Xét $f(t)=\ln t$ khả vi và liên tục với $t\in \left[2n;2n+1 {} \right]$$+f'(t)=\frac{1}{t}$. Theo đlí Lagrange thì $\exists c\in (2n;2n+1)$:
$\frac{f(2n+1)-f(2n)}{2n+1-2n}=f'(c)$
$\Rightarrow \ln(2n+1)-\ln2n=\frac{1}{e}>\frac{1}{2n+1}$
$\Rightarrow \ln\frac{2n+1}{2n}>\frac{1}{2n+1}$
$\Rightarrow \ln(\frac{2n+1}{2n})^2>1$
$\Rightarrow (\frac{2n+1}{2n})^{2n+1}>e$
$\Rightarrow (\frac{2n}{2n+1})^{2n+1}<\frac{1}{e}$ (1)
Xét $f(x)=x^{2n}(1-x), 0<x<1$
$+f'(x)=\left[ 2n-(2n+1x) {} \right]x^{2n-1}=0\Rightarrow x=\frac{2n}{2n+1}$
Từ bbt$\Rightarrow x^{2n}(1-x)\leq (\frac{2n}{2n+1})^{2n}.\frac{1}{2n+1}$
$\Rightarrow 2n.x^{2n}(1-x)<(\frac{2n}{2n+1})^{2n+1}$ (2)
Từ $(1)(2)\Rightarrow đpcm$