Let $n$ be postive integer,I conjecture
$$(1+2n)^n\ge 1^n+2^n+4^n+6^n+\cdots+(2n)^n \tag{1}$$
This problem when I solve this equation $$(1+2n)^n=1^n+2^n+4^n+6^n+\cdots+(2n)^n\tag{2}$$ if this $(1)$ hold,then $(2)$ have only postive integer solution $n=1$
$n=1$ is clear, let $n\geqslant 2$.
Divide by $(2n+1)^n$ and use the estimate $(1-\frac{k}{2n+1})^{n}<e^{-k/2}$ for $k=3,5,\dots$ (for proving this take logarithm, get equivalent inequalities $n\log(1-\frac{k}{2n+1})<-k/2$, $k/2<n(\frac{k}{2n+1}+\frac{k^2}{2 (2n+1)^2}+\dots)$, two first summands already suffice: $$n\left(\frac{k}{2n+1}+\frac{k^2}{2 (2n+1)^2}\right)\geqslant \frac{nk}{2n+1}+\frac{3nk}{2(2n+1)^2}\geqslant \frac{k}2,$$ this last inequality is straightforward.) So, it suffices to check that $(\frac{2n}{2n+1})^n+e^{-3/2}+e^{-5/2}+\dots<1$, summing up geometric progression we rewrite this as $$\left(\frac{2n+1}{2n}\right)^n>\frac{1}{1-e^{-3/2}/(1-e^{-1})}=1.545\dots,$$ but we know that $(1+\frac{1}{2n})^n=\sqrt{(1+\frac1{2n})^{2n}}$ increases, and already for $n=2$ it equals $1.5625$.
