Solving $T(n) = T(n/2) + T (n/3) + n $ with recurrence tree
On the one hand, clearly $T(n) \geq n$ (see detailed proof below). On the other hand, let us find prove by induction that $T(n) \leq Cn$, for large enough $C$. The base case trivially holds for $C \geq T(1)$. For the inductive step, we have
$$
T(n) = T(n/2) + T(n/3) + n \leq C\cdot (n/2) + C\cdot (n/3) + n = [\tfrac{5}{6}C+1]n.
$$
(We're cheating here a bit since $n$ isn't necessarily a multiple of 6.) If $C \geq 6$, then $\tfrac{5}{6}C+1 \leq C$, and so we can conclude that $T(n) \leq Cn$, thus completing the inductive step. In total, if we take $C = \max(T(1),6)$ then the proof goes through, hence
$$ n \leq T(n) \leq \max(T(1),6) n. $$
This implies that $T(n) = \Theta(n)$.



Finally, let us prove that $T(n) \geq n$. We first prove that $T(n) \geq 0$. The proof by induction. The base case clearly holds, and the inductive step is a consequence of
$$ T(n) = T(n/2) + T(n/3) + n \geq 0 + 0 + 0, $$
using the inductive hypothesis twice. Finally,
$$ T(n) = T(n/2) + T(n/3) + n \geq 0 + 0 + n = n. $$