Is increasing the entries of a symmetric matrix has an effect on its largest eigenvalue?

Let a matrix $A = (a_{ij}) \in \mathbb{R}^{n \times n}$ be symmetric, and $\Delta = (\delta_{ij}) \in \mathbb{R}^{n \times n}$ be symmetric and nonnegative (i.e., it has exclusively nonnegative elements). Furthermore, wherever $A$ has an entry $a_{ij} < 0$, let the corresponding element $\delta_{ij} \geq \lvert a_{ij} \rvert$. I will be considering $A + \Delta$, which is, by this construction, symmetric and nonnegative.

Denoting the maximum eigenvalue by $\lambda_{\max}$, is $\lambda_{\max} (A + \Delta) \geq \lambda_{\max}(A)$?

If it indeed is, what is a good starting point for a proof?

Possibly relevant: I believe this to be true due to numerical tests. Similarly, increasing only the nonzero entries of $A$ (i.e., $\delta_{ij} = 0$ if $a_{ij} = 0$) also works. However, I believe dropping the condition $\delta_{ij} \geq \lvert a_{ij} \rvert$ (i.e., letting $A + \Delta$ have negative entries, even though all $\delta_{ij} \geq a_{ij}$), makes it false.

(I am likewise interested in any similar problems or hints.)

The answer is no. As a counterexample, consider $$ A = \pmatrix{0 & -1\\ -1 & 0}, \quad \Delta = \pmatrix{0 & 1\\1 & 0}. $$ $\lambda_{\max}(A + \Delta) = 0 < \lambda_{\max}(A) = 1$.