Existence of matrices such that $\rho(A)+ \rho(B) \lt \rho(A+B)$

Question

Given two matrices $A,B \in \mathcal{M}_n (\mathbb{K})$, and let $\rho(A)$, $\rho(B)$ be the absolute values of its biggest eigenvalues, respectively. It's easy to find examples of matrices such that $\rho(A)+ \rho(B) \geq \rho(A+B)$, but can the converse happen?

Are there matrices $A,B \in \mathcal{M}_n (\mathbb{K})$ such that $\rho(A)+ \rho(B) \lt \rho(A+B)$?

Answer 1

$$A=\pmatrix{0&1\\0&0}$$ $$B=\pmatrix{0&0\\1&0}$$ $\rho(A)=\rho(B)=0$, $\rho(A+B)=1$.

Answer 2

Take $$A = \begin{bmatrix} 4 & 8 \\ 0 & 1\end{bmatrix}$$ Take $$B = \begin{bmatrix} 0 & 4 \\ 3 & 1\end{bmatrix}$$ $$\rho(A) = 4$$ $$\rho(B) = 4$$ $$\rho(A+B) \simeq 9.08$$