Definition: Let $H$ a Hilbert space. An unbounded linear operator $A: D(A) \subseteq H \to H$ is said to be monotone if it satisfies $$\forall v \in D(A),\ (Av, v) \ge 0 $$
It is called maximal monotone if, in addition, $R(I + A) = H$, i.e., $\forall f \in H,\ \exists u \in D(A) \text{ such that }u + Au = f$
If $A $ is maximal monotone then $\lambda A $ is also maximal monotone for every $\lambda > 0$.
However, if $A$ and $B$ are maximal monotone operators, then $A + B$, defined on $D(A) \cap D(B)$, need not be maximal monotone. Why?
A reasonable example is defined on $X=L^{2}[0,\pi]$. Let $H^{2}[0,\pi]$ be the subspace of $X$ consisting of all twice absolutely continuous functions $f \in L^{2}$ for which $f',f'' \in L^{2}$. Define $A$ and $B$ to be $-\frac{d^{2}}{dx^{2}}$ on their respective domains $$ \mathcal{D}(A)= \{ f \in H^{2} : f(0)=f(\pi) = 0 \},\\ \mathcal{D}(B) = \{ f \in H^{2} : f'(0)=f'(\pi)=0 \}. $$ $A$ and $B$ are selfadjoint which satisfy $$ (Af,f) \ge 0,\;\;\; f \in \mathcal{D}(A), \\ (Bf,f) \ge 0, \;\;\; f \in \mathcal{D}(B). $$ However the restriction of $A+B$ to $\mathcal{D}(A)\cap\mathcal{D}(B)$ is not selfadjoint, and $A+B+I$ is not surjective because that would mean it would be possible to solve the following for all $g \in L^{2}$: $$ -2f''+f = g,\\ f(0)=f'(0) = 0,\\ f(\pi)=f'(\pi) = 0. $$ That cannot happen for all $g$.