求证log(n)(n+1)>log(n+1)(n+2),其中n∈N,且n>1

要证：log(n)(n+1)>log(n+1)(n+2),
即要证log(n+1)/logn>log(n+2)/log(n+1)（换底公式）
即要证log(n+1)log(n+1)>logNlog(n+2)
而由基本不等式
lognlog(n+2)<((logn+log(n+2))/2)^2=(0.5logn(n+2))^2
=(0.5log(n+1-1)(n+1+1))^2
<(0.5log(n+1)^2)^2
=log(n+1)log(n+1)
故有题目结论。

证明函数
f(x)=log x (x+1) 在(1,正无穷)上单调减
f'(x)=(ln(x+1)/ln x)'
=(ln x /(x+1)-ln(x+1)/x)/(lnx)^2<0
所以log(n)(n+1)>log(n+1)(n+2)

log(n)(n+1)>log(n+1)(n+2)