A convex $n$-gon can be cut into $n-2$ triangles by just choosing a vertex and drawing all the diagonals from that vertex, so $3n$ obtuse triangles can be reduced to $3n-6$. 

We can do a bit better. E.g., if a convex quadrangle is not a square, then it has at least one obtuse angle, so we can cut off an obtuse triangle incorporating that angle, and just need three more triangles to finish the job, four triangles in all. Every convex $n$-gon, $n\ge5$, has one or more obtuse angles, which we can use to cut off triangles, to reduce the $3n-6$ further.