I recently stumbled upon an inequality involving binomial coefficients. There is reason to suspect that it holds for all $l\in\mathbb{N}$. It states that
$$ (2l+1)^{2l+1} < \sum_{m = 0}^{l} \binom{2l+1}{2m} 2^{2(m-l)}(2l+1)^{2m} $$
I am not sure how to approach this problem (i.e. to prove the inequality) and I am not even sure if the inequality holds for large $l$. Any help on the way to prove this is appreciated. Thanks in advance.
We have:
$$\sum_{m=0}^{l}\binom{2l+1}{2m}(2l+1)^{2m}\left(\frac{1}{2}\right)^{2l-2m}=2\sum_{h\equiv 0\pmod{2}}\binom{2l+1}{h}(2l+1)^h\left(\frac{1}{2}\right)^{2l+1-h}$$ so the RHS equals: $$ \left(\frac{1}{2}+(2l+1)\right)^{2l+1} + \left(\frac{1}{2}-(2l+1)\right)^{2l+1} $$ or: $$\left(2l+\frac{3}{2}\right)^{2l+1}-\left(2l+\frac{1}{2}\right)^{2l+1} $$ and now it is not difficult to show that such a term is greater than $(2l+1)^{2l+1}$.