Is there a relation between $\rho(A) + \rho(B)$ and $\rho(A+B)$, where $\rho$ is the spectral radius?

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Let $A, B \in \mathcal M_n(\mathbb C)$, we want to know if there's any relation between their spectral radius and the spectral radius of $A + B$.

The first part of this exercise is giving examples of the three posibilities: $\rho(A) + \rho(B) < \rho(A+B)$ ; $\rho(A) + \rho(B) > \rho(A+B)$ and $\rho(A)+ \rho(B) =\rho(A+B)$. This part was easy because there are no conditions about A or B.

The second part is proving that if A and B are normal matrixes, (i.e. $ AA^* =A^*A$, where $A^*$ is its conjugate transpose), then $ \rho(A+B) \le \rho(A) + \rho(B) $.

So my first attempt was using the Schur's decomposition, so $A=U^{-1}D_AU$ where $D_A$ is a diagonal matrix with the eigenvalues of $A$ in its diagonal and U is an unitary matrix. Similarly, B is similar to a diagonal matrix $D_B$ with the eigenvalues of B in its diagonal. That way I have $\rho(A) = \rho(D_A)$ and $\rho(B) = \rho(D_B)$ so now since $D_A + D_B$ is a diagonal matrix with $\lambda_i + \mu_i$ for $ 1\le i \le n $, where $\lambda_i$ and $\mu_i$ are the eigenvalues of $A$ and $B$ respectively. From here it's easy to see that $ \rho(D_A+D_B) \le \rho(D_A) + \rho(D_B) $, since $ |\lambda_i+\mu_i|\le|\lambda_i|+|\mu_i| \ \forall i$. But I don't see how to include $\rho(A+B)$ in this.

And the last part of the exercise is proving that if $ \{X_i\}$ is a basis of eigenvectors, which is common to A and B, then $ \rho(A+B) \le \rho(A) + \rho(B) $. Here's my attempt, but I would like to know if it's correct:

Let $\lambda_i$ and $\mu_i$ be eigenvalues of A and B associated to the vector $X_i$, then $AX_i=\lambda_i X_i$ and $BX_i = \mu_iX_i$, and if we sum both equalities, we have $AX_i + BX_i = \lambda_iX_i + \mu_iXi$ so $(A+B)X_i=(\lambda_i+\mu_i)X_i$ so $(\lambda_i+\mu_i)$ is an eigenvalor of $(A+B)$ associated to the eigenvector $X_i$. Then is again clear that $|\lambda_i+\mu_i|\le|\lambda_i|+|\mu_i| \le \rho(A) + \rho(B)\ \forall i$, and particularly, for $\max_i |\lambda_i+\mu_i| = \rho(A+B)$ so $ \rho(A+B)\le \rho(A) + \rho(B)$

Any kind of help is appreciated and thank you very much in advance!

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Maybe we can use this argument. For the operator norm, https://en.wikipedia.org/wiki/Operator_norm,
$$ ||A|| = \sup_{||x||=1} ||Ax|| $$

we have

$$ ||A+B|| \leq ||A|| + ||B|| $$ and

$$ ||AB|| \leq ||A||||B|| $$

So $$ ||(A+B)^k|| \leq (||A|| + ||B||)^k $$

hence the spectral radius (https://en.wikipedia.org/wiki/Spectral_radius) we have $$ \rho(A+B) = \lim_{k\to \infty} ||(A+B)^k||^{1/k} \leq ||A|| + ||B|| $$

For normal matrice $A$, $\rho(A) = ||A||$ QED.