Your proof is faultless. Of course you are assuming that adding the same thing to both sides preserves order, and that dividing both sides by a positive number also preserves order. However, these are by now to be considered standard facts in the course. Perhaps you should have said "Since $2$ is positive, we can $\dots $". But that may be unnecessarily fussy.
The "trick" of adding $b$ to both sides is fairly natural. It only has a faint whiff of magic.
Maybe you did the following before writing down a proof, at least I hope you did. On a "number line", put a dot for $a$, one for $b$, and one for $(a+b)/2$, halfway between them. This finishes things, you now know the result is true. It only remains to write down the details, using the notations and tools of your course. (Theo Buehler writes about the geometry in a comment which is far more important than the question that led to it.)
The geometry suggests the following alternative approach. Recall that $x iff $y-x>0$. So it is enough to prove that $ b-\frac{a+b}{2}>0 $. But we have $b-\frac{a+b}{2}=\frac{2b-(a+b)}{2}=\frac{b-a}{2}>0$ (the inequality $\frac{b-a}{2}>0$ follows from $a.)
A little bit more complicated, but it arises from an understanding of what's happening underneath. And it tells you more than the simple fact that $b>(a+b)/2$. It tells you by how much $b$ is bigger than $(a+b)/2$.
By the way, for complicated expressions $X$ and $Y$, one of the standard strategies for showing that $X is to show that $Y-X>0$.
Comment: Back to your solution. At the manipulational level, maybe this is how I would think. I want to show that $\frac{a+b}{2}. Fractions are unpleasant, they are "broken" numbers (that's the correct etymology). So let's unbreak the fracture.
Note that the desired inequality holds iff $a+b <2b$. And $a+b<2b$ is an obvious consequence of $a. Now we can hide the reasoning that led to the solution by writing the proof backwards. Or not.