I need to proof the following inequality for a research.
$N(N+1) > 2^{\lceil \log_2 N \rceil}(\lceil \log_2 N \rceil+1)$ for $N \ge 8$
I know I have to proof that
$\frac{N(N+1)}{2^{\lceil \log_2 N \rceil}(\lceil \log_2 N \rceil+1)} > 1$
but I don't see any algebra manipulations that I could use. I already able to solve it using the upper bound of the ceil function. But, it's better if I can proof it without using upper bound.
Update: I've managed to prove this using mathematical induction by using two cases: $N = 2^k$ and $N \ne 2^k$, with $k \in \mathbb{Z}$.
Hint: Let $\lceil{\log_{2}N}\rceil=k $. Then we have the inequality $k-1 <\log_{2}N\leq k $ and since $N\geq8$, $k\geq 3$. Then the given inequality becomes
$2^{k}(k+1)<N(2k+2)<N(N+1) $