Question

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

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

Answer 1

证明：令x=log_ab，y=log_bc．
∵1＜a≤b≤c，∴x≥1，y≥1，xy≥1．
则log_ab+log_bc+log_ca≤log_ba+log_cb+log_ac?x+y+

1

xy

≤

1

x

+

1

y

+xy．
右边-左边=

1

x

+

1

y

+xy?x?y?

1

xy

=

(x?1)(y?1)(xy?1)

xy

≥0．
∴右边≥左边．
故原式成立．