It is simple to prove by induction that $$ \label{eq1}\tag{1} n + \left\lfloor \tfrac 3 5 n \right\rfloor + \left\lfloor \tfrac {n+2} 5 \right\rfloor \;=\; \left\lfloor \tfrac 9 5 n \right\rfloor $$ holds for all integers $n \geq 0$: First substitute each $0 \leq n \lt 5$ to verify the 5 'base cases', and then substitute $n := n+5$ and prove that induction step from the induction hypothesis $(\ref{eq1})$.
The same goes for an inequality like $$ n \;\geq\; \left\lfloor \tfrac 3 5 n \right\rfloor + 2 \left\lfloor \tfrac {n+2} 5 \right\rfloor $$
My question: Is there perhaps another more 'direct' way to prove statements like these, avoiding checking multiple individual cases, and perhaps also avoiding induction?