数学不等式证明:n>2时..logn(n-1)

logn(n-1)-log(n+1)n=lg(n-1)/lgn-lgn/lg(n+1)=[lg(n-1)*lg(n+1)-(lgn)^2]/lgn*lg(n+1）而lg(n-1)*lg(n+1)≤{[lg(n-1)+lg(n+1)]/2}^2={[lg(n^2-1)]/2}^20,故logn(n-1)