For example, say I want to analyze $T(n)=3T(\lfloor n/3 \rfloor )+2n$ for $n>2$, and $T(n)=1$ otherwise. This is clearly $O(n\log n)$; however it seems that with induction you can prove it is $O(n)$:
Base case: $T(1)$ is $O(1)$, $T(2)$ is $O(2)$
Inductive step: $T(n) = 3\times O(n/3) + 2n \in O(n)$
Where does this go wrong? Why can I do this? Induction clearly is a valid technique, but what subtlety of $O$ causes me to be able to prove a wrong bound?
You have to carry around the big-O constant. Suppose that we are trying to prove that $T(n) \leq Cn$. In the inductive step, we have $$ T(n) \leq 3C\lfloor n/3 \rfloor + 2n \leq (C+2)n. $$ So we are not able to maintain the constant $C$. Let's have another go, with $T(n) \leq C_nn$. The argument above shows that we can put $C_n = C_{\lfloor n/3 \rfloor} + 2$, and so $C_n = O(\log n)$ works, and we get the correct asymptotics $O(n\log n)$. (To know that this is tight, we would also need a lower bound, proved in much the same way, with some minor technicalities stemming from the floor.)
