已知$b_n=\dfrac{1}{2n-1}$是否存在正数$m$,使得$(1+b_1)(1+b_2)\cdots(1+b_n)\ge m\sqrt{2n+1}$恒成立

分析:记$I_n=(1+b_1)(1+b_2)\cdots(1+b_n)=\dfrac{2}{1}\cdot\dfrac{4}{3}\cdots\dfrac{2n}{2n-1}$
由糖水不等式$\dfrac{k}{k+1}<\dfrac{k+1}{k+2}$知$\dfrac{2n}{2n-1}>\dfrac{2n+1}{2n}>\dfrac{2n+2}{2n+1}$
故$I_n>\dfrac{4}{3}\cdot\dfrac{6}{5}\cdots\dfrac{2n+2}{2n+1}=I_{n+1}$ 即$I_n$关于$n$单调递减,又明显的$\sqrt{2n+1}$关于$n$单调递增.故$f(n)=\dfrac{I_n}{\sqrt{2n+1}}$单调递减,由题意$m\le f(n)$恒成立,只需$m\le f(1)=\dfrac{2\sqrt{3}}{3}$
又$m>0$故$m\in(0,\dfrac{2\sqrt{3}}{3}]$