If I know about the leading eigenvalues and the eigenfunctions of two operators, is there any result about the leading eigenvalue of the sum of the two operators?

You need to add some assumption, otherwise $\begin{pmatrix} 1 & n \\\\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \end{pmatrix}$ add to a matrix with eigenvalues $1 \pm \sqrt{n}$. Maybe your operators are selfadjoint? 


If $A$ is self adjoint, and $\lambda$ its leading eigenvalue, then $\lambda = \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Au \rangle$. If $A$ and $B$ are selfadjoint, we have the obvious consequence $$\mathrm{sup}_{\langle u,u \rangle =1} \langle u, (A+B) u \rangle = \mathrm{sup}_{\langle u,u \rangle =1} \left( \langle u, A u \rangle + \langle u, B u \rangle \right) \leq \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Au \rangle + \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Bu \rangle$$ so the leading eigenvalue of the sum is bounded by the sum of the leading eigenvalues. 


I would guess that the magnitude of the leading eigenvalue of the sum is at most the sum of the magnitudes of the leading eigenvalues of the two operators, because the size of the leading eigenvalue is like a norm and the norms have the triangle inequality. Added: I guess this assumes the operators are self adjoint. 

