Prove logarithmic inequality with greatest integer function. $\left \lfloor n\log_2 n^2 \right \rfloor + \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$
How to show this? I tried using $\left \lfloor k \right \rfloor \leq k$, but I'm not sure that that does anything...
LHS $\leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor  +  \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor $ < $\left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$ because...ugh...
$\left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \log_2(\left \lfloor n\log_2n^2 \right \rfloor) < 1$?
I don't think the last inequality is true.
Help please?
I think it's supposed to hold for all natural numbers or at least all natural numbers after a certain index.

hint:
$\left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1 =\left \lfloor (n\log_2 (n+1)^2)+\log_2 (n+1)^2 \right \rfloor +1\ge \left \lfloor (n\log_2 (n+1)^2) \right \rfloor +\left \lfloor  \log_2 (n+1)^2\right \rfloor+1$
now prove:
$\left\lfloor (n\log_2 (n+1)^2) \right \rfloor \ge \left\lfloor (n\log_2 n^2) \right \rfloor$
$\left \lfloor  \log_2 (n+1)^2\right \rfloor+1 \ge \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor $
