$\frac{1}{2}×\frac{3}{4}×\cdots×\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$

APPSYZY

求证：
\[\frac{1}{2}×\frac{3}{4}×\cdots×\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}.\]

kuing

糖水不等式即可。

来试试加强式：
\[\sqrt{\frac1{2n+1}\cdot\frac2\pi}<\frac12\times\frac34\times\dots\times\frac{2n-1}{2n}<\sqrt{\frac{2n+2}{(2n+1)^2}\cdot\frac2\pi}.\]

isee

源自知乎提问

证明： $\displaystyle\frac 12\cdot\frac 34\cdot \frac 56\cdots \frac{2n-1}{2n}<\frac 1{\sqrt{2n+1}}$ .

可证\[\frac {2n-1}{2n}<\frac {2n}{2n+1}.\]故\[\frac 12\cdot\frac 34\cdot \frac 56\cdots \frac{2n-1}{2n}<\frac 23\cdot \frac 45\cdot \frac 67\cdots \frac{2n}{2n+1},\]两端同乘$\frac 12\cdot\frac 34\cdot \frac 56\cdots \frac{2n-1}{2n}$有\[\left(\frac 12\cdot\frac 34\cdot \frac 56\cdots \frac{2n-1}{2n}\right)^2<\frac{(2n)!}{(2n+1)!}=\frac 1{2n+1},\]两端开方即是.