Collatz Conjecture x+1 and 3x+3 First Question
$\bullet$ For the Collatz Conjecture as $ax+b$, when $a=1$ and $b=1$, is it a safe assumption to say that this variant of the conjecture will always reach $1$, since dividing by $2$ will always lower the number further than it will rise?
Second Question
$\bullet$ Has anyone done any heavy work on the $3x+3$ variant, where all numbers will eventually reach the loop at 3 and 1?
 A: You are right about the $x+1$ variant.  Since the add-$1$ process will always be followed by the divide-by-$2$ process—$x+1$ is even when $x$ is odd—we may formulate the rule as
$$
x\mapsto\begin{cases}(x+1)/2 & \text{if $x$ is odd}\\ x/2 & \text{if $x$ is even}\end{cases}
$$
which is equivalent to the rule $x\mapsto\lceil x/2\rceil$.  Since $\lceil x/2\rceil<x$ for all $x>1$, the sequence of iterates goes to $1$ in a straightforward way, essentially by repeated halving.
The $3x+3$ variant is connected with the usual $3x+1$ problem to such an extent that, in contrast to, say, the $3x+5$ variant, you won't learn much new by studying it. To see this, compare the action of the $3x+1$ process on the starting value $n$ with the action of the $3x+3$ process on the starting value $3n$.  If $n$ is even, then so is $3n$, so in both processes halving will occur.  If $n$ is odd, then so is $3n$.  Applying the $3x+3$ process to $3n$ yields $9n+3=3(3n+1)$, which is precisely $3$ times the result of applying the $3x+1$ process to $n$.  The result is that each of the iterates of the $3x+3$ process applied to $3n$ will be $3$ times the corresponding iterate of the $3x+1$ process applied to $n$.  An example:
$$
\begin{aligned}
(3x+1): & 26\mapsto13\mapsto\ \ 40\mapsto20\mapsto10\mapsto\ \ 5\mapsto16\mapsto\ \ 8\mapsto\ \ 4\mapsto2\mapsto1\mapsto\ \ 4\mapsto2\mapsto\ldots\\
(3x+3): & 78\mapsto39\mapsto120\mapsto60\mapsto30\mapsto15\mapsto48\mapsto24\mapsto12\mapsto6\mapsto3\mapsto12\mapsto6\mapsto\ldots
\end{aligned}
$$
Now combine this with the observation that, regardless of starting value, the $3x+3$ process will eventually produce an iterate that is a multiple of $3$.  This will, in fact, happen the first time the $3x+3$ operation is applied, that is, immediately following the first odd iterate.  In summary, the $3x+3$ process quickly and inevitably yields a multiple of $3$, after which the iterates of the process are the same of those of the $3x+1$ process scaled by a factor of $3$.
