Question

求证a1*a2*a3*a4.*an≥根号2n+1

Answer 1

用数学归纳法证明

证明：n=1时,显然有a1=2≥根号3成立;

假设结论成立,则n+1时,有a(n+1)= 2(n+1)/(2(n+1)-1)=(2n+2)/(2n+1)

从而有1*a2*a3*a4.*an*a(n+1)≥(根号2n+1)*(2n+2)/(2n+1)=(2n+2)/(根号2n+1)

即只需证明(2n+2)/(根号2n+1)≥根号(2(n+1)+1)

也就是证明(2n+2)(2n+2)≥(2n+1)(2(n+1)+1),该等式显然成立,证毕.