I'd like to clarify first that Big O notations aren't tighter bounds and hence,you are free to assume that
f(x) = x^2 + 2x + 4
= O(x^3)... and so on...
As an example, n² + 3n + 4 is O(n²), since n² + 3n + 4 < 2n² for all n > 10 (and many smaller values of n). Strictly speaking, 3n + 4 is O(n²), too, but big-O notation is often misused to mean "equal to" rather than "less than". The notion of "equal to" is expressed by Θ(n).
So,if you're going with the misuse, then your definition doesn't hold as posted by
user 1952500. But,if you're going with the exact definition, then your example is well valid and hence, it satisfies the formula...
Now,coming to actual practice, we should go with the formal definition which will be always verified as
f(n) + g(n) = O(max(f(n) , g(n))) // maximum in terms of power of n.
Also, O(max(f(n) , g(n))) <= O(f(n)* g(n)) // this is a generalised result and will always be correct as understood by your example...
One more thing to make others understand is f(x^-n) is always considered to be O(1).