How to solve an inequality involving the floor function? I am trying to solve the following disequation:
$$\left \lfloor \frac{(n+1)(n+2)}{6}\right\rfloor < \frac{n^2+3n}{8}.$$
Which is the right method to solve that?
My thougths so far: I wrote the term 
$$\left \lfloor \frac{(n+1)(n+2)}{6}\right\rfloor = \frac{(n+1)(n+2)-r}{6}$$
where $r=0, \dots, 5$, but I was wondering if there is another method to solve that.

\begin{align}
   \left \lfloor \frac{(n+1)(n+2)}{6}\right\rfloor &< n^2+4n \\
   \left \lfloor \frac{(n+1)(n+2)}{6} - ( n^2+4n)\right\rfloor &< 0 \\
   \left \lfloor \frac{-5n^2-21n+2}{6} \right\rfloor &< 0 \\
   \frac{-5n^2-21n+2}{6} &< 0 \\
\end{align}
The rest is standard stuff.
