Question

已知：a>1，b>1，c>1，lga+lgb=1，求log(a)(c)+log(b)(c)≥4lgc。注：log(a)(c)表示以a为底c的对数。谢谢

Answer 1

lga+lgb=1
ab=10

logaC+logbC
=lgc(1/lga +1/lgb)
=lgc[(lga+lgb)/(lga*lgb)]
=lgc/(lga*lgb)

lga+lgb=1>=2√(lga*lgb)
所以 1/(lga*lgb)>=4
当且仅当 lga=lgb时取得等号，这时候a=b=5

logaC+logbC=lgc/(lga*lgb)>=4lgc

Answer 2

logac=lgc/lga
logbc=lgc/lgb

log(a)(c)+log(b)(c)=lgc*(1/lga+1/lgb)
=lgc(lga+lgb）/（lgalgb）
又lgalgb《=（lga+lgb)^2/4=1/4
所以lgc(lga+lgb）/（lgalgb）》=4lgc

Answer 3

利用换底公式，log(a)(c)=lgc/lga,则需证明lgc/lga+lgc/lgb≥4lgc，只需证明1/lga+1/lgb≥4，通分只需证明1/(lga*lab)>=4,此式可有lga+lgb>=2√(lga*lgb)得。