Question

利用k^(k+1)>（k+1）^k (k≥3)证明：

Answer 1

这种类型的题通常用取对数来降次

证明：

∵k^(k+1)>（k+1）^k

∴(k+1)lnk>kln(k+1)

∴k[lnk-ln(k+1)]+lnk>0

∴kln[k/(k+1)]+lnk>0 .(1)

又(k+1)^2=k+2k+1>k+2k=k(k+2)

∴(k+1)/(k+2)>k/(k+1)

∴ln[(k+1)/(k+2)]>ln[k/(k+1)].(2)

∴(k+1)ln[(k+1)/(k+2)]+ln(k+1)

=kln[(k+1)/(k+2)]+ln(k+1)+ln[(k+1)/(k+2)]

=kln[(k+1)/(k+2)]+lnk+ln(k+1)-lnk+ln[(k+1)/(k+2)] 第一个对数的真数即式(2)的左边

>{kln[k/(k+1)]+lnk}+ln[(k+1)/k]+ln[(k+1)/(k+2)] 大括号里即式(1)的左边

>ln[(k+1)/k]+ln[(k+1)/(k+2)]

=ln[(k+1)^2/k(k+2)] 因为(k+1)^2=k+2k+1>k+2k=k(k+2),所以真数>1

>0

即(k+2)ln(k+1)-(k+1)ln(k+2)>0

∴(k+2)ln(k+1)>(k+1)ln(k+2)

∴(k+1)^(k+2)>(k+2)^(k+1)

证毕