Does the following result hold? Suppose that $A$ and $B$ are symmetric and non-negative matrices.
Let $\rho(A)$ denote the spectral radius of $A$ and $\rho(B)$ denote the spectral radius of $B$.
Does the following result hold?

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

I tried several examples and found the result true.
But is it true in general?
 A: It is true if $A$ and $B$ are real. For symmetric real matrices, the spectral radius agrees with the operator norm
$$
\|A\|=\max\{\|Ax\|_2:\ \|x\|_2=1\}.
$$
And the sum of real symmetric matrices is real symmetric. Thus
$$
\rho(A+B)=\|A+B\|\leq\|A\|+\|B\|=\rho(A)+\rho(B). 
$$
Non-negative is not necessary for all the above. 
The same result holds for complex matrices if we replace "symmetric" with "selfadjoint". 

Another way to prove it is to use the Min-Max Theorem  (Courant-Fischer-Weyl):  if $\lambda_1\leq\cdots\leq\lambda_n$ are the eigenvalues of $A$ (and $A$ is Hermitian, so real and symmetric qualifies)
$$
\lambda_k=\min_{\dim K=k}\max_{x\in K, \|x\|=1} \langle Ax,x\rangle.
$$
When $k=n$, 
\begin{align}
\lambda_n(A+B)&=\max_{x\in K, \|x\|=1} \langle (A+B)x,x\rangle\\
&\leq \max_{x\in K, \|x\|=1} \langle Ax,x\rangle+\langle Bx,x\rangle\\
&\leq \max_{x\in K, \|x\|=1} \langle Ax,x\rangle+\max_{x\in K, \|x\|=1}\langle Bx,x\rangle\\
&=\lambda_n(A)+\lambda_n(B)\\
&\leq\rho(A)+\rho(B). 
\end{align}
Since $\rho(A+B)$ is either $|\lambda_1(A+B)|$ or $|\lambda_n(A+B)|$, we get
$$
\rho(A+B)\leq \rho(A)+\rho(B). 
$$
Again, "non-negative" is not needed. 
