We know that the complexity of $T(n)=2T(n/3 + 1) + n$ is $\Theta(n)$, as has been proved on this exchange before. However, what about proving it inductively? I believe that this method might work.
Guess $T(n) \leq cn-d$. Assume true for $m < n$. Specifically, $m = n/3 < n$ Then, $T(n) \leq cn-d$.
$T(n) \leq 2(cn/3 - d + 1) + n$ $= 2cn/3 - 2d + 2 + n$ $=n(2c/3+1)+2(1-d)$. Then for values of $c \geq 1$ and $d \geq2$ we have $T(n) \leq cn - d$. Then $T(n)=\Theta(n)$.
What does everyone think?