a>1,0

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a>1,0a>1,0a>1,0logab=1/logbalogab=2logab*logba=2logab+logba用对数基本性质分解原式得lgb/lga+lga/lgb=-[(-lgb/lga)+

a>1,0
a>1,0

a>1,0
logab=1/logba
logab=2logab*logba=2
logab+logba

用对数基本性质分解原式得
lgb/lga+lga/lgb=-[(-lgb/lga)+(-lga/lgb)]
(-lgb/lga)+(-lga/lgb)>=2(用不等式的基本性质a^2+b^2>=2ab)
所以原式<=-2