- 高二不等式已知n>1且n∈正整数集，求证：以n为底的（n+
- 已知n>1且n∈正整数集，求证：以n为底的（n+1）的 >以(n+1)为底的(n+2)的对数。
- 证：log(n)(n+1)-log(n+1)(n+2)
=1/log(n+1)(n)-log(n+1)(n+2)【换成以n+1为底的对数，项目省略底数n+1】
=[1-lognlog(n+2)]/logn
lognlog(n+2)<{[logn+log(n+2)]/2}^2
={log[n(n+2)]/2}^2
=[log(n^2+2n)/2]^2
<[log(n+1)^2/2]^2=[2log(n+1)/2]^2=(2*1/2)^2=1
因此1-lognlog(n+2)/2>0,又logn>0
所以差[1-lognlog(n+2)]/logn>0
故log(n)(n+1)>log(n+1)(n+2)成立。