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Sum of the absolute eigenvalues of A>=B

Kindly help me to prove/disprove the following statement.
Let $A$ be a symmetric matrix of order $n \times n$ with all the diagonal entry equal to $0$, and other non-diagonal entry equal to $k$ (where $k$ is a fixed positive integer). Similarly,
Let $B$ be a symmetric matrix of order $n \times n$ with all the diagonal entry equal to $0$, and other non-diagonal entry equal to $\ell$ (where $1\leq \ell \leq k<n$).
For example:$ A= \begin{pmatrix} \begin{array}{cccc} 0 & 3 & 3 & 3 \\ 3 & 0 & 3 & 3 \\ 3 & 3 & 0 & 3 \\ 3 & 3 & 3 & 0 \\ \end{array}\end{pmatrix}$, $B=\begin{pmatrix} \begin{array}{cccc} 0 & 3 & 1 & 2 \\ 3 & 0 & 2 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 3 & 1 & 0 \\ \end{array} \end{pmatrix}$.
Prove/disprove the following statement:
Suppose $\lambda_1, \lambda_2, ..., \lambda_n$ are eigenvalues of $A$ and $\mu_1, \mu_2, ..., \mu_n$ are eigenvalues of $B$. Then $$\sum_{i=1}^{n}{|\lambda_i|}\geq \sum_{i=1}^{n}{|\mu_i|}.$$
Note that $A$ and $B$ are the distance matrices of some vertices(diameteral) of $G_1$ and $G_2$ respectively.