First, if the matrices are not symmetric it is not true. If $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, then the maximum eigenvalue of $A$ is $1$ and the eigenvalues of $\frac{1}{2}(A+A^T)$ are $\frac{1}{2},\frac{3}{2}$ (hence $A$ is positive definite).
Now set $B = A^T$. Then the maximum eigenvalue of $A+B=2 \frac{1}{2}(A+A^T)$ is $3$, but the sum of the maximum eigenvalues is $2$.
However, if the matrices are symmetric, then it is clear from the diagonal form that $\lambda_{\max}(A) = \|A\| $, and since $\|A+B\|\leq \|A\|+\|B\|$, you have $\lambda_{\max}(A+B) \leq \lambda_{\max}(A) + \lambda_{\max}(B)$.