Matrix rescaling increases lowest eigenvalue? Consider the set $\mathbf{N}:=\left\{1,2,....,N \right\}$ and let $$\mathbf M:=\left\{ M_i; M_i \subset \mathbf N \text{ such that } \left\lvert M_i \right\rvert=2 \text{ or }\left\lvert M_i \right\rvert=1 \right\}$$
be the set of all subsets of $\mathbf{N}$ that are of cardinality $1$ or $2.$
The cardinality of the set $\mathbf M$ itself is $\binom{n}{1}+\binom{n}{2}=:K$
We can then study for $y \in (0,1)$ the $K \times K$ matrix
$$A_N = \left(  \frac{\left\lvert M_i \cap M_j \right\rvert}{\left\lvert M_i \right\rvert\left\lvert M_j \right\rvert}y^{-\left\lvert M_i \cap M_j \right\rvert} \right)_{i,j}$$
and 
$$B_N = \left(  \left\lvert M_i \cap M_j \right\rvert y^{-\left\lvert M_i \cap M_j \right\rvert} \right)_{i,j}.$$
Question
I conjecture that $\lambda_{\text{min}}(A_N)\le \lambda_{\text{min}}(B_N)$ for any $N$ and would like to know if one can actually show this?
As a first step, I would like to know if one can show that $$\lambda_{\text{min}}(A_N)\le C\lambda_{\text{min}}(B_N)$$ for some $C$ independent of $N$?
In fact, I am not claiming that $A_N \le B_N$ in the sense of matrices. But it seems as if the eigenvalues of $B_N$ are shifted up when compared with $A_N.$
Numerical evidence:
For $N=2$ we can explicitly write down the matrices 
$$A_2 =\left(
\begin{array}{ccc}
 \frac{1}{y} & 0 & \frac{1}{2 y} \\
 0 & \frac{1}{y} & \frac{1}{2 y} \\
 \frac{1}{2 y} & \frac{1}{2 y} & \frac{1}{2 y^2} \\
\end{array}
\right) \text{ and }B_2 = \left(
\begin{array}{ccc}
 \frac{1}{y} & 0 & \frac{1}{y} \\
 0 & \frac{1}{y} & \frac{1}{y} \\
 \frac{1}{y} & \frac{1}{y} & \frac{2}{y^2} \\
\end{array}
\right)$$
We obtain for the lowest eigenvalue of $A_2$ (orange) and $B_2$(blue) as a function of $y$

For $N=3$ we get qualitatively the same picture, i.e. the lowest eigenvalue of $A_3$ remains below the lowest one of $B_3$:
In this case: 
$$A_3=\left(
\begin{array}{cccccc}
 \frac{1}{y} & 0 & 0 & \frac{1}{2 y} & 0 & \frac{1}{2 y} \\
 0 & \frac{1}{y} & 0 & \frac{1}{2 y} & \frac{1}{2 y} & 0 \\
 0 & 0 & \frac{1}{y} & 0 & \frac{1}{2 y} & \frac{1}{2 y} \\
 \frac{1}{2 y} & \frac{1}{2 y} & 0 & \frac{1}{2 y^2} & \frac{1}{4 y} & \frac{1}{4 y} \\
 0 & \frac{1}{2 y} & \frac{1}{2 y} & \frac{1}{4 y} & \frac{1}{2 y^2} & \frac{1}{4 y} \\
 \frac{1}{2 y} & 0 & \frac{1}{2 y} & \frac{1}{4 y} & \frac{1}{4 y} & \frac{1}{2 y^2} \\
\end{array}
\right)\text{ and } B_3=\left(
\begin{array}{cccccc}
 \frac{1}{y} & 0 & 0 & \frac{1}{y} & 0 & \frac{1}{y} \\
 0 & \frac{1}{y} & 0 & \frac{1}{y} & \frac{1}{y} & 0 \\
 0 & 0 & \frac{1}{y} & 0 & \frac{1}{y} & \frac{1}{y} \\
 \frac{1}{y} & \frac{1}{y} & 0 & \frac{2}{y^2} & \frac{1}{y} & \frac{1}{y} \\
 0 & \frac{1}{y} & \frac{1}{y} & \frac{1}{y} & \frac{2}{y^2} & \frac{1}{y} \\
 \frac{1}{y} & 0 & \frac{1}{y} & \frac{1}{y} & \frac{1}{y} & \frac{2}{y^2} \\
\end{array}
\right)$$


The matrices are of the form
$$
A=\begin{pmatrix} 1 & C \\ C^* & D \end{pmatrix}, \quad\quad
B= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} A \begin{pmatrix} 1 & 0\\ 0&2\end{pmatrix} ,
$$
with the blocks corresponding to the sizes of the sets $M_j$ involved.
Let $v=(x,y)^t$ be a normalized eigenvector for the minimum eigenvalue $\lambda $ of $B$. Then
$$
(x,2y)A\begin{pmatrix} x \\ 2y \end{pmatrix} = \lambda
$$
also, but this modified vector has larger norm.
So the desired inequality $\lambda_j(A)\le\lambda_j(B)$ (for all eigenvalues, not just the first one) will follow if we can show that $B\ge 0$. This is true for $y=1$ because in this case we can interpret
$$
|M_j\cap M_k|=\sum_n \chi_j(n)\chi_k(n)
$$
as the scalar product in $\ell^2$ of the characteristic functions, and this makes
$v^*Bv$ equal to $\|f\|_2^2\ge 0$, with $f=\sum v_j\chi_j$.
For general $y>0$, we have $B(y)=(1/y)B(1) + D$, for a diagonal matrix $D$ with non-negative entries.
