相关logca的文学知识

设1<a≤b≤c,*:logab+logbc+logca≤logba+logcb+logac.

设1<a≤b≤c,*:logab+logbc+logca≤logba+logcb+logac.

问题详情:设1<a≤b≤c,*:logab+logbc+logca≤logba+logcb+logac.【回答】 (2)令logab=x,logbc=y,则logca=,logba=,logcb=,logac=xy.由1<a≤b≤c得x=logab≥1,y=logbc≥1.由(1)亦即得到所*的不等式成立.知识点:不等式题型:解答题...