Stuck on the formula of spectral radius I am having two questions on spectral radius:

We denote spectral radius of a matrix $A$ by $\rho(A)$

Question 1: Is it true that for any two matrices $A,B$,

$\rho(A+B)\le \rho(A)+\rho(B)$

When does it hold?

Question 2:

Is it true that $\rho(A)=\max_{||x||=1} x^TAx$?

If it holds then is it  not true that $\rho(A+B)= \rho(A)+\rho(B)$?

because  $\rho(A+B)=\max_{||x||=1} x^T(A+B)x=\max_{||x||=1} 
x^T(A)x+\max_{||x||=1} x^T(B)x=\rho(A)+\rho(B)$

I am stuck on these two questions. 
Can someone please help me out?
If some more information is needed please reach me.
Question 1 holds when;


$A$ and $B$ are normal. Because in this case $\rho(A) = \lVert A \rVert_2$, which means
$$ \rho(A+B) \leq \lVert A + B \rVert_2 \leq \lVert A \rVert_2 + \lVert B \rVert_2 = \rho(A) + \rho(B) $$
Note: $A+B$ may not be normal, but the first inequality holds for all matrices.
$A$ and $B$ are simultaneously upper-triangulizable, i.e. there exists an invertible $T$ such that $U_A:=T^{-1}AT$ and $U_B:=T^{-1}BT$ are upper triangular, in which case equality occurs.
$$ \rho(A+B) = \rho(T^{-1} (A+B) T) = \rho(U_A + U_B) = \rho(A) + \rho(B)$$
$AB = BA$ (special case of 2)


For Question 2, both claims are not true. However, for a symmetric matrix $A$
$$\lambda_\max(A)=\max_{\lVert x \rVert=1} x^T A x$$
holds. Note that this is not equivalent to $\rho(A)$, which is the maximum absolute value of the eigenvalues. So, for symmetric matrices
$$\lambda_\max(A+B)=\max_{\lVert x \rVert=1} x^T (A+B) x \leq \max_{\lVert x \rVert=1} x^T A x + \max_{\lVert x \rVert=1} x^T B x = \lambda_\max(A)+\lambda_\max(B)$$
holds.
For example, try $A = \pmatrix{0 &amp; 1 \cr 0 &amp; 0\cr}$ and $B = A^\top$.