已知f(x)=lnx-[a(x-1)/(x+1)],设m,n属于正有理数,求证:[(m-n)/(lnm-lnn)]

Question

已知f(x)=lnx-[a(x-1)/(x+1)],设m,n属于正有理数,求证:[(m-n)/(lnm-lnn)]

数学

作业帮用户2017-11-03

Answer 1

（一）函数f(x)=㏑x-[2(x-1)/(x+1)].(x≥1)求导得：f′(x)=(1/x)-[4/(x+1)²]=(x-1)²/[x(x+1)²]≥0.(x≥1).∴函数f(x)在[1,+∞)上递增.∴当x＞1时,f(x)＞f(1)=0.即当x＞1时,有f(x)＞0.即当x＞1时,有(㏑x)/2＞(x-1)/(x+1).（二）当m,n＞0且m≠n时,不妨设m＞n＞0.则m/n＞1.由前面的结论可知,[㏑(m/n)]/2＞[(m/n)-1]/[(m/n)+1]=(m-n)/(m+n).===>(m+n)/2＞(m-n)/(㏑m-㏑n).证毕.