Xét:$f(x)=x^{2n}(1-x)=x^{2n}-x^{2n+1},x\in (0,1)$
$f'(x)=2n.x^{2n-1}-(2n+1)x^{2n}$ (vì $x\in (0,1)$)
$f'(x)=0\Leftrightarrow x=\frac{2n}{2n+1}$
Bảng biến thiên:
$\Rightarrow x^{2n}(1-x)\leq (\frac{2n}{2n+1})^{2n}.\frac{1}{2n+1},\forall x\in (0,1)$
Do đó để chứng minh:
$x^{n}\sqrt{1-x}<\frac{1}{\sqrt{2ne}} \Leftrightarrow x^{2n}(1-x)<\frac{1}{2ne}$
Ta sẽ chứng minh:
$(\frac{2n}{2n+1})^{2n}.\frac{1}{2n+1}<\frac{1}{2ne}$
$\Leftrightarrow (1+\frac{1}{2n})^{2n+1}>e$
$\Leftrightarrow (1+\frac{1}{k})^{k+1}>e (k=2n>0) (*)$
Xét: $g(x)=\ln x,x>0$
$g'(x)=\frac{1}{x}$
Theo định lý Lagrange: $\exists c\in (k,k+1)$:
$g(k+1)-g(k)=g'(c).(k+1-k)=g'(c)=\frac{1}{c}>\frac{1}{k+1}$
$\Rightarrow \ln \frac{k+1}{k}>\frac{1}{k+1} $
$\Rightarrow \ln (1+\frac{1}{k})^{k+1}>1$
$\Rightarrow (1+\frac{1}{k})^{k+1}>e$
$\Rightarrow (*)$ đúng.$\Rightarrow$ (ĐPCM)