已知$b_n=dfrac{1}{2n-1}$是否存在正数$m$,使得$(1+b_1)(1+b_2)cdots(1+b_n)ge msqrt{2n+1}$恒成立
分析:记$I_n=(1+b_1)(1+b_2)cdots(1+b_n)=dfrac{2}{1}cdotdfrac{4}{3}cdotsdfrac{2n}{2n-1}$
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$I_n>dfrac{4}{3}cdotdfrac{6}{5}cdotsdfrac{2n+2}{2n+1}=I_{n+1}$ 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$mle f(n)$恒成立,只需$mle f(1)=dfrac{2sqrt{3}}{3}$
又$m>0$故$min(0,dfrac{2sqrt{3}}{3}]$