There's no procedure for creating a context-free grammar for the complement of a context-free language, because the complement of a context-free language might not be context-free, and the question of whether a language is context-free is not decidable.
Of course, particular context-free languages (and some languages which are not context-free) do have context-free complements, and this is an example of such a language. Indeed, as @Tonita observes in her excellent answer, $L^C$ is a deterministic context-free grammar, which are closed under complement. So that's not entirely a dead end.
But on the whole, it's rarely helpful to try to start with a grammar for $L$ and transform that into a grammar for $L^C$, since the latter might not even exist, and even if it dies, mechanically generating it could be a lot of work.
Here's another approach which often works for problems like this. When trying to deal with languages whose descriptions involve inequalities, it's often useful to remember that $i\ne j$ is the union of the two predicates $i<j$ and $i>j$, and furthermore that $i>j$ is the same as $\exists k>0 \mid i = j + k$. Finally, $a^{j+k} = a^ja^k$. That should be enough to quickly find the solution.