some $C$ independent of $N$ and all $x_i,x_j$?

EDIT:

It holds true if $x_i=1$, since then the left hand side is $\binom{N}{2}^2/N=\frac{1}{4}(N-1)^2N$ and the right hand side is, using the hypergeometric distribution, $$2\binom{N}{2}+ \binom{N}{2}^2\frac{4}{N}=N(N-1)+N(N-1)^2.$$

So an obvious scaling argument does not disprove the estimate. However, it is not clear to me whether this estimate is true in general?

I emphasize that the left-hand side $\frac{1}{N} \sum_{i,j=1}^{\binom{N}{2}} x_i x_j$ can be interpreted as $$\frac{1}{N} \langle (x_1,..,x_{\binom{N}{2}}),\mathbb 1 (x_1,..,x_{\binom{N}{2}})\rangle $$ where $\mathbb 1$ is the matrix with all entries equal to one. This one has one non-zero eigenvalue with eigenvector $(1,...1)$ and all other eigenvalues are zero. That's why I was particularly curious to study that case separately.

So does this inequality hold true in general?