相關logcb的文學知識

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

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

問題詳情:設1<a≤b≤c,*:logab+logbc+logca≤logba+logcb+logac.【回答】 (2)令logab=x,logbc=y,則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.知識點:不等式題型:解答題...