利用均值不等式证明一道题

依n+1元基本不等式得
(1+1/n)^n
=1·(1+1/n)·(1+1/n)·...·(1+1/n)
<[(1+(1+1/n)+(1+1/n)+...+(1+1/n))/(n+1)]^(n+1)
=[(1+n·(1+1/n))/(n+1)]^(n+1)
=[(1+n+1)/(n+1)]^(n+1)
=[1+1/(n+1)]^(n+1)
∴(1+1/n)^n<[1+1/(n+1)]^(n+1)
故原不等式得证咐做亏胡困。衡神