How to solve an inequality involving the floor function?

Question

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.

Answer 1

hint

$$\lfloor X-(n^2+4n) \rfloor =\lfloor X\rfloor -(n^2+4n)$$

$$\lfloor Y \rfloor <0 \iff Y <0$$

Your disequation is then equivalent to

$$-5n^2-23n+2 <0$$

Answer 2

\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.