You need $A$, $D$ symmetric to guarantee that the eigenvalues are real. Then you have the elementary estimate
\begin{multline*}
\max\{\lambda_{\max}(A)+\lambda_{\min}(D),\lambda_{\min}(A)+\lambda_{\max}(D)\} \\ \leq \lambda_{\max}(A+D) \leq \lambda_{\max}(A)+\lambda_{\max}(D),
\end{multline*}
and this estimate is sharp as easy examples with $A$ being diagonal show.

EDIT: For instance,
\begin{multline*}
\lambda_{\max}(A+D) = \max_{\|x\|_2=1}\langle (A+D)x,x\rangle \\ \geq
\max_{\|x\|_2=1}\langle Ax,x\rangle + \min_{\|x\|_2=1}\langle
Dx,x\rangle =\lambda_{\max}(A) + \lambda_{\min}(D).
\end{multline*}